{"success":true,"data":{"id":"hash.68dfc6d515c47.a2d4878df9ae892988c779fe731a0983","title":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","description":null,"count":39,"maxScore":40,"items":[{"identifier":"i6834714f7388c20250526154903010fe827","guid":"i6834714f7388c20250526154903010fe827","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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2025 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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i686279fd431fa202506301350216a9d61a8","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":[],"openQuestion":0}},{"identifier":"i683471830c22620250526154955b2fa95dc","guid":"i683471830c22620250526154955b2fa95dc","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 arba 2 taškais.\n

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Spauskite apačioje esančią rodyklę.

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\n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a031f13e20250630135027b0834f40","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":[],"openQuestion":0}},{"identifier":"i680032b7f18182025041700440747cb24e3","guid":"i680032b7f18182025041700440747cb24e3","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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  • \n SVARBU! Mygtuką \"ivad_s\" spauskite tik tada, kai būsite tikri, jog norite pateikti visus savo atsakymus vertinti. Paspaudę šį mygtuką, savo atsakymų daugiau taisyti nebegalėsite.\n
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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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a08e19652025063013503271fd5d94","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":[],"openQuestion":0}},{"identifier":"i67ca167cb337920250306224116bb04e8cb","guid":"i67ca167cb337920250306224116bb04e8cb","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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a0ecf8c02025063013503833a75b4a","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":[],"openQuestion":0}},{"identifier":"i6621b77df0dbb127550ad72183703bbc2","guid":"i6621b77df0dbb127550ad72183703bbc2","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 \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 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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\n
\n
\n

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
\n
\n

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

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
IAIIAIIIAIVAVAVIAVIIAVIIIA
1H
\n\t\t\t2,2		 He
2\n

Li

\n

1,0

\n 		\n

Be

\n

1,6

\n 		\n

B

\n

2,0

\n 		\n

C

\n

2,6

\n 		\n

N

\n

3,0

\n 		\n

O

\n

3,4

\n 		\n

F

\n

4,0

\n 		\n

Ne

\n

\n
3\n

Na

\n

0,9

\n 		\n

Mg

\n

1,3

\n 		\n

Al

\n

1,6

\n 		\n

Si

\n

1,9

\n 		\n

P

\n

2,2

\n 		\n

S

\n

2,6

\n 		\n

Cl

\n

3,2

\n 		\n

Ar

\n

\n
4\n

K

\n

0,8

\n 		\n

Ca

\n

1,0

\n 		\n

Ga

\n

1,8

\n 		\n

Ge

\n

2,0

\n 		\n

As

\n

2,2

\n 		\n

Se

\n

2,6

\n 		\n

Br

\n

3,0

\n 		\n

Kr

\n

\n
5\n

Rb

\n

0,8

\n 		\n

Sr

\n

1,0

\n 		\n

In

\n

1,8

\n 		\n

Sn

\n

2,0

\n 		\n

Sb

\n

2,1

\n 		\n

Te

\n

2,1

\n 		\n

I

\n

2,7

\n 		\n

Xe

\n

2,6

\n
6\n

Cs

\n

0,8

\n 		\n

Ba

\n

0,9

\n 		\n

Tl

\n

1,8

\n 		\n

Pb

\n

1,8

\n 		\n

Bi

\n

1,9

\n 		\n

Po

\n

2,0

\n 		\n

At

\n

2,2

\n 		\n

Rn

\n

\n
7\n

Fr

\n

0,7

\n 		\n

Ra

\n

0,9

\n
\n

\n
\n
\n

* 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

\n

\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\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

\n

\n 7. Rūgščių jonizacijos (disociacijos) 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 \n \n \n \n \n \n \n \n
\n

Rūgšties vandeninis tirpalas

\n 		HFHClHBrHIHNO3H2CO3H2SH2SO3H2SO4HCOOHCH3COOH
Jonizacijos konstanta\n 6,8 ∙ 10−4\n Labai didelėLabai didelėLabai didelėLabai didelė\n

Ka1 = 4,4 ∙ 10−7

\n

Ka2 = 5,6 ∙ 10−11

\n

Ka1 = 5,7 ∙ 10−8

Ka2 = 1,3 ∙ 10−13

Ka1 = 1,7 ∙ 10−2

Ka2 = 6,2 ∙ 10−8

Ka1 – labai didelė

Ka2 = 1,2 ∙ 10−2		\n 1,8 ∙ 10−4\n \n 1,8 ∙ 10−5\n
\n

\n

\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
\n

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\n

\n 9. Periodinė cheminių elementų lentelė\n

\n

\n \"Chemini\n

\n

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

\n

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.

