{"success":true,"data":{"id":"hash.6842b8ac93220.0cccc050cb86b93c6a60453c090bf1b6","title":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","description":null,"count":40,"maxScore":40,"items":[{"identifier":"i67ca16685488e202503062240564d206b23","guid":"i67ca16685488e202503062240564d206b23","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. 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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67ca16685488e202503062240564d206b23","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":[],"openQuestion":0}},{"identifier":"i67ca166f2fdae20250306224103e4b4912c","guid":"i67ca166f2fdae20250306224103e4b4912c","label":"INFO_2","title":"INFO_ 2","submissionMode":null,"xml":"\n\n \n \n \n
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\n Turėsite atlikti įvairias užduotis:\n

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ㅤ

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

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Spauskite apačioje esančią rodyklę.

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\n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67ca166f2fdae20250306224103e4b4912c","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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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    \n
  • 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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i680032b7f18182025041700440747cb24e3","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67ca167cb337920250306224116bb04e8cb","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i6806af0a14f4b20250421224810f2083724","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":0},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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
\n
\n

\n I dalis\n

\n
\n
\n
\n
\n

\n 1.\n

\n
\n
\n

Pažymėkite paveikslą, kuriame pavaizduotos sp³ hibridizacijos anglies atomo teorinės molekulinės orbitalės.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n
\n \n \n \"1A\"\n \n \n \"1B\"\n \n \n \"1C\"\n \n \n \"1D\"\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631a5f153e2025021920314925753e7a","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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

\n 2.\n

\n
\n
\n

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

\n

\n (1 taškas)\n

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

2-chlor-3-metilbut-2-enas

\n \n 2-chlor-3-metilbutandienas\n \n

3-chlor-2-metilbutenas

\n \n \n

2-chlor-2,3-dimetilpropenas

\n \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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631aae4bdb20250219203154daa0f39d","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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

\n 3.\n

\n
\n
\n

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

\n

\n (1 taškas)\n

\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631afb66652025021920315921e771ec","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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
\n
\n

\n 4.\n

\n
\n
\n

Gliukozė vandeniniame tirpale būna pusiausvyroje tarp ciklinės ir aldehidinės formų. Pavaizduotoje struktūrinėje formulėje pažymėkite cheminę jungtį¹ (a, b, c ar d), kuri nutrūksta, gliukozės ciklinei formai virstant aldehidine.

\n

\n (1 taškas)\n

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

¹ jungtis – wiązanie – cвязь – зв'язок

\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631b4a300e202502192032046250e5eb","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["hotspotInteraction"],"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

\n 5.\n

\n
\n
\n

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

\n

\n \"5\"\n
\n Kurios organinės medžiagos yra šis spektras? Pažymėkite teisingą atsakymą.\n

\n

\n (1 taškas)\n

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

CH3OH       

\n \n CH3OCH3     \n \n

CH3COOCH2CH3

\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631b9c477d20250219203209e064472b","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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
\n
\n

\n 6.\n

\n
\n
\n

Chemikas, atlikdamas riebalų hidrolizę, naudojo natrio šarmo vandeninį tirpalą. Įvykus reakcijai, į gautą mišinį jis įpylė druskos rūgšties (perteklius¹). Pažymėkite produktą, kuris, įpylus rūgšties, liks ištirpęs vandeniniame tirpale.

\n

\n (1 taškas)\n

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

Etilenglikolis

\n \n \n

Riebalų rūgštys

\n \n \n

Glicerolis

\n \n \n

Riebalų rūgščių natrio druskos

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

¹ perteklius – nadmiar – избыток – надлишок

\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631beba91b20250219203214eb551e27","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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
\n
\n

\n 7.\n

\n
\n
\n

Kuri iš šių aminorūgščių geriausiai tirpsta vandenyje? Pažymėkite jos struktūrinę formulę.

\n

\n (1 taškas)\n

\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631c37e62f202502192032199106dd03","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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
\n
\n

\n 8.\n

\n
\n
\n

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

\n

\n (1 taškas)\n

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

Jodido anijonas (I⁻) organinėse reakcijose gali dalyvauti kaip radikalas.elektrofilas.nukleofilas.tirpiklis¹.

