{"success":true,"data":{"id":"hash.6a4e35f7f10af.e9fad5ad9879b6f8a1ec04d8ee16e3d5","title":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","description":null,"count":40,"maxScore":40,"items":[{"identifier":"i6a30b1858b74d202606160414299fac4f31","guid":"i6a30b1858b74d202606160414299fac4f31","label":"INFO_1","title":"INFO_ 1","submissionMode":null,"xml":"\n\n \n \n \n
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CHEMIJOS VALSTYBINIO BRANDOS EGZAMINO
\nI DALIS

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2026 m. pakartotinė sesija

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Užduotį sudaro dvi dalys. Užduoties taškų suma – 40.

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Atidžiai perskaitykite kiekvieną užduoties klausimą ir pasistenkite kuo geriau atsakyti.

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Jei kiltų techninių nesklandumų arba sistema „pakibtų“, prašytume šiek tiek palaukti ir tęsti darbą, kai tik sistema tai leis.

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\n Trukmė – 90 min. \n
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Spauskite apačioje esančią rodyklę.

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\n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a30b1858b74d202606160414299fac4f31","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":[],"openQuestion":0}},{"identifier":"i69f12abd59dde20260428234637a9dcd9f6","guid":"i69f12abd59dde20260428234637a9dcd9f6","label":"INFO_2","title":"INFO_ 2","submissionMode":null,"xml":"\n\n \n \n \n
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\n Turėsite atlikti įvairias užduotis:\n

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ㅤ

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  • atsakyti į klausimus su pasirenkamaisiais atsakymais, t. y. pasirinkti vieną ar kelis, jūsų nuomone, teisingus atsakymus;
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  • įrašyti skaičių nurodytoje vietoje; skaičius reikia rašyti taisyklingai, t. y. nedėti papildomų tarpų, nerašyti nereikalingų simbolių;
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  • sudėlioti korteles, t. y. į tam tikras vietas įkelti korteles;
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  • paveiksle pažymėti reikiamą atsakymą, t. y. kompiuterio ekrane pažymėti aktyvų tašką arba paveikslo dalį; tai galima padaryti, paspaudus reikiamą vietą pele.
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\n Kiekviena teisingai atlikta užduotis bus vertinama 1 tašku.\n

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Spauskite apačioje esančią rodyklę.

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\n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i69f12abd59dde20260428234637a9dcd9f6","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":[],"openQuestion":0}},{"identifier":"i69f12ac3ab9e5202604282346433765e62b","guid":"i69f12ac3ab9e5202604282346433765e62b","label":"INFO_3","title":"INFO_3","submissionMode":null,"xml":"\n\n \n \n
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\n Atkreipkite dėmesį!\n

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  • Kai kurios užduotys gali netilpti į kompiuterio langą, todėl peržiūrėkite užduotį, naudodamiesi vertikalia slankjuoste ar kompiuterio pele.
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  • Jei norite naudotis skaičiuotuvu, paspauskite viršuje esantį skaičiuotuvo \"ivad_skaic\" ženklą.
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  • Prieš pirmą užduotį rasite chemijos lentelių rinkinį. Juo galite naudotis per visą egzamino laiką.
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  • Pereiti prie tolesnės užduoties galite spausdami arba apačioje esančią rodyklę \"FR_info_tolyn\" , arba lango apačioje esantį reikiamą skrituliuką su skaičiumi.
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  • Atlikus užduotį, jos skrituliukas lango apačioje patamsės. Neatliktos užduoties skrituliukas liks baltas.
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  • Jeigu dėl savo atsakymo abejojate arba užduoties neatlikote, galite paspausti mygtuką \"ivad_p\" ir pažymėti šią užduotį, kad prie jos galėtumėte grįžti vėliau.
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  • Visada galite grįžti prie bet kurios ankstesnės užduoties ir, jei reikia, atsakymą pataisyti.
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  • Paspaudę mygtuką \"ivad_s\" galite bet kuriuo testo atlikimo metu pasitikrinti, kurias užduotis jau atlikote, o kurių dar ne.
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  • SVARBU! Atlikę užduotis, nepamirškite paspausti mygtuko\"ivad_s\" iki užduotims atlikti skirto laiko pabaigos, kad būtų įskaityti jūsų atsakymai. Paspaudę šį mygtuką, savo atsakymų daugiau taisyti nebegalėsite.
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\n ㅤ\n

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\n Norėdami pradėti, spauskite apačioje esančią rodyklę.\n

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\n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i69f12ac3ab9e5202604282346433765e62b","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":[],"openQuestion":0}},{"identifier":"i69f12acb6ef66202604282346518e63924f","guid":"i69f12acb6ef66202604282346518e63924f","label":"INFO_4","title":"INFO_4","submissionMode":null,"xml":"\n\n \n \n \n
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ㅤ

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CHEMIJOS VALSTYBINIO BRANDOS EGZAMINO
\nI DALIES UŽDUOTIS

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Trukmė – 90 min.

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Linkime sėkmės!

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Spauskite apačioje esančią rodyklę.

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\n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i69f12acb6ef66202604282346518e63924f","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":[],"openQuestion":0}},{"identifier":"i69f12adccbf5120260428234708cadbd4a2","guid":"i69f12adccbf5120260428234708cadbd4a2","label":"INFO_5","title":"LENTELIŲ RINKINYS","submissionMode":null,"xml":"\n\n \n \n \n
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\n LENTELIŲ RINKINYS\n