\n

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.

\n
\n
\n
\n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a14b59272025063013504492094ad8","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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

\n
\n
\n
\n
\n

1. Koks yra kampas laipsniais tarp sp² hibridizacijos elektronų orbitalių anglies atome? Pažymėkite teisingą atsakymą.

\n

\n (1 taškas)\n

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

90°

\n \n \n

109° 28'

\n \n \n

120°

\n \n \n

180°

\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a1accc9b2025063013505048f4f27c","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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
\n
\n

2. Koks pavaizduoto junginio pavadinimas pagal IUPAC nomenklatūrą? Pažymėkite teisingą atsakymą.

\n

\n (1 taškas)\n

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

5-dichlor-1-pentinas

\n \n 5,5-dichlorpent-1-inas\n \n

5,5-dichlorpent-1-enas

\n \n 5-dichlorpentenas\n \n
\n
\n

\n \"2\"\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a20991fc202506301350562487d13b","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"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
\n
\n

3. Kurį polimerą galima gauti iš pavaizduoto monomero? Pažymėkite teisingą atsakymą.

\n

\n (1 taškas)\n

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

Polieteną

\n \n \n

Polipropeną

\n \n \n

Polistireną

\n \n \n

Polichloreteną

\n \n \n
\n
\n

\n \"3\"\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a27ee42f202506301351034f143467","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė 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
\n
\n

4. Vyko cheminė reakcija:

\n

C15H29COOH + Cl2 →

\n

Pažymėkite šios cheminės reakcijos produktus (arba produktą).

\n

\n (1 taškas)\n

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

C15H27Cl2COOH + H2

\n \n \n

C15H29COOCl + HCl

\n \n \n

C15H29Cl2COOH

\n \n \n

C15H28ClCOOH + HCl

\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a2dd043420250630135109f2c8a766","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė 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
\n
\n

5. Paveiksle pavaizduotas tam tikros organinės medžiagos vandenilio branduolių magnetinio rezonanso (¹H BMR) spektras; tirpiklio signalas nėra pavaizduotas.

\n

\n \"5\"\n \n

\n

\n Kurios organinės medžiagos šis spektras? Pažymėkite teisingą atsakymą.\n

\n

\n (1 taškas)\n

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

CH3CH2CH2CH2OH

\n \n CH3COCH2CH3\n \n

CH3COCH3

\n \n CH3CH2OCH2CH3\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a338f017202506301351151eda2b5c","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė 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
\n
\n

6. Pasirinkdami tinkamą žodį, pabaikite teiginį, kad jis būtų teisingas.

\n

Metano rūgščiai reaguojant su sidabro(I) oksidu amoniakiniame tirpale, sidabro(I) oksidas šioje reakcijoje yra nukleofilaselektrofilasoksidatoriusreduktorius.

\n

\n (1 taškas)\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a3960e2b20250630135121aee5c707","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["inlineChoiceInteraction"],"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
\n
\n

7. Į mėgintuvėlį, kuriame yra įpilta vario(II) sulfato ir natrio hidroksido, įlašinama glicerolio. Mišinys suplakamas. Pažymėkite mėgintuvėlyje vykstančios cheminės reakcijos požymį.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n \"7A\"\n \n \n \"7B\"\n \n \n \"7C\"\n \n \n \"7D\"\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a3f27f5b202506301351271ebe1e90","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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 \n
\n
\n

8. Kuo negalima gesinti¹ užsidegusio benzino? Pažymėkite teisingą atsakymą.

\n

\n (1 taškas)\n

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

Žemėmis

\n \n Vandeniu\n \n

Milteliniu gesintuvu

\n \n \n

Nedegiu audiniu

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

¹ gesinti – gasić – тушить – гасити

\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a451a48220250630135133f65087d5","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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 \n \n 1\n \n \n \n \n \n \n
\n
\n

9. Atsitiktine tvarka pateiktos bromo elektrofilinio prijungimo AE prie eteno cheminės reakcijos mechanizmo pradinių medžiagų, tarpinių būsenų¹ ir produktų struktūros.

