{"success":true,"data":{"id":"hash.6a4e35d73c17c.ae3cdec7cfd315fe3bdb49f9994bdc76","title":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","description":null,"count":40,"maxScore":40,"items":[{"identifier":"i69f12ab75081920260428234631b7dd1529","guid":"i69f12ab75081920260428234631b7dd1529","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. pagrindinė 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/vbe1che6-1779280455-redaguotas/items/i69f12ab75081920260428234631b7dd1529","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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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  • 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/vbe1che6-1779280455-redaguotas/items/i69f12abd59dde20260428234637a9dcd9f6","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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/vbe1che6-1779280455-redaguotas/items/i69f12ac3ab9e5202604282346433765e62b","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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/vbe1che6-1779280455-redaguotas/items/i69f12acb6ef66202604282346518e63924f","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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

\n
\n
\n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/vbe1che6-1779280455-redaguotas/items/i69f12adccbf5120260428234708cadbd4a2","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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
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\n I dalis\n

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\n

1. Paveiksle pavaizduota ciklinio angliavandenilio struktūra ir nurodytas cheminių ryšių ilgis (nm). Vandenilio atomai nepavaizduoti.

\n

Pažymėkite raidę (A, B, C arba D), rodančią, kur turėtų būti pavaizduotas C=C dvigubasis ryšys šioje struktūroje.

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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/vbe1che6-1779280455-redaguotas/items/i69f0fb36205f72026042820235011dda096","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["hotspotInteraction"],"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. Tolueno (metilbenzeno) nesutrumpintojoje struktūrinėje formulėje pažymėkite anglies atomą, kurio hibridizacija yra sp3.

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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 \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/vbe1che6-1779280455-redaguotas/items/i69f0fb3b302ef20260428202355165a5f27","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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. Paveiksle pateikta chloroformo (trichlormetano, CHCl3) butelio etiketės dalis. Joje pavaizduotos šios cheminės medžiagos ženklinimo piktogramos.

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Pažymėkite teiginį, kuriuo teisingai paaiškintos saugaus darbo su chloroformu sąlygos.

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\n (1 taškas)\n

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\n
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\n \n Nepurkšti į atvirą liepsną arba kitus degimo šaltinius.\n \n

Neįkvėpti garų (arba dulkių, dūmų, dujų, aerozolio).

\n \n \n

Mūvėti nuo šalčio apsaugančias pirštines.

\n \n \n

Laikyti tik originalioje pakuotėje.

\n \n \n
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\n

\n \"K03\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fb4010c7620260428202400ef3a5656","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"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. Paveiksluose pavaizduotos keturių aminų struktūrinės formulės. Pažymėkite aminą, kuris yra silpniausia bazė.

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\n (1 taškas)\n

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\n
\n
\n
\n \n \n \"K04A\"\n \n \n \"K04B\"\n \n \n \"K04C\"\n \n \n \"K04D\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fb44d669f20260428202404fc6d4b23","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"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 1\n \n \n \n \n
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5. Paveiksle pavaizduota gamtinės cheminės medžiagos fragmento struktūrinė formulė. Pažymėkite, kokia tai cheminė medžiaga.

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\n (1 taškas)\n

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\n
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\n \n \n

Baltymai

\n \n \n

Deoksiribonukleorūgštis (DNR)

\n \n \n

Ribonukleorūgštis (RNR)

\n \n \n

Krakmolas

\n \n \n
\n
\n

\n

\n \"K05\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fb499fe35202604282024098d463c3e","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"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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\n

6. Paveiksle pavaizduota angliavandenilio cheminė formulė. Koks polimeras susidaro polimerizuojant šią cheminę medžiagą? Pažymėkite teisingą atsakymą.

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\n (1 taškas)\n

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\n
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\n \n Polietilenas\n Polipropilenas\n Polivinilchloridas\n Polistirenas\n \n
\n
\n

 

\n

       \n\"K06\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fb4e652b2202604282024142a332f5a","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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. Koks yra paveiksle pavaizduoto angliavandenilio pavadinimas pagal IUPAC nomenklatūrą? Pažymėkite teisingą atsakymą.