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\n 1. Medžiagų tirpumo vandenyje lentelė\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
\n \n \n \n \n \n Jonai\n \n \n \n \\textrm{Jonai}\n \n \n \n \n \n \n \n \n \n H\n \n +\n \n \n \n \\textrm{H}^+\n \n \n \n \n \n \n \n \n \n \n Na\n \n +\n \n \n \n \\textrm{Na}^+\n \n \n \n \n \n \n \n \n \n NH\n \n 4\n +\n \n \n \n \\textrm{NH}_4^+\n \n \n \n \n \n \n \n \n \n K\n \n +\n \n \n \n \\textrm{K}^+\n \n \n \n \n \n \n \n \n \n Li\n \n +\n \n \n \n \\textrm{Li}^+\n \n \n \n \n \n \n \n \n \n Ag\n \n +\n \n \n \n \\textrm{Ag}^+\n \n \n \n \n \n \n \n \n \n Mg\n \n \n 2\n +\n \n \n \n \n \\textrm{Mg}^{2+}\n \n \n \n \n \n \n \n \n \n Ca\n \n \n 2\n +\n \n \n \n \n \\textrm{Ca}^{2+}\n \n \n \n \n \n \n \n \n \n Ba\n \n \n 2\n +\n \n \n \n \n \\textrm{Ba}^{2+}\n \n \n \n \n \n \n \n \n \n Zn\n \n \n 2\n +\n \n \n \n \n \\textrm{Zn}^{2+}\n \n \n \n \n \n \n \n \n \n Cu\n \n \n 2\n +\n \n \n \n \n \\textrm{Cu}^{2+}\n \n \n \n \n \n \n \n \n \n Ni\n \n \n 2\n +\n \n \n \n \n \\textrm{Ni}^{2+}\n \n \n \n \n \n \n \n \n \n Pb\n \n \n 2\n +\n \n \n \n \n \\textrm{Pb}^{2+}\n \n \n \n \n \n \n \n \n \n Sn\n \n \n 2\n +\n \n \n \n \n \\textrm{Sn}^{2+}\n \n \n \n \n \n \n \n \n \n Fe\n \n \n 2\n +\n \n \n \n \n \\textrm{Fe}^{2+}\n \n \n \n \n \n \n \n \n \n Fe\n \n \n 3\n +\n \n \n \n \n \\textrm{Fe}^{3+}\n \n \n \n \n \n \n \n \n \n Al\n \n \n 3\n +\n \n \n \n \n \\textrm{Al}^{3+}\n \n \n
\n \n \n \n \n \n \n Br\n \n −\n \n \n \n \\textrm{Br}^-\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n
\n \n \n \n \n \n \n CH\n \n 3\n \n \n \n COO\n \n −\n \n \n \n \\textrm{CH}_3\\textrm{COO}^-\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n
\n \n \n \n \n \n \n CO\n \n 3\n \n 2\n −\n \n \n \n \n \\textrm{CO}^{2-}_3\n \n \n \n \n \n \n \n \n \n CO\n \n 2\n \n \n \n \\textrm{CO}_2\n \n \n
\n
\n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n 		\n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n
\n \n \n \n \n \n \n Cl\n \n −\n \n \n \n \\textrm{Cl}^-\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n
\n \n \n \n \n \n \n F\n \n −\n \n \n \n \\textrm{F}^-\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n
\n \n \n \n \n \n \n I\n \n −\n \n \n \n \\textrm{I}^-\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n
\n \n \n \n \n \n CuI\n \n \n \n \\textrm{CuI}\n \n \n
\n
\n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n 		\n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n –\n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n
\n \n \n \n \n \n \n NO\n \n 3\n \n −\n \n \n \n \n \\textrm{NO}_3^{-}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n
\n \n \n \n \n \n \n OH\n \n −\n \n \n \n \\textrm{OH}^-\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n \n NH\n \n 3\n \n \n \n \\textrm{NH}_3\n \n \n
\n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n 		\n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n \n Ag\n \n 2\n \n \n O\n \n \n \n \\textrm{Ag}_2\\textrm{O}\n \n \n
\n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n 		\n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n
\n \n \n \n \n \n \n PO\n \n 4\n \n 3\n −\n \n \n \n \n \\textrm{PO}^{3-}_4\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n
\n \n \n \n \n \n \n S\n \n \n 2\n −\n \n \n \n \n \\textrm{S}^{2-}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n
\n \n \n \n \n \n \n SO\n \n 3\n \n 2\n −\n \n \n \n \n \\textrm{SO}^{2-}_3\n \n \n \n \n \n \n \n \n \n SO\n \n 2\n \n \n \n \\textrm{SO}_2\n \n \n
\n
\n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n 		\n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n
\n \n \n \n \n \n \n SO\n \n 4\n \n 2\n −\n \n \n \n \n \\textrm{SO}^{2-}_4\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n m\n \n \n \n \\textrm{m}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n
\n \n \n \n \n \n \n SiO\n \n 3\n \n 2\n −\n \n \n \n \n \\textrm{SiO}^{2-}_3\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n r\n \n \n \n \\textrm{r}\n \n \n \n \n \n \n \n \n t\n \n \n \n \\textrm{t}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n \n \n \n \n \n \n n\n \n \n \n \\textrm{n}\n \n \n
\n

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t\\textrm{t} – tirpus, m\\textrm{m} – mažai tirpus, n\\textrm{n} – netirpus, r\\textrm{r} – susidarymo metu reaguoja su vandeniu (hidrolizuojasi). Brūkšnys rodo, kad tokio junginio nėra. Jeigu junginys nepatvarus, lentelėje nurodytas galutinis skilimo produktas.

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\n 2. Pagrindinių periodų ir grupių elementų elektriniai neigiamumai santykiniais vienetais*\n

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\n \"F2\"\n \n

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* Parengta pagal David R. Lide, ed., „CRC Handbook of Chemistry and Physics“, 90th Edition (CD-ROM Version 2010), CRC Press/Taylor and Francis, Boca Raton, FL.

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\n 3. Metalų aktyvumo eilė\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
LiKBaCaNaMgAlTiMnZnCrFeCoNiSnPbH2CuAgHgPtAu
\n

\n

\n 4. Kai kurios pagrindinės konstantos\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
Dujų molio tūrisVM = 22,7 dm3/mol arba L/mol*
Oro molinė masėM(oro) = 28,96 g/mol
Avogadro konstantaNA = 6,02 · 1023 mol−1
Standartinis slėgis100 kPa arba 1 bar
Standartinė temperatūra0 ­­°C arba 273 K
\n

* Tūrio matavimo vienetą litrą galima trumpinti dvejopai: l arba L.