\n

 

\n

 

\n

¹ tirpiklis – rozpuszczalnik – растворитель – розчинник

\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631c84542a20250219203224eaa9da11","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["inlineChoiceInteraction"],"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 9.\n

\n
\n
\n

Kairėje pusėje atsitiktine tvarka pateiktos etanalio ir vandenilio cianido nukleofilinio prisijungimo AN cheminės reakcijos mechanizmo pradinių medžiagų, tarpinių būsenų ir produktų struktūros.

\n

Į tuščius laukelius įkelkite trūkstamas struktūras, kad būtų teisingai pavaizduotas nukleofilinio prisijungimo AN 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 \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631cd0a4cd2025021920322977f05c29","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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 1\n \n \n \n \n
\n
\n

\n 10.\n

\n
\n
\n

Kuris iš šių bromalkenų yra trans- geometrinis izomeras? Pažymėkite teisingą atsakymą.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n
\n \n \n \"10A\"\n \n \n \"10B\"\n \n \n \"10C\"\n \n \n \"10D\"\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631d1c8389202502192032338113479b","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_11","guid":"I_11","label":"I_11","title":"I_11","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

\n 11.\n

\n
\n
\n

Pradinei medžiagai reaguojant su bromu (perteklius), vyksta elektrofilinio jungimosi reakcija ir gaunamas produktas, kurio cheminė formulė C15H30Br2. Pažymėkite pradinės medžiagos molekulinę formulę.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n
\n \n C15H30\n C15H32\n C30H60\n C15H28\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631d6b7f552025021920323860398b95","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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 1\n \n \n \n \n
\n
\n

\n 12.\n

\n
\n
\n

Vyno rūgšties struktūrinė formulė yra HOOC–CHOH–CHOH–COOH.
\nPažymėkite produktą, kurį gausime, į vyno rūgšties vandeninį tirpalą įbėrę NaOH (perteklius).

\n

\n (1 taškas)\n

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

NaOOC–CHOH–CHOH–COOH

\n \n \n

NaOOC–CHOH–CHOH–COONa

\n \n \n

HOOC–CHONa–CHONa–COOH

\n \n \n

NaOOC–CHONa–CHONa–COONa

\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631db6fa7e2025021920324305673842","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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

\n 13.\n

\n
\n
\n

Šildomas mėgintuvėlis, kuriame yra gliukozės ir sidabro(I) oksido amoniakinis tirpalas. Pažymėkite paveikslą, kuriame pavaizduotas vykusios cheminės reakcijos požymis.

\n

\n (1 taškas)\n

\n
\n
\n
\n
\n
\n \n \n \"13A\"\n \n \n \"13B\"\n \n \n \"13C\"\n \n \n \"13D\"\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631e03176d202502192032488ae1608d","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_14","guid":"I_14","label":"I_14","title":"I_14","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
\n
\n

\n 14.\n

\n
\n
\n

Pasirinkite tinkamus atsakymus, kad teiginiai būtų teisingi.

\n

\n (1 taškas)\n

\n

Nustatant, ar bulvėje yra krakmolo, pakanka ant jos užlašinti CuSO₄fenolftaleinodruskos rūgštiesI₂ tirpalo lašą. Toje vietoje, kur užlašinta, atsiras mėlynarožinėžaliaraudona spalva.

\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 2\n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67c8cd63ab3fa20250305231707bce1e283","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["inlineChoiceInteraction"],"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

\n 15.\n

\n
\n
\n

Ką reikia daryti, įtarus, kad patalpoje yra gamtinių dujų nuotėkis¹? Pažymėkite teisingą atsakymą.

\n

\n (1 taškas)\n

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

Uždegti degtuką ar šviesą, ieškoti gamtinių dujų nuotėkio ir užsukti sklendę².

\n \n \n

Uždaryti visus langus ir duris, kad gamtinės dujos nepatektų į išorę.