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\n (1 taškas)\n

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2-etil-4-pentinas

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3-metil-5-heksinas

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4-metil-1-heksinas

\n \n 4-etil-1-pentinas\n \n
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\n

 

\n

    \n\"K07\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fb531d9302026042820241919d58096","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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. Kuris alkanas susidaro brometanui reaguojant su natriu? Pažymėkite teisingą atsakymą.

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\n (1 taškas)\n

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\n
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\n \n Metanas\n Etanas\n Propanas\n \n

Butanas

\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/vbe1che6-1779280455-redaguotas/items/i69f0fb57c1b1520260428202423d24a9a0b","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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. Paveiksle pavaizduota esterio struktūrinė formulė. Pasirinkite tinkamą žodį, kad teiginys apie esterį būtų teisingas.

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\n (1 taškas)\n

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Šis esteris susidaro karboksirūgščiai reaguojant su alkoholiu, kuriame hidroksigrupė yra prie pirminioantriniotretinioketvirtinio anglies atomo.

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\n
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\n \"K09\"\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","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/vbe1che6-1779280455-redaguotas/items/i69f0fb5c7ca4c20260428202428ce0c28a9","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["inlineChoiceInteraction"],"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. Lentelėje nurodytos skruzdžių, acto, citrinų ir vyno rūgščių jonizacijos (disociacijos) konstantų (Ka) reikšmės. Lentelėje pažymėkite stipriausios rūgšties pavadinimą.

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\n (1 taškas)\n

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1 lentelėje pateiktos pirmosios pakopos citrinų ir vyno rūgščių jonizacijos (disociacijos) konstantos

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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/vbe1che6-1779280455-redaguotas/items/i69f0fb6134a9b20260428202433af4c2549","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["hotspotInteraction"],"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 1\n \n \n \n \n
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11. Kolboje buvo sumaišyti A ir B cheminių medžiagų tirpalai ir įvyko cheminė reakcija. Paveiksle pavaizduota plonasluoksnė chromatografijos plokštelė1, kurią naudojant buvo analizuojami pradiniai A ir B cheminių medžiagų tirpalai bei tirpalas po cheminės reakcijos.

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Remdamiesi paveikslu, pažymėkite teisingą teiginį apie tirpalo sudėtį po cheminės reakcijos.

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\n (1 taškas)\n

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Tirpalą po cheminės reakcijos sudarė B cheminė medžiaga ir dvi naujos medžiagos.

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Tirpalą po cheminės reakcijos sudarė A cheminė medžiaga ir dvi naujos medžiagos.

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Tirpalą po cheminės reakcijos sudarė B cheminė medžiaga ir viena nauja medžiaga.

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Tirpalą po cheminės reakcijos sudarė A ir B cheminės medžiagos, naujų medžiagų nesusidarė.

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1 plokštelė – płytka – пластина – пластинка

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\n
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\n \"K11\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fb661f0852026042820243860129b81","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"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 \n \n \n \n \n 1\n \n \n \n \n
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12. Paveiksle pavaizduotos dviejų organinių junginių struktūrinės formulės.

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\n \"K12\"\n \n

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Pasirinkite tinkamus žodžius, kad teiginys apie šių organinių junginių panašumą1 būtų teisingas.

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\n (1 taškas)\n

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Organiniai junginiai turi vienodąskirtingą sigma (σ) ir pi (π) cheminių ryšių skaičių ir yra vienas kito izomeraihomologai.

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1 panašumas – podobieństwo – сходство – подібність

\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/vbe1che6-1779280455-redaguotas/items/i69f10af3cdbfc20260428213059dd6fb922","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["inlineChoiceInteraction"],"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 \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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13. Lentelėje aprašyti bandymai, atlikti su vandeniniais gliukozės ir sacharozės tirpalais, įpiltais į atskirus mėgintuvėlius. Pasirinkite tinkamus cheminių reakcijų požymius.

\n

\n (1 taškas)\n

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
Reagentai ir bandymo eigaCheminių reakcijų požymiai
\n\t\t\tgliukozės tirpale		Cheminių reakcijų požymiai
\n\t\t\tsacharozės tirpale
\n

Į tirpalą įpilta Ag2O amoniakinio
\n\t\t\ttirpalo ir mėgintuvėlis pašildytas.