\n

Į tuščius laukelius įkelkite trūkstamas struktūras, kad būtų teisingai pavaizduotas elektrofilinio prijungimo cheminės reakcijos mechanizmas.

\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

¹ tarpinė būsena – stan przejściowy – промежуточное состояние – перехідний стан

\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a4adf8622025063013513806725376","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["graphicGapMatchInteraction"],"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 \n \n 1\n \n \n \n \n \n
\n
\n

10. Lentelėje į tinkamas vietas įkelkite vandenyje tirpius celiuliozės, baltymų ir riebalų hidrolizės produktus.

\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 \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a50d76d52025063013514477876db5","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["graphicGapMatchInteraction"],"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
\n
\n

11. Kas yra nebūdinga organinėms medžiagoms? Pažymėkite teisingą atsakymą.

\n

\n (1 taškas)\n

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

Organinėse medžiagose anglies atomai yra keturvalenčiai.

\n \n Organinės medžiagos sudaro homologines eiles.\n \n

Organines medžiagas sudaro tik anglies atomai.

\n \n \n

Organinėse medžiagose tarp anglies atomų galimos viengubosios, dvigubosios ir trigubosios jungtys.

\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a56dc8132025063013515039eddf63","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė 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 1\n \n \n \n \n
\n
\n

12. Kuris iš šių alkenų yra cis- geometrinis izomeras? Pažymėkite teisingą atsakymą.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n \"12A\"\n \n \n \"12B\"\n \n \n \"12C\"\n \n \n \"12D\"\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a5c8e76f202506301351567f0c8f89","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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
\n
\n

13. Adenino deoksiribonukleotide vieną vandenilio atomą pakeitę hidroksigrupe gausime adenino ribonukleotidą. Pažymėkite šį vandenilio atomą.

\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 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a64201582025063013520467506071","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["hotspotInteraction"],"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
\n
\n

14. Pažymėkite antrinį aminą.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n \"14A\"\n \n \n \"14B\"\n \n \n \"14C\"\n \n \n \"14D\"\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a6a105a220250630135210d7d7c9a7","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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
\n
\n

15. Tam tikras angliavandenilis turi devynias sigma (σ) ir dvi pi (π) chemines jungtis. Pažymėkite šio angliavandenilio molekulinę formulę.

\n

\n (1 taškas)\n

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

C2H6

\n \n \n

C4H6

\n \n \n

C4H8

\n \n \n

C9H16

\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a6fce55320250630135215c228533a","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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 1\n \n \n \n \n \n
\n
\n

\n II DALIS\n

\n

\n 1 struktūrinis klausimas\n

\n
\n
\n
\n
\n

16. Žieminiuose automobilių langų plovikliuose¹ naudojami įvairių alkoholių vandeniniai tirpalai. Diagramoje pavaizduota etanolio, metanolio, propan-2-olio, etilenglikolio ir vandens mišinių užšalimo temperatūrų priklausomybė nuo šių medžiagų koncentracijos tūrio dalimis vandenyje. Remdamiesi diagrama, nustatykite žemiausioje temperatūroje užšąlantį medžiagos ir vandens mišinį, kuris pagamintas, sumaišius vienodus medžiagos ir vandens tūrius. Pažymėkite teisingą atsakymą.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n Metanolio ir vandens mišinys\n \n

Etanolio ir vandens mišinys

\n \n \n

Propan-2-olio ir vandens mišinys

\n \n \n

Etilenglikolio ir vandens mišinys

\n \n \n
\n
\n

\n \"16\"\n \n

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

¹ langų ploviklis – płyn do mycia szyb – моющее средство для окон – засіб для миття скла

\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a759d54c2025063013522156a45ac2","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė 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. Pažymėkite propan-2-olio izomerą.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n \n \n \"17A\"\n \n \n \"17B\"\n \n \n \"17C\"\n \n \n \"17D\"\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a7b602412025063013522753a35dc8","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"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 \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

18. Vykstant dviem skirtingoms cheminėms reakcijoms, iš medžiagos X galima gauti etanolį ir etilenglikolį. Užrašykite medžiagos X molekulinę formulę.