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\n 5. Anglies–anglies ir anglies–vandenilio ryšių ilgiai ir energija\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
Sutrumpinta
\n\t\t\tstruktūrinė
\n\t\t\tformulė		C – C ryšio
\n\t\t\tilgis (pm)		C – C ryšio
\n\t\t\tenergija (kJ/mol)		C – H ryšio
\n\t\t\tilgis (pm)		C – H ryšio
\n\t\t\tenergija (kJ/mol)
CH3 – CH3154377110339
CH2 = CH2134728108466
CH ≡ CH120967106548
\n

\n

\n 6. Bazių jonizacijos (disociacijos) konstantos (Kb, mol/L) 25 °C temperatūroje \n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
Bazės vandeninis tirpalasNH3CH3NH2(CH3)2NH(CH3)3N
Jonizacijos konstanta\n

1,8 ∙ 10−5

\n 		4,6 ∙ 10−45,4 ∙ 10−46,5 ∙ 10−5
\n

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\n 7. Rūgščių jonizacijos (disociacijos) konstantos (Ka, mol/L) 25 °C temperatūroje\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
\n

Rūgšties vandeninis tirpalas

\n 		HFHClHBrHIHNO3H2CO3H2SH2SO3H2SO4HCOOHCH3COOH
Jonizacijos konstanta\n 6,8 ∙ 10−4\n 7,94  ∙  1056,31  ·  1083,16  ·  1091  ·  101\n

Ka1 = 4,4 ∙ 10−7

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Ka2 = 5,6 ∙ 10−11

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Ka1 = 5,7 ∙ 10−8

Ka2 = 1,3 ∙ 10−13

Ka1 = 1,7 ∙ 10−2

Ka2 = 6,2 ∙ 10−8

Ka1 = 1  ·  103

Ka2 = 1,2 ∙ 10−2		\n 1,77 ∙ 10−5\n \n 1,75 ∙ 10−5\n
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\n 8. Kai kurios fizikinių dydžių formulės\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
Fizikinis dydisRyšys tarp fizikinių dydžiųFizikinių dydžių žymėjimas
Tankis, ρ\\rho (kg/m3)\n \n \n \n \n ρ\n =\n \n m\n V\n \n \n \n \\rho=\\frac{m}{V}\n \n \n mm – medžiagos masė
\n\t\t\tVV – medžiagos tūris
Tirpalo molinė koncentracija, cc (mol/L)\n \n \n \n \n c\n =\n \n n\n \n V\n \n \n (tirpalo)\n \n \n \n \n \n \n c=\\frac{n}{V_{\\textrm{(tirpalo)]]]}\n \n \n nn – medžiagos kiekis
\n\t\t\tV(tirpalo)V_{\\textrm{(tirpalo)]]] – tirpalo tūris
Tirpalo procentinė koncentracija, ω\\omega (%)\n \n \n \n \n \n ω\n \n \n (A)\n \n \n \n =\n \n \n m\n \n \n (A)\n \n \n \n \n m\n \n \n (tirpalo)\n \n \n \n \n \n \n \\omega_{\\textrm{(A)]]]=\\frac{m_{\\textrm{(A)]]]}{m_{\\textrm{(tirpalo)]]]}\n \n \n m(A)m_{\\textrm{(A)]]] – grynos medžiagos masė
\n\t\t\tm(tirpalo)m_{\\textrm{(tirpalo)]]] – mišinio ar tirpalo masė
Tirpalo masės koncentracija, cwc_w (g/L arba g/dm3)\n \n \n \n \n \n c\n w\n \n =\n \n m\n \n V\n \n \n (tirpalo)\n \n \n \n \n \n \n c_w=\\frac{m}{V_{\\textrm{(tirpalo)]]]}\n \n \n mm – grynos medžiagos masė
\n\t\t\tV(tirpalo)V_{\\textrm{(tirpalo)]]] – tirpalo tūris
Išeiga, η\\eta (%)\n \n \n \n \n η\n =\n \n \n n\n \n \n (praktinis)\n \n \n \n \n n\n \n \n (teorinis)\n \n \n \n \n ⋅\n 100\n *ATTRIBUTE_NEWLINE*\n %\n ;\n *ATTRIBUTE_NEWLINE*\n η\n =\n \n \n V\n \n \n (praktinis)\n \n \n \n \n V\n \n \n (teorinis)\n \n \n \n \n ⋅\n 100\n *ATTRIBUTE_NEWLINE*\n %\n ;\n *ATTRIBUTE_NEWLINE*\n η\n =\n \n \n m\n \n \n (praktinis)\n \n \n \n \n m\n \n \n (teorinis)\n \n \n \n \n ⋅\n 100\n *ATTRIBUTE_NEWLINE*\n %\n \n \n \\eta = \\frac{n_{\\textrm{(praktinis)]]]}{n_{\\textrm{(teorinis)]]]}\\cdot100\\ \\%;\\ \\eta = \\frac{V_{\\textrm{(praktinis)]]]}{V_{\\textrm{(teorinis)]]]}\\cdot100\\ \\%;\\ \\eta = \\frac{m_{\\textrm{(praktinis)]]]}{m_{\\textrm{(teorinis)]]]}\\cdot100\\ \\%\n \n \n nn – medžiagos kiekis
\n\t\t\tVV – medžiagos tūris
\n\t\t\tmm – medžiagos masė
Medžiagos kiekis, nn (mol)\n \n \n \n \n n\n =\n \n m\n M\n \n ;\n *ATTRIBUTE_NEWLINE*\n n\n =\n \n N\n \n N\n A\n \n \n ;\n *ATTRIBUTE_NEWLINE*\n n\n =\n \n \n V\n \n \n (dujų)\n \n \n \n \n V\n M\n \n \n \n \n n=\\frac{m}{M};\\ n = \\frac{N}{N_A};\\ n=\\frac{V_{\\textrm{(dujų)]]]}{V_M}\n \n \n V(dujų)V_{\\textrm{(dujų)]]] – dujų tūris
\n\t\t\tmm – grynos medžiagos masė
\n\t\t\tNN – dalelių skaičius
\n\t\t\tMM – medžiagos molinė masė
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\n 9. Periodinė cheminių elementų lentelė\n

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\n \"Chemini\n

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IUPAC rekomenduoja grupes numeruoti arabiškaisiais skaitmenimis. Skliaustuose nurodyti tradiciniai grupių numeriai.