\n 		\n \n Cheminės reakcijos požymis nepasireiškė.\n Blizgi danga² ant vidinės mėgintuvėlio sienelės.\n \n \n \n Cheminės reakcijos požymis nepasireiškė.\n Blizgi danga² ant vidinės mėgintuvėlio sienelės.\n \n
Į tirpalą įdėta sacharazės1, paskui
\n\t\t\tįpilta Ag2O amoniakinio tirpalo ir
\n\t\t\tmėgintuvėlis pašildytas.		\n \n Cheminės reakcijos požymis nepasireiškė.\n Blizgi danga² ant vidinės mėgintuvėlio sienelės.\n \n \n \n Cheminės reakcijos požymis nepasireiškė.\n Blizgi danga² ant vidinės mėgintuvėlio sienelės.\n \n
\n

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1 sacharazė – fermentas, katalizuojantis glikozidinio ryšio hidrolizę

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2 blizgi danga – błyszcząca powłoka – глянцевое покрытие – глянцеве покриття

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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 \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/vbe1che6-1779280455-redaguotas/items/i69f10d1594e0b202604282140050be5d282","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["inlineChoiceInteraction"],"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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\n

14. Paveiksle pavaizduota baltymo polipeptido grandinės struktūrinė formulė. Pažymėkite polipeptido grandinėje aminorūgšties fragmentą, kuris esant pH < 7 įgyja teigiamąjį krūvį.

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\n (1 taškas)\n

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\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/vbe1che6-1779280455-redaguotas/items/i69f0fb7434afb20260428202452b757d12b","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["hotspotInteraction"],"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
\n
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15. Laborantas renka stiklinę skysčių distiliavimo įrangą. Paveiksle pavaizduotos jo paimtos šios įrangos dalys.

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\n \"K15\"\n \n

\n

Be kurios dar vienos dalies laborantui nepavyks surinkti stiklinės skysčių distiliavimo įrangos? Pažymėkite trūkstamą dalį.

\n

\n (1 taškas)\n

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\n
\n
\n
\n \n \n \"K15A\"\n \n \n \"K15B\"\n \n \n \"K15C\"\n \n \n \"K15D\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fb78d22512026042820245618b93151","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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 1\n \n \n \n \n
\n
\n

\n II dalis\n

\n

\n 1 struktūrinis klausimas\n

\n
\n
\n
\n
\n

Aušinimo skystis1 yra mišinys, saugantis variklį nuo perkaitimo ir užšalimo. Šį mišinį sudaro etilenglikolio (1,2‑etandiolio) arba propilenglikolio (1,2‑propandiolio) vandeninis tirpalas ir specialūs priedai.

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1 aušinimo skystis – płyn chłodzący – охлаждающая жидкость – охолоджувальна рідина

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\n
\n
\n
\n

16. Pažymėkite propilenglikolio (1,2‑propandiolio) struktūrinę formulę.

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\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n \"K16A\"\n \n \n \"K16B\"\n \n \n \"K16C\"\n \n \n \"K16D\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fb7d9b19120260428202501fc9f3ed9","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"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 1\n \n \n \n \n
\n
\n

17. Paveiksle pavaizduoti etilenglikolio, propilenglikolio ir glicerolio vandeninių tirpalų užšalimo temperatūros priklausomybės nuo šių cheminių medžiagų masės dalies tirpaluose grafikai.

\n

Pažymėkite tašką (A, B, C arba D), rodantį temperatūrą, kurioje užšąla 50 % etilenglikolio vandeninis tirpalas.

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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 \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/vbe1che6-1779280455-redaguotas/items/i69f0fb825791a202604282025065a753652","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["hotspotInteraction"],"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. Paveiksle pateikti etilenglikolio, propilenglikolio ir glicerolio vandeninių tirpalų užšalimo temperatūros priklausomybės nuo šių cheminių medžiagų masės dalies tirpaluose grafikai.