\n \n \n

Atidaryti visus langus ir duris, išjungti visus elektros prietaisus ir užsukti sklendę.

\n \n \n

Nieko nedaryti, nes gamtinės dujos yra nenuodingos ir nesprogios.

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

¹ nuotėkis – wyciek – утечка – витік

\n

² sklendė – zawór – клапан – засувка

\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631e9b916f2025021920325701ab0920","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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

\n 16.\n

\n
\n
\n

Gamtines dujas patogiausia transportuoti suskystintas. Lentelėje pateiktos gamtinių dujų pagrindinių komponentų virimo ir lydymosi temperatūros, esant 1 ∙ 10⁵ Pa slėgiui.

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

Kuri iš šių medžiagų yra skystosios būsenos, esant −50 °C temperatūrai ir 1 ∙ 10⁵ Pa slėgiui? Pažymėkite teisingą atsakymą.

\n
\n
\n \n Metanas\n Etanas\n Propanas\n \n
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MedžiagaVirimo temperatūra, °CLydymosi temperatūra, °C
Metanas–162–183
Etanas–89–172
Propanas–42–188
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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67e9b14a6e089202503302302024210d7af","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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
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\n 17.\n

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Atskirai deginamos metano, etano, propano ir gamtinės dujos (vienodas tūris). Remdamiesi gamtinių dujų sudėties diagrama ir lentelėje pateiktais duomenimis apie metano ir jo homologų degimo šilumą, pažymėkite dujas, kurias deginant išsiskiria daugiausia šilumos. Į kitų homologų degimo šilumą neatsižvelkite.

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\n \"17\"\n

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Medžiaga\n

Šilumos kiekis,
\n\t\t\tišsiskiriantis sudeginus vieną molį medžiagos standartinėmis sąlygomis, kJ/mol

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Metanas890
Etanas1560
Propanas2220
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\n (1 taškas)\n

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Metanas

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Etanas

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Propanas

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Gamtinės dujos

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67e9b2a02793320250330230744ac085711","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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 1\n \n \n \n \n \n
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\n 18.\n

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Išlyginkite etano dujų degimo lygtį. Įrašykite reikiamus skaičius.

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Atsakymas:C2H6 + 7O2 → CO2 + H2O

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67c8d113852a9202503052332515399abe9","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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 1\n \n \n \n \n
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\n 19.\n

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Remdamiesi lentelėje pateikta informacija, užrašykite teisingą metano degimo termocheminės reakcijos standartinės entalpijos pokyčio reikšmę.

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Medžiaga\n

Šilumos kiekis,
\n\t\t\tišsiskiriantis sudeginus vieną molį
\n\t\t\tmedžiagos standartinėmis sąlygomis, kJ/mol

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Metanas890
Etanas1560
Propanas2220
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\n (1 taškas)\n

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Atsakymas: kJ

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b631fd88541202502192033174b3b9c00","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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 \n \n \n \n 1\n \n \n \n \n
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\n 20.\n

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Kiek molių yra 100 gramų metano? Metano molinė masė yra 16,0 g/mol. Užrašykite skaičių.

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Atsakymas: mol

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b632027b7f120250219203322cbae65ee","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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 \n \n \n 1\n \n \n \n \n
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\n 21.\n

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Remdamiesi lentelėje pateikta informacija, įrašykite šilumos kiekį, kuris išsiskirs, sudeginus 10 molių propano.

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Medžiaga\n

Šilumos kiekis,
\n\t\t\tišsiskiriantis sudeginus vieną molį
\n\t\t\tmedžiagos standartinėmis sąlygomis, kJ/mol

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Metanas890
Etanas1560
Propanas2220
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\n (1 taškas)\n

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Atsakymas: kJ

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67c8d1dc05f6c20250305233612a4b21777","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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
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\n 22.\n

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Diagramoje pavaizduota metano slėgio priklausomybė nuo temperatūros, taip pat agregatinės būsenos.

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Diagramoje pažymėkite tašką (A, B, C ar D), kuriame metanas tuo pačiu metu¹ yra ir skystosios, ir kietosios, ir dujinės būsenos.