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Parengta pagal:
\nJ. Meija et al. Atomic weights of the elements, 2013 (IUPAC Technical Report). Pure Appl. Chem. 88 (2016) 265–291. Skliaustuose nurodyta stabiliausio izotopo masė.

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L Öhrström et al. Names and symbols of the elements with atomic numbers 113, 115, 117 and 118 (IUPAC Recommendations 2016). Pure Appl. Chem. 88 (2016) 1 225–1 229.

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\n 10. Gamtinių aminorūgščių tirpumas (g/100 g H2O) vandenyje 25 °C temperatūroje*\n

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\n \"F10\"\n

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* Parengta pagal Michel Fleck, Aram M. Petrosyan, „Salts of Amino Acids Crystallization, Structure and Properties Springer“, 2014.

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\n 11. Dujų tirpumas (g/100 g H2O) vandenyje skirtingose temperatūrose\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
T, °CO2N2H2CO2COSO2NOHClNH3CH4C2H6C2H4C2H2
00,0069450,0029420,00019220,33460,00439722,830,00983382,389,50,0039590,013170,02810,200
100,0053680,0023120,00017400,23180,00347916,210,00756077,268,40,0029550,008700,02000,150
200,0043390,0019010,00016030,16880,00283811,280,00617372,152,90,0023190,006200,01490,117
300,0035880,0016240,00014740,12570,0024057,800,00516567,341,00,0019040,004680,01180,094
400,0030820,0013910,00013840,09730,0020755,410,00439463,331,60,0015860,00366
500,0026570,0012160,00012870,07610,001797 0,00375856,123,50,0013590,00294
600,0022740,0010520,00011780,05760,001522 0,003237 16,80,0011440,00239
700,0018560,0008510,000102 0,001276 0,002668 11,10,0009260,00185
800,0013810,0006600,000079 0,000980 0,001984 6,50,0006950,00134
900,000790,000380,000046 0,00057 0,00113 3,00,000400,00080
1000,000000,000000,000000 0,00000 0,00000 0,00,000000,00000
\n

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\n
\n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i69f12adccbf5120260428234708cadbd4a2","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":[],"openQuestion":0}},{"identifier":"I_01","guid":"I_01","label":"I_01","title":"I_01","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

\n I dalis\n

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1. Vieną iš naftos perdirbimo procesų galima užrašyti tokia cheminės reakcijos lygtimi:

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Alkanas → Mažesnės molinės masės alkanas + Alkenas (+ kartais vandenilio dujos)

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Pažymėkite šio proceso pavadinimą.

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\n (1 taškas)\n

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\n
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\n \n \n

Distiliavimas

\n \n \n

Alkilinimas

\n \n \n

Krekingas

\n \n \n

Hidrinimas

\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f1870dfdf0202606142309040fa73ae8","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_02","guid":"I_02","label":"I_02","title":"I_02","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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2. Cheminės medžiagos struktūrinėje formulėje pažymėkite sp hibridizacijos anglies atomą.

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\n (1 taškas)\n

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\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f1875d10c8202606142309098bb3d658","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["hotspotInteraction"],"openQuestion":0}},{"identifier":"I_03","guid":"I_03","label":"I_03","title":"I_03","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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3. Pateikta etanalio struktūrinė formulė. Pažymėkite cheminį ryšį (raidę a, b, c arba d), kuris nutrūks etanalį oksiduojant iki etano rūgšties.

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\n (1 taškas)\n

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\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f187a7c5a020260614230914ec305126","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["hotspotInteraction"],"openQuestion":0}},{"identifier":"I_04","guid":"I_04","label":"I_04","title":"I_04","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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4. Reaguodamas į stresą, žmogaus organizmas išskiria adrenaliną.

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\n \"K04\"\n \n

\n

Remdamiesi pateikta adrenalino struktūrine formule, pasirinkite tinkamą žodį, kad teiginys apie šią cheminę medžiagą būtų teisingas.

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\n (1 taškas)\n

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Adrenalinas yra pirminisantrinistretinisketvirtinisaminas.

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\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f187f0f67c20260614230919f4b703e0","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["inlineChoiceInteraction"],"openQuestion":0}},{"identifier":"I_05","guid":"I_05","label":"I_05","title":"I_05","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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5. Tam tikros savybės būdingos tik organinėms medžiagoms. Įkelkite į laukelius dvi tinkamas savybių korteles.

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\n (1 taškas)\n

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\n
\n
\n
\n \n Sudaro homologines eiles.\n Anglėja kaitinamos be deguonies.\n Lydosi aukštoje temperatūroje.\n Tirpsta etanolyje arba benzine.\n Lydalai yra laidūs elektros srovei.\n Nedega deguonies atmosferoje.\n

​Savybės, būdingos tik organinėms medžiagoms:

\n

​

\n

​

\n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 2\n \n \n 1\n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a3096e1e62aa20260616022049bfec402f","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["gapMatchInteraction"],"openQuestion":0}},{"identifier":"I_06","guid":"I_06","label":"I_06","title":"I_06","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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6. Remdamiesi pateiktais teiginiais, nustatykite cheminę medžiagą ir pažymėkite jos struktūrinę formulę.

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62,02 % šios cheminės medžiagos masės sudaro anglis.

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Medžiagą pakaitinus su vario(II) hidroksidu, susidaro raudonos nuosėdos.

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\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n

CH3COCH3

\n \n \n

CH3CH2CHO

\n \n \n

CH3CHO

\n \n \n

CH3COOCH2CH3

\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18884d4282026061423092849c7a839","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_07","guid":"I_07","label":"I_07","title":"I_07","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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7. Pateiktoje nuosekliai vykdomų kitimų eilėje raidė X  žymi tam tikrą cheminę medžiagą. Kokią? Pažymėkite teisingą atsakymą.