\n

Remdamiesi paveikslu, pažymėkite, kokia turi būti etilenglikolio masės dalis (%) tirpale, kad tirpalas užšaltų žemiausioje temperatūroje.

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\n (1 taškas)\n

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

66–68 %

\n \n \n

0–2 %

\n \n \n

98–100 %

\n \n \n

63–65 %

\n \n \n
\n
\n

 

\n

\n \"K18\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fb8715b93202604282025117dd4b6b6","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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 \n 1\n \n \n \n \n
\n
\n

19. Aušinimo skysčiui dažniausiai naudojamas 50 % (masės dalimis) etilenglikolio vandeninis tirpalas. Apskaičiuokite, kiek litrų etilenglikolio reikėtų įpilti į 10 L vandens, norint gauti šį etilenglikolio vandeninį tirpalą. Vandens tankis1 – 1,0 g/ml, etilenglikolio tankis – 1,1 g/ml.

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Atsakymą pateikite dviejų reikšminių skaitmenų tikslumu.

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\n (1 taškas)\n

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\n
\n
\n
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Atsakymas: reikalingas etilenglikolio tūris L.

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\n

1 tankis – gęstość – плотность – густина

\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/vbe1che6-1779280455-redaguotas/items/i69f0fb8bb91ee20260428202515fb717100","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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 \n \n \n \n 1\n \n \n \n \n
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\n

20. Paveiksle pavaizduoti 50 % etilenglikolio vandeninio tirpalo ir gryno vandens tankio priklausomybės nuo temperatūros grafikai.

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Remdamiesi šiais duomenimis, pasirinkite tinkamus žodžius, kad teiginys, paaiškinantis, kodėl aušinimo skysčiui naudojamas 50 % etilenglikolio vandeninis tirpalas, būtų teisingas.

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\n (1 taškas)\n

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\n
\n
\n
\n

50 % etilenglikolio vandeninis tirpalas užšąla
\n žemesnėjeaukštesnėje temperatūroje nei grynas vanduo
\nir užšaldamas staigiai nesiplečiaplečiasi.

\n
\n
\n

\n \"K20\"\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/vbe1che6-1779280455-redaguotas/items/i69f115ad9866520260428221645aa2807bc","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["inlineChoiceInteraction"],"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 \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 II struktūrinis klausimas\n

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\n
\n
\n
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Raumenyse kaupiamas kreatinas yra vienas iš žmogaus organizmo energijos šaltinių.

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\n
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21. Pateikta kreatino struktūrinė formulė. Baikite rašyti kreatino empirinę formulę įrašydami reikiamus skaičius.

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\n (1 taškas)\n

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\n
\n
\n
\n

C   H  N  O 

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\n
\n

\n \"K21\"\n \n

\n
\n
\n \n \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/vbe1che6-1779280455-redaguotas/items/i69f117afd06ef2026042822251992d89239","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["textEntryInteraction"],"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. Kreatinas vandeniniame tirpale gali būti kaip neutralios molekulės ir kaip jonai. Pažymėkite paveikslą, kuriame teisingai pavaizduota kreatino vandeniniame tirpale struktūrinė formulė, kai vandeninio tirpalo pH > 7.

\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/vbe1che6-1779280455-redaguotas/items/i69f0fb9a1fe6920260428202530c63e15c9","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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 \n \n \n \n 1\n \n \n \n \n
\n
\n

23. Kreatino yra žuvų raumenų mėsoje. 100 g sverianti silkė turi 0,83 g kreatino. Raumenys (filė) sudaro 63 % visos žuvies masės. Apskaičiuokite, kokią silkės raumenų masės dalį (procentais) sudaro kreatinas. Atsakymą pateikite dviejų reikšminių skaitmenų tikslumu.

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\n (1 taškas)\n

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\n
\n
\n
\n

Atsakymas: %

\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/vbe1che6-1779280455-redaguotas/items/i69f0fb9ebe9a3202604282025348f07e21f","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["textEntryInteraction"],"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 1\n \n \n \n \n
\n
\n

24. Paveiksle pavaizduotas kreatino 1H BMR spektras: viršuje nurodyti signalų cheminiai poslinkiai, apačioje vandenilio atomų signalų integralai (santykiniai kiekiai). Tirpiklio signalas sutampa su vandenilio atomų, esančių prie azoto ir deguonies atomų, signalais.