\n

CH3−CH2−OH ← X → CH2(OH)−CH2(OH)

\n

\n (1 taškas)\n

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

\n Atsakymas:\n \n \n \n \n \n \n

\n
\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 1\n \n \n \n \n \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 4\n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a8145c1c20250630135233f2a89ef6","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"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 \n \n \n \n \n 1\n \n \n \n \n
\n
\n

19. Diagramoje pavaizduota etanolio, metanolio, propan-2-olio, etilenglikolio ir vandens mišinių užšalimo temperatūrų priklausomybė nuo šių medžiagų koncentracijos tūrio dalimis vandenyje. Remdamiesi diagrama, užrašykite vandens tūrio dalį (%) žemiausioje temperatūroje užšąlančiame etilenglikolio ir vandens mišinyje. Užrašykite skaičių.

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\n (1 taškas)\n

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Vandens tūrio dalis yra %

\n
\n
\n

\n \"16\"\n \n

\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a870b8452025063013523993402ec6","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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 \n \n 1\n \n \n \n \n \n
\n
\n

20. Metanolis yra nuodingas. Šios medžiagos mirtinoji dozė¹ 1 kg žmogaus kūno masės yra 0,20 ml. Apskaičiuokite mirtinąją metanolio dozę (ml) 80 kg kūno masės žmogui. Užrašykite atsakymą.

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\n (1 taškas)\n

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\n
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Atsakymas: ml.

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\n
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¹ mirtinoji dozė – dawka śmiertelna – смертельная доза – смертельна доза

\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a8cd5e95202506301352441dfaa216","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["textEntryInteraction"],"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

21. Paveiksle pavaizduota įranga, kuri naudota metanolio, etanolio, propan-2-olio ir etilenglikolio mišiniui distiliuoti. Kaitinimo vonelės temperatūra yra 100 °C. Remdamiesi pateikta informacija, lentelėje pažymėkite alkoholį, kuris pasibaigus distiliavimui liks apvaliadugnėje kolboje¹.

\n

\n (1 taškas)\n

\n
\n
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\n \n \n \n \n \n \n \n \n
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\n \"21s\"\n \n

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\n \n
\n
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¹ apvaliadugnė kolba – kolba okrągłodenna – круглодонная колба – круглодонна колба

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\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a92b492220250630135250bfb173ba","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["hotspotInteraction"],"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. Iš kurios medžiagos, atlikus vieną cheminę reakciją, galima gauti propan-2-olį? Pažymėkite šios medžiagos cheminę formulę.

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\n (1 taškas)\n

\n
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\n \n \n

CH3CH2OH

\n \n \n

CH3CH2CHO

\n \n \n

CH3COOCH3

\n \n \n

CH3COCH3

\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627a9888b8e20250630135256208aceef","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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 \n \n \n \n \n \n \n \n \n \n \n \n 2\n \n \n \n \n
\n
\n

23. Chemikas, sumaišęs 1,00 L etanolio (tankis – 0,790  g/ml) ir 1,00  L distiliuoto vandens (tankis – 1,00  g/ml), gavo mišinį, kurio tankis yra 0,930  g/ml. Apskaičiuokite gauto mišinio masę gramais ir tūrį litrais. Užrašykite skaičius.

\n

\n (2 taškai)\n

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\n
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Mišinio masė yra g.

\n

Mišinio tūris yra L.

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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 \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627aa01148b202506301353048b1996da","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":2},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė 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. Vandeniniame tirpale etanolio ir vandens molekulės sudaro vandenilinius ryšius. Pažymėkite paveikslą, kuriame teisingai pavaizduotas vandenilinis ryšys tarp šių molekulių.

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\n (1 taškas)\n

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\n
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\n \n \n \"24A\"\n \n \n \"24B\"\n \n \n \"24C\"\n \n \n \"24D\"\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627aa612616202506301353103aba78a8","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"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
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\n 2 struktūrinis klausimas\n

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25. Chemikas, sintetindamas jazminų kvapo¹ organinę medžiagą, į apvaliadugnę kolbą supylė etano rūgštį ir benzilo alkoholį, įlašino kelis lašus koncentruotos sieros rūgšties. Tada šį reakcijos mišinį virino. Vyko pavaizduota cheminė reakcija.