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Etenas → X  → Etilenglikolis

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\n (1 taškas)\n

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\n
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\n \n Chloretanas\n Etinas\n 1,1-dichloretanas\n 1,2-dichloretanas\n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f188ccd86c20260614230932bb78f746","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_08","guid":"I_08","label":"I_08","title":"I_08","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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8. Kurių dujų tankis standartinėmis sąlygomis yra lygus 1,15 g/L? Pažymėkite teisingą atsakymą.

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\n (1 taškas)\n

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\n
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\n \n \n

Eteno

\n \n \n

Etino

\n \n Propeno\n Propino\n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f1891636be20260614230937c475ae6b","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_09","guid":"I_09","label":"I_09","title":"I_09","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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9. Pavaizduota riebalų molekulė. Kurie trys cheminiai ryšiai nutrūks šiai molekulei visiškai hidrolizuojantis? Pažymėkite teisingą atsakymą.

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\n (1 taškas)\n

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\n
\n
\n
\n

\n \"K09\"\n \n

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\n
\n \n 1, 4, 9\n 2, 6, 8\n 2, 6, 9\n 3, 6, 7\n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18960191420260614230942df80b1d0","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_10","guid":"I_10","label":"I_10","title":"I_10","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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10. Pažymėkite cis-alkeno struktūrinę formulę.

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\n (1 taškas)\n

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\n
\n
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\n \n \n \"K10A\"\n \n \n \"K10B\"\n \n \n \"K10C\"\n \n \n \"K10D\"\n \n \n
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\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f189a9586e2026061423094632b65cc4","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_11","guid":"I_11","label":"I_11","title":"I_11","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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11. Chemikas iš dalies hidrolizavo celiuliozę ir krakmolą, o tada iš jų pagamino vandeninius A ir B tirpalus. Į tirpalus jis įpylė reagentų ir stebėjo cheminių reakcijų požymius.

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
\n Reagentai ir sąlygos\n \n A tirpalas\n \n B tirpalas\n
Ką tik paruoštas vario(II) hidroksidasNuosėdos ištirpo ir tirpalas nusidažė mėlyna spalva.Nuosėdos ištirpo ir tirpalas nusidažė mėlyna spalva.
Jodo tirpalasTirpalas nusidažė tamsiai mėlyna spalva.Cheminės reakcijos požymis nepasireiškė.
\n

Remdamiesi lentelės duomenimis, nustatykite, kuriam tirpalui buvo panaudota celiuliozė, kuriam – krakmolas. Lentelės langeliuose pasirinkite reikiamos cheminės medžiagos pavadinimą.

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\n (1 taškas)\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n
\n A tirpalui\n \n B tirpalui\n
Panaudota cheminė medžiaga\n \n Celiuliozė\n Krakmolas\n \n \n \n Celiuliozė\n Krakmolas\n \n
\n

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\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n \n \n \n 2\n \n \n 1\n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a309ac5c49f120260616023725a14ed68c","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["inlineChoiceInteraction"],"openQuestion":0}},{"identifier":"I_12","guid":"I_12","label":"I_12","title":"I_12","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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12. Etiletanoato (CH3COOCH2CH3) saugos duomenų lape rašoma, kad tai degus saldaus kvapo bespalvis skystis, stipriai dirginantis akis ir galintis sukelti galvos svaigimą. Etiketėse šios medžiagos pavojingumas nurodomas dviem piktogramomis. Pažymėkite šią piktogramų porą.

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\n (1 taškas)\n

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\n
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\n \n \n \"K12A\"\n \n \n \"K12B\"\n \n \n \"K12C\"\n \n \n \"K12D\"\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18a4230d820260614230956451b0c2d","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_13","guid":"I_13","label":"I_13","title":"I_13","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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13. Jodmetanas (CH3I) reagavo su natrio šarmo (NaOH) vandeniniu tirpalu ir susidarė metanolis (CH3OH). Pažymėkite šios cheminės reakcijos mechanizmo pavadinimą.

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\n (1 taškas)\n

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Radikalinis pakaitų

\n \n Elektrofilinis jungimosi\n Nukleofilinis pakaitų\n Nukleofilinis jungimosi\n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18a8bc3d5202606142310009be4d33d","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_14","guid":"I_14","label":"I_14","title":"I_14","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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14. Pažymėkite paveikslą, kuriame teisingai pavaizduoti vandeniliniai ryšiai tarp guanino ir citozino azotinių bazių.

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\n (1 taškas)\n

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\n
\n
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\n \n \n \"K14A\"\n \n \n \"K14B\"\n \n \n \"K14C\"\n \n \n \"K14D\"\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18ad6e2b9202606142310054635a83e","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_15","guid":"I_15","label":"I_15","title":"I_15","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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15. Kurios gamtinės medžiagos struktūroje yra peptidinis ryšys? Pažymėkite teisingą atsakymą.

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\n (1 taškas)\n

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\n
\n
\n
\n \n Riebalų\n \n

Baltymų

\n \n Celiuliozės\n Deoksiribonukleorūgšties (DNR)\n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18b20d50620260614231010a2127134","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_16","guid":"I_16","label":"I_16","title":"I_16","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n
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\n II DALIS\n

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\n 1 struktūrinis klausimas\n

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\n
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Citrinų rūgštis (M(C6H8O7) = 192,12 g/mol) plačiai naudojama maisto pramonėje. Ji gaunama fermentuojant gliukozę (M(C6H12O6) = 180,16 g/mol).

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16. Išlyginkite gliukozės fermentacijos į citrinų rūgštį cheminės reakcijos lygtį. Įrašykite reikiamus skaičius.