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¹ tuo pačiu metu – w tym samym czasie – в одно и то же время – у той же час

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b6320c217fc20250219203332e9fc7280","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["hotspotInteraction"],"openQuestion":0}},{"identifier":"I_23","guid":"I_23","label":"I_23","title":"I_23","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 23.\n

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Saturno palydovo Titano paviršiuje vidutinė temperatūra yra −175 °C, o slėgis yra 1 ∙ 10⁵ Pa. Kokia yra metano agregatinė būsena¹ Titano paviršiuje? Diagramoje pažymėkite teisingą atsakymą.

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¹ būsena – stan – состояние – стан

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b63210cffa720250219203336a4b29956","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["hotspotInteraction"],"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
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\n 24.\n

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Kuris iš išvardytų nešakotos grandinės metano homologų turi šešis anglies atomus? Pažymėkite teisingą atsakymą.

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n-butanas

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n-heptanas

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n-heksanas

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n-dekanas

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b63215989b520250219203341c6949960","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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
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\n 25.\n

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Biodujos išsiskiria, vykstant anaerobinio (bedeguonio) mikrobiologinio rūgimo ir atliekų irimo cheminėms reakcijoms. Biodujas sudaro apie 70 % metano ir 30 % anglies dioksido dujų. Jos gaminamos iš nuotekų dumblo, žemės ūkio ir maisto pramonės atliekų, specialiai auginamų augalų masės. Kuriuo teiginiu teisingai pagrįsta, kodėl biodujos yra alternatyvus atsinaujinantis energijos šaltinis? Pažymėkite teisingą atsakymą.

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Nors didžioji dalis biodujų yra metano dujos, tačiau jose metano yra mažiau negu gamtinėse dujose, todėl mažiau teršiama aplinka.

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Biodujose yra daug anglies dioksido, todėl reikia mažiau energijos, kad jose esančias metano dujas paverstume degimo produktais.

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Biodujos susidaro savaime iš neorganinių medžiagų, kurias galima iškasti, todėl nereikia eikvoti¹ energijos joms gaminti.

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Biodujos yra žmogaus veiklos rezultatas, o racionaliai jas perdirbant gaunama energijos, padedančios tausoti gamtinius energijos išteklius.

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¹ eikvoti – marnować – расходовать – витрачати

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b6321a66b4b20250219203346f8ef1c34","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 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
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\n 2 struktūrinis klausimas\n

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\n 26.\n

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Chemikė gavo užduotį iš anilino susintetinti 2,4,6-tribromaniliną.

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Į organinės sintezės veiksmų seką įkelkite tinkamą praleistą veiksmą.

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\n \n Gauta grynoji medžiaga suberiama į buteliuką.\n Sumaišomos pradinės ir galutinės medžiagos.\n Išgryninamas produktas.\n Įrodoma pradinės medžiagos struktūra.\n

\n Organinės sintezės veiksmų eiliškumo seka\n

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Išgryninamos pradinės medžiagos.

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        🢃

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Vyksta cheminė reakcija.

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        🢃

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        🢃

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Įrodoma produkto struktūra.

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b6321f28b5a20250219203351b64adc70","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["gapMatchInteraction"],"openQuestion":0}},{"identifier":"I_27","guid":"I_27","label":"I_27","title":"I_27","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 27.\n

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Pažymėkite anilino struktūrinę formulę.

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\n \n \n \"27A\"\n \n \n \"27B\"\n \n \n \"27C\"\n \n \n \"27D\"\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b63223d75d9202502192033554de67da0","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:16","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_28","guid":"I_28","label":"I_28","title":"I_28","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \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
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\n 28.\n

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Chemikė, norėdama iš anilino susintetinti 2,4,6-tribromaniliną, paėmė anilino, ištirpinto tam tikrame tirpiklyje. Yra žinoma, kad šis tirpiklis sudarytas tik iš anglies ir vandenilio atomų, anglies masės dalis jame yra 92,26 % ir šios medžiagos ¹H BMR spektre stebimas tik vienas signalas. Užrašykite šio tirpiklio molekulinę formulę.