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\n

Pažymėkite kreatino struktūroje vandenilio atomus, kurių signalas yra ties 2,95 ppm.

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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 \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/vbe1che6-1779280455-redaguotas/items/i69f0fba36fca920260428202539b489bb74","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["hotspotInteraction"],"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. Vykstant kondensacijai iš kreatino (M = 131,14 g/mol) susidaro kreatininas (M = 113,12 g/mol), iš organizmo pašalinamas su šlapimu1.

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\n \"K25\"\n \n

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Pažymėkite paveikslą, kuriame pavaizduota kreatinino struktūrinė formulė.

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\n (1 taškas)\n

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\n
\n
\n
\n \n \n \"K25A\"\n \n \n \"K25B\"\n \n \n \"K25C\"\n \n \n \"K25D\"\n \n \n
\n
\n
\n
\n

1 šlapimas – mocz – моча – сеча

\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/vbe1che6-1779280455-redaguotas/items/i69f0fba826f6220260428202544b495aac2","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė 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 1\n \n \n \n \n
\n
\n

\n III struktūrinis klausimas\n

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\n

Riebalai yra svarbus žmogaus organizmo energijos šaltinis.

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26. Pažymėkite funkcinę grupę, esančią riebalų struktūroje.

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\n (1 taškas)\n

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\n
\n
\n
\n \n \n \"K26A\"\n \n \n \"K26B\"\n \n \n \"K26C\"\n \n \n \"K26D\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fbacd9bd2202604282025487ff1441a","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"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 \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n
\n
\n

27. Žvakėms gaminti naudojama riebalų rūgštis (C18H36O2). Išlyginkite riebalų rūgšties degimo reakcijos lygtį. Įrašykite reikiamus skaičius.

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\n (1 taškas)\n

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C18H36O2(k) + O2(d) → CO2(d) + H2O(d)

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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 \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/vbe1che6-1779280455-redaguotas/items/i69f11f1c617372026042822570020645751","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["textEntryInteraction"],"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. Kiek skirtingų medžiagų susidarys visiškai hidrolizavus paveiksle pavaizduotą riebalų molekulę? Įrašykite skaičių.

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\n (1 taškas)\n

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\n
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\n Atsakymas:\n \n

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\n
\n

\n \"K28\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fbb62b52e20260428202558cad39df4","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["textEntryInteraction"],"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 1\n \n \n \n \n
\n
\n

29. Kurių riebalų lydymosi temperatūra yra žemiausia? Pažymėkite jų struktūrinę formulę.

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\n (1 taškas)\n

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\n
\n
\n
\n \n \n \"K29A\"\n \n \n \"K29B\"\n \n \n \"K29C\"\n \n \n \"K29D\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fbbb6464c20260428202603b51cdb7d","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"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 1\n \n \n \n \n
\n
\n

30. Paveiksle pavaizduota riebalų molekulės skeletinė formulė. Kuri riebalų rūgšties liekana, reaguodama su bromo (Br2) vandeniu, prisijungs daugiausia bromo molekulių? Pažymėkite šią riebalų rūgšties liekaną žyminčią raidę (A, B arba C).

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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 \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2026/07/08/70/vbe1che6-1779280455-redaguotas/items/i69f0fbc01493c202604282026088fabab1a","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["hotspotInteraction"],"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 1\n \n \n \n \n
\n
\n

31. Kuris iš alkoholių gaunamas hidrolizavus riebalus? Pažymėkite jo struktūrinę formulę.

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\n (1 taškas)\n

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\n
\n
\n
\n \n \n \"K31A\"\n \n \n \"K31B\"\n \n \n \"K31C\"\n \n \n \"K31D\"\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/vbe1che6-1779280455-redaguotas/items/i69f0fbc4b40b62026042820261288b71916","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2026-07-08 14:34:47","counterId":"ID27332","name":"2026 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"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 \n \n \n 1\n \n \n \n \n
\n
\n

32. Kiek skirtingų bromalkanų pavaizduota šiomis struktūrinėmis formulėmis? Įrašykite jų skaičių.

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\n (1 taškas)\n

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\n
\n
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Atsakymas: skirtingų bromalkanų skaičius .