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\n (1 taškas)\n

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C6H12O6 + 3O2 → C6H8O7 + H2O

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\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n \n \n \n 3\n \n \n 1\n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a309e7fe95a820260616025319fb686f09","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_17","guid":"I_17","label":"I_17","title":"I_17","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

17. Remdamiesi citrinų rūgšties struktūrine formule, nustatykite, kiek molių CO2 susidarys visiškai sudeginus 1 mol citrinų rūgšties. Įrašykite atsakymą.

\n

\n (1 taškas)\n

\n

\n \"K17\"\n \n

\n

Atsakymas: susidarys mol CO2.

\n
\n
\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18bb4ab60202606142310199bda7504","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_18","guid":"I_18","label":"I_18","title":"I_18","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

18. Remdamiesi citrinų rūgšties struktūrine formule, nustatykite, kokių funkcinių grupių ji turi. Pažymėkite teisingą atsakymą.

\n

\n (1 taškas)\n

\n

\n \"K17\"\n \n

\n
\n
\n
\n
\n \n \n

Amino ir hidroksilo

\n \n Hidroksilo ir karboksilo\n Karboksilo ir aldehidinę\n \n

Esterio ir amino

\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18bfeef11202606142310232bf627e7","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_19","guid":"I_19","label":"I_19","title":"I_19","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n
\n
\n

19. Obuolių rūgšties ir citrinų rūgšties molekulių santykis apelsinų sultyse buvo išmatuotas naudojantis 1H  BMR spektroskopija.

\n

Obuolių rūgšties ir citrinų rūgšties struktūrinėse formulėse pažymėtų vandenilio atomų signalų cheminiai poslinkiai ir tų signalų integralai (plotai) apelsinų sulčių 1H  BMR spektre yra pateikti lentelėje. Šie integralai yra tiesiogiai proporcingi molekulių skaičiui tirpale.

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
Organinė rūgštisStruktūrinė formulėPažymėtų vandenilio
\n\t\t\tatomų signalų cheminis
\n\t\t\tposlinkis, ppm		Pažymėtų vandenilio
\n\t\t\tatomų signalų integralas
Citrinų\n \"K19_1\"\n \n 2,9516
Obuolių\n \"K19_2\"\n \n 2,882
\n

Apskaičiuokite citrinų rūgšties ir obuolių rūgšties molekulių santykį apelsinų sultyse.

\n

\n (1 taškas)\n

\n

 

\n

Atsakymas: citrinų rūgšties ir obuolių rūgšties molekulių santykis yra : 1.

\n
\n
\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18c49b8b020260614231028f3a45ede","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_20","guid":"I_20","label":"I_20","title":"I_20","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

20. Karboksirūgšties funkcinės grupės atomams priskirkite dalinį teigiamąjį (σ+) krūvį. Į tinkamus langelius įkelkite dalinio teigiamojo krūvio korteles.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18c98fff520260614231033b628690f","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["graphicGapMatchInteraction"],"openQuestion":0}},{"identifier":"I_21","guid":"I_21","label":"I_21","title":"I_21","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

\n 2 struktūrinis klausimas\n

\n
\n
\n
\n
\n

Anilinas – organinė medžiaga, naudojama gaminant dažiklius.

\n
\n
\n
\n
\n

21. Pažymėkite anilino struktūrinę formulę.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n \"K22A\"\n \n \n \"K22B\"\n \n \n \"K22C\"\n \n \n \"K22D\"\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18d2bf54f202606142310422b23336b","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_22","guid":"I_22","label":"I_22","title":"I_22","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

22. Anilinas gaminamas redukuojant nitrobenzeną, todėl cheminės reakcijos produkte visada būna nitrobenzeno pėdsakų. Siekiant išgryninti aniliną, taikoma distiliacija. Pasirinkite distiliavimo įrangą, kurioje termometro padėtis yra teisinga.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n \"K23A\"\n \n \n \"K23B\"\n \n \n \"K23C\"\n \n \n \"K23D\"\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18d763dca20260614231047ac947e1d","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_23","guid":"I_23","label":"I_23","title":"I_23","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

23. Paveiksle pateikti nitrobenzeno ir anilino virimo temperatūros priklausomybės nuo slėgio distiliavimo įrangoje grafikai.

\n

\n \"K24\"\n \n

\n

Remdamiesi grafiku, pasirinkite anilino virimo temperatūrą 10 kPa slėgyje.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n 200–202 °C\n \n

182–184 °C

\n \n \n

118–120 °C

\n \n \n

108–110 °C

\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18dc05360202606142310526c9e3001","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_24","guid":"I_24","label":"I_24","title":"I_24","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

24. Paveiksle pateikti nitrobenzeno ir anilino virimo temperatūros priklausomybės nuo slėgio distiliavimo įrangoje grafikai.

\n

\n \"K24\"\n \n

\n

Remdamiesi grafiku, pasirinkite tinkamus žodžius, kad teiginys apie anilino ir nitrobenzeno mišinio distiliavimą būtų teisingas.

\n

\n (1 taškas)\n

\n

Esant 10 kPa slėgiui, pirmasis distiliuojasi anilinasnitrobenzenas, nes šios cheminės medžiagos virimo temperatūra yra žemesnėaukštesnė.

\n
\n
\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a30a7008593220260616032936a6556304","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["inlineChoiceInteraction"],"openQuestion":0}},{"identifier":"I_25","guid":"I_25","label":"I_25","title":"I_25","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

25. Distiliuojant surinkto produkto grynumas buvo tiriamas plonasluoksnės chromatografijos būdu. Tyrimui naudota chromatografinė plokštelė yra pavaizduota paveiksle.

\n

\n \"K26\"\n \n

\n

\n Remdamiesi plokštelės duomenimis, pažymėkite teisingą teiginį apie distiliacijos produkto grynumą.\n

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n

Distiliacijos produktas negrynas, nes jame yra nežinomų priemaišų.

\n \n \n

Distiliacijos produktas grynas, nes jį sudaro tik anilinas.

\n \n \n

Distiliacijos produktas negrynas, nes jame yra nitrobenzeno ir anilino.

\n \n \n

Distiliacijos produktas grynas, nes jį sudaro tik nitrobenzenas.