\n

\n \"25\"\n \n

\n

\n Pažymėkite organinių junginių klasę, kuriai priskiriamas gautas produktas. \n

\n

\n (1 taškas)\n

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\n
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\n \n \n

Ketonai

\n \n \n

Aldehidai

\n \n \n

Esteriai

\n \n \n

Karboksirūgštys

\n \n \n
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\n
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¹ jazminų kvapas – zapach jaśminu – аромат жасмина – запах жасмину

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\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627aabd01d320250630135315b221647e","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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 \n \n \n \n 1\n \n \n \n \n
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\n

26. Baikite rašyti susintetintos jazminų kvapo organinės medžiagos molekulinę formulę. Įrašykite reikiamus skaičius.

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\n (1 taškas)\n

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CHO

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\n \"26\"\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 \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 \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627ab1dc504202506301353218516f096","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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 \n \n 1\n \n \n \n \n \n
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27. Ant indų su etano rūgšties, benzilo alkoholio ir koncentruotos sieros rūgšties reagentais yra užklijuotos atitinkamos lentelėje pavaizduotos piktogramos.

\n

Remdamiesi pateikta informacija, į lentelę įkelkite saugumo taisykles¹, kurių reikia laikytis, dirbant su kiekviena iš šių cheminių medžiagų.

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\n (1 taškas)\n

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¹ saugumo taisykles – zasady bezpieczeństwa – правила безопасности – правила безпеки

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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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627ab7a058d20250630135327bc33d4ff","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["graphicGapMatchInteraction"],"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
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\n

28. Chemikas, sintetindamas jazminų kvapo organinę medžiagą, į apvaliadugnę kolbą supylė etano rūgštį ir benzilo alkoholį, įlašino kelis lašus koncentruotos sieros rūgšties. Tada šį reakcijos mišinį virino. Pažymėkite įrangą, kuri naudojama šiai organinei sintezei atlikti.

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\n (1 taškas)\n

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\n \n \n \"28A\"\n \n \n \"28B\"\n \n \n \"28C\"\n \n \n \"28D\"\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627abd80fc9202506301353338ff02a56","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 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 1\n \n \n \n \n \n
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\n

29. Chemikas atliko jazminų kvapo cheminės medžiagos organinę sintezę – jos metu vyko etano rūgšties ir benzilo alkoholio cheminė reakcija. Kas 10 minučių jis imdavo mėginį iš reakcijos mišinio ir užlašindavo ant plonasluoksnės chromatografijos plokštelės¹. Paveiksle pavaizduota ši plokštelė po analizės, kai buvo užlašintas pradinės medžiagos (benzilo alkoholio) mėginys² (laiko momentas – 0 min.) ir mėginiai iš reakcijos mišinio kas 10 min. Ant plokštelės reakcija nevyksta, nes lakios³ medžiagos (etano rūgštis) išgaruoja.

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Remdamiesi pateikta informacija, nustatykite cheminės reakcijos mišinio sudėtį, kai mėginys buvo paimtas 30-ąją minutę nuo cheminės reakcijos pradžios. Pažymėkite teisingą atsakymą.

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\n (1 taškas)\n

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\n
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\n \n Mišinyje yra tik benzilo alkoholis.\n \n

Mišinyje yra benzilo alkoholis ir reakcijos produktas.

\n \n \n

Mišinyje yra produktas ir tarpinis reakcijos produktas.

\n \n \n

Mišinyje yra tik produktas.

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\n \"29\"\n \n

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¹ plokštelė – płytka – пластина – пластина

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² mėginys – próbka – проба – зразок

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³ laki – lotny – летучий – летка

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\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627ac368bd42025063013533961a5a495","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė 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. Chemikas atliko jazminų kvapo cheminės medžiagos organinę sintezę. Jai naudojo etano rūgštį (perteklius), benzilo alkoholį ir kelis lašus koncentruotos sieros rūgšties.

\n

\n \"25\"\n \n

\n

\n Pasibaigus cheminei reakcijai, chemikas mišinį neutralizavo, įpildamas sočiojo vandeninio NaHCO3 tirpalo. Kuri medžiaga susidaro iš etano rūgšties po neutralizacijos? Pažymėkite teisingą atsakymą.\n

\n

\n (1 taškas)\n

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\n
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\n \n \n

CH3CH2COONa

\n \n \n

C6H5CH2ONa

\n \n \n

C6H5ONa

\n \n CH3COONa\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/2025/10/03/19323/pvbe1che5-emokykla-1759412387/items/i68627ac94e11a20250630135345fd0fb488","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-10-03 15:51:33","counterId":"ID19821","name":"2025 m. chemijos VBE I dalis. Pakartotinė sesija","transformed":false,"interactions":["choiceInteraction"],"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
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\n