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

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\n 29.\n

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Naudodamiesi lentelėje pateiktais duomenimis apie anilino, bromo, n-heksano ir nežinomo tirpiklio virimo ir lydymosi temperatūras, esant 1 ∙ 10⁵ Pa slėgiui, pažymėkite medžiagą, kuri pradės pirmoji garuoti, verdant šį mišinį.

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MedžiagaLydymosi temperatūra, °CVirimo temperatūra, °C
Tirpiklis680
Anilinas–6184
Bromas–759
n-heksanas–9569
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\n \n Tirpiklis\n Anilinas\n Bromas\n n-heksanas\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67e9b658e73c620250330232336153df595","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 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
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\n 30.\n

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Pažymėkite anilino ir druskos rūgšties cheminės reakcijos produktą.

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\n \n \n \"30A\"\n \n \n \"30B\"\n \n \n \"30C\"\n \n \n \"30D\"\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b63232ed56d2025021920341080843028","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė 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 \n \n \n 1\n \n \n \n \n
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\n 31.\n

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Grynas anilinas buvo sumaišytas su bromu. Vykstant cheminei reakcijai, susidarė 2,4,6-tribromanilinas.

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Kiek bromo molekulių reagavo su viena anilino molekule? Užrašykite skaičių.

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\n (1 taškas)\n

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\n Atsakymas:\n \n

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\n \"31\"\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b63237bc4c6202502192034152f5b6787","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_32","guid":"I_32","label":"I_32","title":"I_32","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 32.\n

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Chemikė atliko anilino brominimo reakciją. Kas 10 minučių ji imdavo mėginį iš reakcijos mišinio ir užlašindavo ant plonasluoksnės chromatografijos plokštelės¹. Paveiksle pavaizduota ši plokštelė po analizės, kai buvo užlašintas pradinės medžiagos (anilino) mėginys² (laiko momentas – 0 min.) ir mėginiai iš reakcijos mišinio kas 10 min. Ant plokštelės reakcija nevyksta, nes lakios³ medžiagos (bromas) išgaruoja.

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\n \"32\"\n \n

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\n Užrašykite laiko momentą, kai reakcija pasibaigė ir susidarė tik galutinis produktas.\n

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\n (1 taškas)\n

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Atsakymas: min.

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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 \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n \n\n","folderRoot":"/upload/tasks/collections/zip/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b6323c7cf7120250219203420d79a29fb","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_33","guid":"I_33","label":"I_33","title":"I_33","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 33.\n

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Pasibaigus cheminei reakcijai, chemikė mišinį atskiedė vandeniu ir atliko ekstrahavimą, naudodama organinį tirpiklį. Pažymėkite priemonę, kuri naudojama mišiniams ekstrahuoti.

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\n (1 taškas)\n

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\n \n \n \"33A\"\n \n \n \"33B\"\n \n \n \"33C\"\n \n \n \"33D\"\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b6324143d3e202502192034259d8d70e2","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_34","guid":"I_34","label":"I_34","title":"I_34","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 34.\n

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Atlikus ekstrahavimą, atsiskyrė vandeninis ir organinis sluoksniai. Kur bus 2,4,6-tribromanilinas po ekstrahavimo? Pažymėkite teisingą atsakymą.

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\n (1 taškas)\n

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Nuosėdose

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Vandeniniame sluoksnyje

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Organiniame sluoksnyje

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Medžiaga išgaruos

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b632460629020250219203430e3b3b02d","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_35","guid":"I_35","label":"I_35","title":"I_35","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 35.\n

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Kurį metodą naudosime, įrodydami išgryninto 2,4,6-tribromanilino struktūrą? Pažymėkite teisingą atsakymą.

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\n (1 taškas)\n

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Plonasluoksnę chromatografiją

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Vandenilio branduolių magnetinio rezonanso spektroskopiją

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Distiliavimą

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Ekstrahavimą

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b6324abdeaf20250219203434194a89eb","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_36","guid":"I_36","label":"I_36","title":"I_36","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 3 struktūrinis klausimas\n

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\n 36.\n

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Endorfinas yra natūrali nuskausminamoji¹ medžiaga.