\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18e53c6d320260614231101943b662e","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_26","guid":"I_26","label":"I_26","title":"I_26","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n
\n
\n

\n 3 struktūrinis klausimas\n

\n
\n
\n
\n
\n

Vandenilio dujos – ekologiškas, bet sunkiai transportuojamas kuras. Siekiant užtikrinti veiksmingą jo transportavimą ir saugojimą, kuriama speciali technologija. Jai naudojamos skystosios organinės medžiagos, galinčios prisijungti ir išlaisvinti vandenilio dujas vykstant cheminei reakcijai.

\n

Viena iš tokių organinių medžiagų yra dekalinas (C10H18).

\n
\n
\n
\n
\n

26. Dekalino homologinės eilės nariai yra izomerai angliavandenilių, tarp anglies atomų turinčių vieną trigubąjį ryšį. Baikite rašyti šių angliavandenilių homologinės eilės bendrąją formulę.

\n

\n (1 taškas)\n

\n

 

\n

Bendroji formulė CnH 

\n
\n
\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a30a94abae7f2026061603392220fcaff8","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_27","guid":"I_27","label":"I_27","title":"I_27","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

27. Dekalinas gaunamas iš naftaleno, kurio struktūrinė formulė:

\n

\n \"K28\"\n \n

\n

Žinoma, kad naftalenas neblukina vandeninių kalio permanganato ir bromo tirpalų. Pažymėkite, kaip klasifikuojamas šis angliavandenilis.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n

Sotusis

\n \n \n

Nesotusis

\n \n \n

Aromatinis

\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18ee780b320260614231110ce66eb7a","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_28","guid":"I_28","label":"I_28","title":"I_28","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

28. Paveiksle pavaizduota diagrama rodo naftaleno būsenas (kietąją, skystąją arba dujinę) tam tikrame slėgyje ir temperatūroje.

\n

\n \"K29\"\n \n

\n

Remdamiesi diagrama, pažymėkite teiginį, kuriuo apibūdinami naftaleno būsenų virsmai, jį kaitinant nuo 20 °C iki 100 °C ir esant 1 atm slėgiui.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n

Tokiomis sąlygomis naftalenas lydosi ir garuoja.

\n \n \n

Tokiomis sąlygomis naftalenas tik lydosi.

\n \n \n

Tokiomis sąlygomis naftalenas tik garuoja.

\n \n \n

Tokiomis sąlygomis naftaleno būsena nesikeičia.

\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18f36dfd02026061423111597953406","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_29","guid":"I_29","label":"I_29","title":"I_29","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

29. Dekalinas (M = 138,28 g/mol) gaunamas vandenilio dujomis hidrinant naftaleną (M = 128,18 g/mol) ir naudojant katalizatorių.

\n

Vyksta cheminė reakcija:

\n

\n \"K30\"\n \n

\n

Apskaičiuokite, kiek litrų vandenilio dujų standartinėmis sąlygomis sureaguos su 1,00 kg naftaleno. Atsakymą pateikite trijų reikšminių skaitmenų tikslumu.

\n

\n (1 taškas)\n

\n

Atsakymas: vandenilio dujų tūris L.

\n
\n
\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18f80692e2026061423112040cd52af","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_30","guid":"I_30","label":"I_30","title":"I_30","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

30. Kaitinant dekaliną ir naudojant katalizatorių, pirmiausia susidaro tarpinė cheminė medžiaga tetralinas, o galiausiai – naftalenas.

\n

\n \"K31\"\n \n

\n

Kaitinant dekaliną ir naudojant katalizatorių, gautas angliavandenilių mišinys, kuriame pradinio dekalino nebuvo likę. Mišinį sudarė 20 mmol tetralino ir 15 mmol naftaleno. Apskaičiuokite pradinį dekalino kiekį (mmol). Įrašykite atsakymą.

\n

\n (1 taškas)\n

\n

Atsakymas: dekalino kiekis mmol.

\n
\n
\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f18fcb5577202606142311245e0cace0","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_31","guid":"I_31","label":"I_31","title":"I_31","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

31. Pateikta naftaleno (C10H8) hidrinimo vandenilio dujomis (H2) termocheminės reakcijos lygtis:

\n

C10H8(k) + 5H2(d) → C10H18(k); ΔHo = –260 kJ.

\n

Apskaičiuokite, kiek sureagavo vandenilio dujų (mol), jeigu per cheminę reakciją išsiskyrė 3640 kJ šilumos. Įrašykite atsakymą.

\n

\n (1 taškas)\n

\n

Atsakymas: vandenilio dujų kiekis mol.

\n
\n
\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f19016338f202606142311298ce1b856","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_32","guid":"I_32","label":"I_32","title":"I_32","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

32. Vandeniliui saugoti ir transportuoti naudojamos ir kitos cheminės medžiagos.

\n

Pažymėkite cheminės medžiagos, kurios 1 mol gali prisijungti daugiausia vandenilio dujų, struktūrinę formulę.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n \"K33A\"\n \n \n \"K33B\"\n \n \n \"K33C\"\n \n \n \"K33D\"\n \n \n
\n
\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f1905f294a2026061423113397b69e92","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_33","guid":"I_33","label":"I_33","title":"I_33","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

33. 2-metilbut-1-enas reaguoja su vandenilio bromidu (HBr).

\n

\n \"K33\"\n \n

\n

Pažymėkite, kurio cheminės reakcijos produkto susidarys daugiausia.

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\n (1 taškas)\n

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\n \n \n \"K33A\"\n \n \n \"K33B\"\n \n \n \"K33C\"\n \n \n \"K33D\"\n \n \n
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\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a34fe2599543202606191030295135a2e3","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_34","guid":"I_34","label":"I_34","title":"I_34","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 4 struktūrinis klausimas\n

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Daugybė buityje naudojamų daiktų yra pagaminti iš polimerinių medžiagų.

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34. Vienas paprasčiausių polimerų yra polietilenas (polietenas). Jo struktūrinė formulė:

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(–CH2–CH2–)n;

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čia n – polimerizacijos laipsnis.