31. Pasibaigus jazminų kvapo medžiagos organinės sintezės cheminei reakcijai, neutralus cheminės reakcijos mišinys buvo sumaišytas su sočiuoju NaCl tirpalu ir supiltas į dalijamąjį piltuvą¹. Yra žinoma, kad cheminės reakcijos produktas netirpus vandenyje ir jo tankis yra 1,0 g/ml, cheminės reakcijos pradinės medžiagos tirpios vandenyje, o NaCl vandeninio tirpalo tankis yra 1,3 g/ml. Dalijamajame piltuve atsiskyrė sluoksniai² A ir B (žr. pav.). Remdamiesi pateikta informacija, nustatykite, kuri medžiaga bus sluoksnyje A. Pažymėkite teisingą atsakymą.

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\n (1 taškas)\n

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\n \n \n

Sotusis NaCl tirpalas

\n \n \n

Benzilo alkoholis

\n \n \n

Reakcijos produktas

\n \n \n

Sotusis NaHCO3 tirpalas

\n \n \n
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\n \"31\"\n \n

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¹ dalijamasis piltuvas – lejek rozdzielczy – делительная воронка – ділильна лійка

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² sluoksnis – warstwa – слой – шар

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32. Jazminų kvapo organinė medžiaga, gauta įvykus organinei sintezei, džiovinama¹ ir vėliau distiliuojama, esant sumažintam slėgiui. Paveiksle pavaizduotoje distiliavimo įrangoje pažymėkite šaldytuvą (a, b, c arba d).

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\n (1 taškas)\n

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¹ džiovinimas – suszenie – сушка – сушіння

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\n 3 struktūrinis klausimas\n

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Sacharozė (C12H22O11) – saldūs bespalviai kristalai, gerai tirpstantys vandenyje.

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33. Veikiant fermentams, sacharozė virsta etanoliu ir anglies dioksidu. Išlyginkite sacharozės fermentacijos reakcijos lygtį. Įrašykite reikiamus skaičius.

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C12H22O11 + H2O [Fermentai]→^\\underrightarrow{[Fermentai]} CO2 + C2H6O

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34. Pateikta sacharozės struktūrinė formulė. Kas susidaro, hidrolizuojantis sacharozei? Pažymėkite teisingą atsakymą.

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Po vieną gliukozės ir manozės molekulę

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Po vieną gliukozės ir galaktozės molekulę

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Po vieną gliukozės ir fruktozės molekulę

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Dvi gliukozės molekulės

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\n \"34\"\n \n

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35. Remdamiesi pateikta sacharozės struktūrine formule, pažymėkite fruktozės struktūrinę formulę.

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\n \n \n \"35A\"\n \n \n \"35B\"\n \n \n \"35C\"\n \n \n \"35D\"\n \n \n
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\n \"34\"\n \n

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36. Pasirinkite tinkamą angliavandenį, kad tekstas būtų teisingas.

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Angliavandeniai yra gyvųjų organizmų energijos šaltinis. Ląstelėje energijai gaminti tiesiogiai naudojama medžiaga yra gliukozėsacharozėkrakmolasceliuliozė.

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37. Sacharozės degimo reakcijos termocheminė lygtis:

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C12H22O11 + 12O2 → 12CO2 + 11H2O     ΔH° = −5650 kJ

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Užrašykite šilumos kiekį (kJ), kuris išsiskirs, sudeginus 20 mmol sacharozės.

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Išsiskirs kJ.

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38. Apskaičiuokite anglies masės dalį (%) sacharozės molekulėje (C12H22O11). Sacharozės molekulinė masė yra 342,34 g/mol. Atsakymą pateikite keturių reikšminių skaitmenų tikslumu.

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Anglies masės dalis yra %.

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39. Chemikė trijuose mėgintuvėliuose vandenyje ištirpino tris skirtingas medžiagas – gliukozę, sacharozę ir krakmolą. Po to į gautus tirpalus supylė lentelėje pateiktus reagentus ir stebėjo vykstančių cheminių reakcijų požymius. Į lentelę reikiamose vietose įkelkite korteles su medžiagomis, kurios buvo ištirpintos mėgintuvėliuose.

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