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\n \"K32\"\n \n

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33. Pažymėkite cheminės reakcijos mechanizmą, pagal kurį nesotieji riebalai ir alkenai reaguoja su bromu.

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\n (1 taškas)\n

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Elektrofilinis jungimosi

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Elektrofilinis pakaitų

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Nukleofilinis pakaitų

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Nukleofilinis jungimosi

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34. Linolo rūgščiai1 reaguojant su šaltu šarminiu kalio permanganato tirpalu susidaro junginys, turintis vieną karboksigrupę ir keturias hidroksilo grupes. Linolo rūgšties formulėje pažymėkite keturis anglies atomus, prie kurių po reakcijos bus prisijungusios −OH grupės.

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1 linolo rūgštis – kwas linolowy – линолевая кислота – лінолева кислота

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35. Į tuščius laukelius įkelkite medžiagų struktūrines formules tokia tvarka, kad gautumėte teisingai užrašytą biodyzelino sintezės iš riebalų reakcijos lygtį.

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\n IV struktūrinis klausimas\n

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Fotocheminė reakcija – tai cheminė reakcija, vykstanti dėl šviesos poveikio.

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36. Paveiksle pateikta bromo (Br2) fotocheminės reakcijos energinė diagrama1.

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Pažymėkite dvikryptę2 rodyklę, vaizduojančią bromo fotocheminės reakcijos aktyvacijos energiją.

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\n (1 taškas)\n

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1 energinė diagrama – diagram energetyczny – энергетическая диаграмма – енергетична діаграма

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2 dvikryptė – dwukierunkowa – двунаправленная – двонаправлена

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37. Bromo (Br2) homolitinio skilimo1 šviesoje termocheminė reakcijos lygtis:

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Br2 → Br· + Br·, ΔH° = +194 kJ/mol.

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Apskaičiuokite, kiek energijos (kJ) reikia sunaudoti, kad susidarytų 1 mol bromo radikalų.

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\n (1 taškas)\n

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Atsakymas: reikia kJ energijos.

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1 homolitinis skilimas – rozkład homolityczny – гомолитический разрыв – гомолітичний розрив

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38. Paveiksle pavaizduota metano (CH4) ir bromo (Br) radikalų cheminės reakcijos energinė diagrama.

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Pažymėkite dvikryptę rodyklę, žyminčią šilumos kiekį, išsiskyrusį įvykus reakcijai, kurioje dalyvavo 1 mol metano ir 1 mol bromo radikalų. Šios cheminės reakcijos išeiga1 100 %.

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1 išeiga – wydajność – выход – вихід

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39. 0,5 mol metano (CH4) sureagavo su 1 mol bromo (Br2). Pabaikite rašyti per reakciją susidariusio bromalkano molekulinę formulę, kai reakcijos išeiga 100 %. Įrašykite reikiamus skaičius.

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\n CH Br \n

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40. Apskaičiuokite bromo masės dalį (%) anglies tetrabromide (CBr4). Atsakymą pateikite keturių reikšminių skaitmenų tikslumu.

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Anglies tetrabromido molinė masė M(CBr4) = 331,61 g/mol. Bromo (Br) santykinė atominė masė yra 79,90.

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Atsakymas: bromo masės dalis %.

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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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1. CRC Handbook of Chemistry and Physics, 95th ed.; Haynes, W. M., Ed.; CRC Press: Boca Raton, 2014. https://doi.org/10.1201/b17118.

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2. L. B. Lane, Freezing Points of Glycerol and Its Aqueous Solutions. Ind. Eng. Chem. 1925, 17 (9), 924–924. https://doi.org/10.1021/ie50189a017.

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3. Creatine monohydrate (6020-87-7) 1H NMR spectrum. https://www.chemicalbook.com/SpectrumEN_6020-87-7_1HNMR.htm.

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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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Taip

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Vidutiniškai

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Ne

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Spauskite „Baigti darbą“

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