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Endorfino struktūrinėje formulėje pažymėkite peptidinę cheminę jungtį.

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\n (1 taškas)\n

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¹ nuskausminamoji – przeciwbólowa – обезболивающее – знеболююча

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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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b6324f7b067202502192034396461f8bc","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["hotspotInteraction"],"openQuestion":0}},{"identifier":"I_37","guid":"I_37","label":"I_37","title":"I_37","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 37.\n

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Kiek aminorūgščių bus gauta, visiškai hidrolizavus vieną endorfino molekulę? Užrašykite skaičių.

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\n (1 taškas)\n

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\n Atsakymas:\n \n

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\n \"kl37\"\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b6325452b6220250219203444a6b24559","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_38","guid":"I_38","label":"I_38","title":"I_38","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 38.\n

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Kiek vienodos molekulinės masės dipeptidų gali sudaryti dvi skirtingos aminorūgštys? Užrašykite skaičių.

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\n (1 taškas)\n

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\n Atsakymas:\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b6325954e0320250219203449b2991867","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["textEntryInteraction"],"openQuestion":0}},{"identifier":"I_39","guid":"I_39","label":"I_39","title":"I_39","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 39.\n

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Pažymėkite schemą, kurioje teisingai pavaizduoti vandeniliniai ryšiai tarp aminorūgščių fragmentų polipeptiduose.

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\n (1 taškas)\n

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\n \n \n \"39A\"\n \n \n \"39B\"\n \n \n \"39C\"\n \n \n \"39D\"\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b6325e1a4b620250219203454e6e8504a","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"I_40","guid":"I_40","label":"I_40","title":"I_40","submissionMode":null,"xml":"\n\n \n \n \n \n \n \n \n \n 1\n \n \n \n \n
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\n 40.\n

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Ant indo su paruoštu reagentu baltymams nustatyti yra pavaizduotos piktogramos.

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Pažymėkite piktogramą, įspėjančią, jog šiame inde esantį skystį galima pilti tik į cheminiams reagentams skirtą konteinerį ir saugoti, kad nepatektų į aplinką.

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\n (1 taškas)\n

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\n \n \n \"40A\"\n \n \n \"40B\"\n \n \n \"40C\"\n \n \n \"40D\"\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/2025/06/06/70/vbe1che5-emokykla-1749203003/items/i67b63262ce3f82025021920345804859c84","sessionControl":{"showFeedback":true,"validateResponses":true,"submissionMode":"individual"},"assessmentItem":{"maxScore":1},"details":{"dateAdded":"2025-06-06 12:45:17","counterId":"ID17946","name":"2025 m. chemijos VBE I dalis. Pagrindinė sesija","transformed":false,"interactions":["choiceInteraction"],"openQuestion":0}},{"identifier":"i67ca1688b4186202503062241280af938b9","guid":"i67ca1688b4186202503062241280af938b9","label":"INFO_6","title":"Literatūra","submissionMode":null,"xml":"\n\n \n \n \n
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Naudota literatūra

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1. A. Nomoto, S. Nishinami, K. Shiraki, „Solubility Parameters of Amino Acids on Liquid–Liquid Phase Separation and Aggregation of Proteins“, Frontiers in Cell and Developmental Biology, 2021, Vol. 9, https://doi.org/10.3389/fcell.2021.691052.

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2. A. Demirbas, N. Cek, S. Acar, „Chemical Analyses of Shale Gas and Conventional Natural Gas“, Petroleum Science and Technology, 2018, Vol. 36, https://doi.org/10.1080/10916466.2018.1504070.

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3. U. Ranieri, F. Formisano, F. A. Gorelli, M. Santoro, M. M. Koza, A. De Francesco, L. E. Bove, „Crossover from Gas-like to Liquid-like Molecular Diffusion in a Simple Supercritical Fluid“, Nature Communications, 2024, 15(1), 4142, https://doi.org/10.1038/s41467-024-47961-7.

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