Khi thủy phân hoàn toàn 65 gam một peptit X thu được 22,25 gam alanin và 56,25 gam glyxin. X là
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Lời giải:
Báo sainAla = 0,25; nGly = 0,75 → Ala : Gly = 1 : 3
Bảo toàn khối lượng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGibWaaSbaaWqaaiaaikdaaeqaaSGaam4taaqa % baGccqGH9aqpcaaIWaGaaiilaiaaiEdacaaI1aaaaa!3F8B! \to {n_{{H_2}O}} = 0,75\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaadaWada % qaamaabmaabaGaamyqaiaadYgacaWGHbaacaGLOaGaayzkaaWaaeWa % aeaacaWGhbGaamiBaiaadMhaaiaawIcacaGLPaaadaWgaaWcbaGaaG % 4maaqabaaakiaawUfacaGLDbaadaWgaaWcbaGaamOBaaqabaGccqGH % RaWkdaqadaqaaiaaisdacaWGUbGaeyOeI0IaaGymaaGaayjkaiaawM % caaiaadIeadaWgaaWcbaGaaGOmaaqabaGccaWGpbGaeyOKH4QaamOB % amaaBaaaleaacaWGbbGaamiBaiaadggaaeqaaOGaey4kaSIaaG4mai % aad6gadaWgaaWcbaGaam4raiaadYgacaWG5baabeaaaOqaaiaac6ca % caGGUaGaaiOlaiaac6cacaGGUaGaaiOlaiaac6cacaGGUaGaaiOlai % aac6cacaGGUaGaaiOlaiaac6cacaGGUaGaaiOlaiaac6cacaGGUaGa % aiOlaiaac6cacaGGUaGaaiOlaiaac6cacaGGUaGaaiOlaiaac6caca % GGUaGaaiOlaiaac6cacaGGUaGaaiOlaiaac6cacaGGUaGaaGimaiaa % cYcacaaI3aGaaGynaiaac6cacaGGUaGaaiOlaiaac6cacaGGUaGaai % Olaiaac6cacaGGUaGaaiOlaiaac6cacaGGUaGaaiOlaiaac6cacaGG % UaGaaGimaiaacYcacaaIYaGaaGynaaaaaa!7BC7! \begin{gathered} {\left[ {\left( {Ala} \right){{\left( {Gly} \right)}_3}} \right]_n} + \left( {4n - 1} \right){H_2}O \to {n_{Ala}} + 3{n_{Gly}} \hfill \\ ......................0,75..............0,25 \hfill \\ \end{gathered} \)
→ 0,25(4n – 1) = 0,75n → n = 1
→ X là (Ala)(Gly)3 (Tetrapeptit)
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