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Apskaičiuokite mažo tankio polietileno polimerizacijos laipsnį, jeigu jo molekulinė masė lygi 30 000 g/mol.

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\n (1 taškas)\n

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Atsakymas: n = .

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\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f190a9e5db2026061423113826519c23","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_35","guid":"I_35","label":"I_35","title":"I_35","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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35. Didelio tankio polietileno (polieteno) molekulės yra ilgos ir nešakotos. Šio polimero polimerizacijos laipsnis n = 10 000. Monomerinės grandies ilgis yra 0,26 nm. Apskaičiuokite didelio tankio polietileno molekulės ilgį mikrometrais (μm). 1 μm = 1000 nm. Įrašykite atsakymą.

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\n (1 taškas)\n

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Atsakymas: polietileno molekulės ilgis μm.

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\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f190f45b152026061423114394b11309","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_36","guid":"I_36","label":"I_36","title":"I_36","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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36. Tiriant polimerų pailgėjimo priklausomybę nuo įtempio (tempimo jėgos, tenkančios bandinio plotui), polipropileno (polipropeno), didelio bei mažo tankio polietilenų (polietenų) ir polivinilchlorido (polichloreteno) vienodų matmenų bandiniai buvo tempiami, kol nutrūko. Gauti duomenys pavaizduoti grafike.

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\n \"K36\"\n \n

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Remdamiesi grafiku, pažymėkite polimerą, kuris pailgėjo daugiausia.

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\n (1 taškas)\n

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\n \n Polipropilenas\n Didelio tankio polietilenas\n \n

Mažo tankio polietilenas

\n \n Polivinilchloridas\n \n
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\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f1913db46d20260614231147fc79bb84","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_37","guid":"I_37","label":"I_37","title":"I_37","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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37. Remdamiesi polimerų pailgėjimo priklausomybės nuo įtempio grafiku, pažymėkite polimerą, kuriam tempti reikia mažiausios jėgos, kad jis pailgėtų 20 %.

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\n (1 taškas)\n

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\n \"K37\"\n \n

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\n \n Polipropilenas\n \n

Didelio tankio polietilenas

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Mažo tankio polietilenas

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Polivinilchloridas

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\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f19187bde620260614231152bf412ab1","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_38","guid":"I_38","label":"I_38","title":"I_38","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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38. Pažymėkite polipropileno (polipropeno) fragmento skeletinę formulę.

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\n (1 taškas)\n

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\n \n \n \"K38A\"\n \n \n \"K38B\"\n \n \n \"K38C\"\n \n \n \"K38D\"\n \n \n
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\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f191d87b392026061423115797c297e4","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_39","guid":"I_39","label":"I_39","title":"I_39","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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39. Degant polivinilchloridui (polichloretenui) išsiskiria nuodingosios dujos. Į polivinilchlorido degimo lygtį įrašykite nuodingųjų dujų cheminę formulę.

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\n (1 taškas)\n

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2(–CH2–CHCl–)n + 5nO2 → 4nCO2 + 2nH2O + 2n

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\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f19222899220260614231202b957652c","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_40","guid":"I_40","label":"I_40","title":"I_40","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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40. Iš polietileno (polieteno) tereftalato (PET) polimero gaminami gaiviųjų gėrimų buteliai ir skaidrios plėvelės. Polietileno tereftalato struktūrinė formulė:

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\n \"K40\"\n \n

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Susmulkintas polietileno tereftalato bandinys buvo virinamas natrio hidroksido (NaOH) tirpale (perteklius), kol visiškai ištirpo. Pažymėkite šios cheminės reakcijos produktus.

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\n (1 taškas)\n

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\n \n \n \"K40A\"\n \n \n \"K40B\"\n \n \n \"K40C\"\n \n \n \"K40D\"\n \n \n
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\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/pvbe1che6-1782367450-redaguotas/items/i6a2f1926ca8cb202606142312062c56c398","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:35:20","counterId":"ID27335","name":"2026 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"i69f1315a73fe02026042900145040abd131","guid":"i69f1315a73fe02026042900145040abd131","label":"INFO_41","title":"INFO_41","submissionMode":null,"xml":"\n\n \n \n
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DĖMESIO! Atlikę užduotis, nepamirškite paspausti mygtuko \"ivad_u\" iki užduotims atlikti skirto laiko pabaigos, kad būtų įskaityti jūsų atsakymai. Paspaudę šį mygtuką, savo atsakymų daugiau taisyti nebegalėsite.

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Mygtuką \"ivad_u\" rasite, paspaudę apačioje esančią rodyklę.

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Naudota literatūra

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CRC Handbook of Chemistry and Physics, 95th ed.; W. M. Haynes, Ed.; CRC Press: Boca Raton, 2014.

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G. del Campo; I. Berregi; R. Caracena; J. I. Santos. Quantitative Analysis of Malic and Citric Acids in Fruit Juices Using Proton Nuclear Magnetic Resonance Spectroscopy. Anal. Chim. Acta 2006, 556 (2), 462–468.

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N. A. Gordeeva; A. A. Shesterkina; K. V. Vikanova; A. L. Kustov. Naphthalene and Its Derivatives Hydrogenation for Hydrogen Storage: Comparative Analysis of the Role of Noble and Non-Noble Metal Catalysts – A Review. Int. J. Hydrog. Energy 2024, 69, 113–121.

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\n EGZAMINO I DALIS BAIGTA.\n

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Prašytume skirti laiko atsakyti į kelis klausimus. Jūsų atsakymai padės geriau pasirengti būsimiems pasiekimų patikrinimams.

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\n Atsakę į klausimus, būtinai paspauskite mygtuką „Baigti darbą“.\n

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Ar užduotys Jums patiko?

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Taip

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Vidutiniškai

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Ne

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\n Ar užteko laiko?\n
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Taip

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Vidutiniškai

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Ne

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\n Ar buvo lengva viską atlikti?\n
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Taip

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Vidutiniškai

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Ne

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Spauskite „Baigti darbą“

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