Trong phản ứng xảy ra: sự oxi hoá Fe và sự khử Cu²⁺                   

Beri không tác dụng với nước ở điều kiện thường

“Nước đá khô” là CO₂ rắn.

Sản phẩm của phản ứng este hóa giữa ancol metylic và axit propionic là metyl propionat.          

Cu không tác dụng với dung dịch H₂SO₄ loãng

 axit α-aminoglutaric làm quỳ tím chuyển thành màu hồng

AlCl₃ không có tính lưỡng tính

Trong FeCO₃ sắt có số oxi hóa +2      

Thủy tinh hữu cơ là chất rắn trong suốt, có khả năng cho ánh sáng truyền qua, rất cứng và bền với nhiệt.

Phản ứng của Al với Fe₂O₃ gọi là phản ứng nhiệt nhôm

Saccarozơ - đisaccarit

Phương trình phản ứng điện phân dung dịch

2KCl + 2H₂O → 2KOH + H₂ + Cl₂

Na₂CO₃ làm mềm mọi loại nước cứng

2NaHSO₄ + Ba(OH)₂ → BaSO₄ + Na₂SO₄ + 2H₂O

Muối là chất điện li mạnh

Công thức của triolein là: (CH₃[CH₂]₇CH=CH[CH₂]₇COO)₃C₃H₅             

Canxi hiđroxit có công thức hoá học là Ca(OH)₂

Polietilen không chứa liên kết đôi trong các mắt xích

Valin có CTPT: C₅H₁₁NO₂

Quặng hematit đỏ có thành phần chính là Fe₂O₃            

số mol Fe = 0,01 < mol AgNO₃ = 0,05

→ sản phẩm tạo Fe³⁺

→ n_Ag = 3n_Fe = 0,03 

→ m_Ag = 3,24 gam

pH = 13 → pOH = 1 → [OH^-] = 0,1 → mol OH^- = 0,1

n_Al = 0,15

2Al + 2OH^- + 2H₂O → 2AlO₂^- + 3H₂

→ mol H₂ = 0,15 → V = 3,36

(c), (a), (b). 

Trong dung dịch, glyxin tồn tại chủ yếu ở dạng ion lưỡng cực H₃N⁺CH₂COO^-.

sơ đồ hợp thức:

         180       →        92

          300................  ?

→ m_ancol = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaGaaGimaiaaicdacqGHxdaTcaaI5aGaaGOmaaqaaiaaigdacaaI % 4aGaaGimaaaaaaa!3E01! \frac{{300 \times 92}}{{180}}\)= \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI0aGaaGOnaiaaicdaaeaacaaIZaaaaaaa!38F8! \frac{{460}}{3}\)

→ H = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI5aGaaGOmaiabgEna0kaaigdacaaIWaGaaGimaaqaamaalaaabaGa % aGinaiaaiAdacaaIWaaabaGaaG4maaaaaaGaeyypa0JaaGOnaiaaic % daaaa!414D! \frac{{92 \times 100}}{{\frac{{460}}{3}}} = 60\) %

28,48 gam Ala → n_Ala = 0,32

32 gam Ala-Ala → n_Ala-Ala = 0,2

27,72 gam Ala-Ala-Ala → n_Ala-Ala-Ala = 0,12

 → n_tetrapeptit = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIWaGaaiilaiaaiodacaaIYaGaey4kaSIaaGimaiaacYcacaaIYaGa % aiOlaiaaikdacqGHRaWkcaaIWaGaaiilaiaaigdacaaIYaGaaiOlai % aaiodaaeaacaaI0aaaaiabgEna0kaacIcacaaI4aGaaGyoaiaac6ca % caaI0aGaeyOeI0IaaGymaiaaiIdacaGGUaGaaG4maiaacMcacqGH9a % qpcaaI4aGaaGymaiaacYcacaaI1aGaaGinaaaa!523A! \frac{{0,32 + 0,2.2 + 0,12.3}}{4} \times (89.4 - 18.3) = 81,54\)

số C = 6

số H = 12

A có CT: C₆H₁₂O₆ 

mà A ngọt hơn đường mía → fructozơ

Nhiệt độ nóng chảy của kim loại kiềm thổ tăng dần từ Be đến Ba

Phát biểu đúng là (b), (c), (d)

tơ nhân tạo (bán tổng hợp) là tơ visco, tơ xenlulozơ triaxetat.

6,4 gam kim loại không tan là Cu → Cu đã phản ứng là 0,1 mol

Theo BT e: 2n_Fe + 2n_Cu = 2n_O + \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xb9qqFj0d % Xdbba91qpepeI8k8fiI+fsY-rqaqpepae9pg0FirpepesP0xe9Fve9 % Fve9qapdbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaad6 % gadaWgaaWcbaGaamisamaaBaaameaacaaIYaaabeaaaSqabaaaaa!3C96! 2{n_{{H_2}}}\) → n_O = 0,35 mol

→ m = 0,4.56 + 0,2.64 + 0,35.16 = 40,8 g

E gồm các este của ancol A (x mol) và các este của phenol B (y mol)

     A + NaOH → muối + ROH

     B + 2NaOH → muối + H₂O

Ta có: x + 2y = n_NaOH và x + y = 0,12 (1) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaa4ajaSqaai % aadkeacaWGubGaam4saiaadYeaaeqakiaawkziaiaaigdacaaI2aGa % aiilaiaaiodacaaIYaGaey4kaSIaaGinaiaaicdacaGGUaGaaiikai % aacIhacqGHRaWkcaaIYaGaaiyEaiaacMcacqGH9aqpcaaMc8UaaGym % aiaaiIdacaGGSaGaaG4naiaaiIdacqGHRaWkcaWGTbWaaSbaaSqaai % aadIfaaeqaaOGaey4kaSIaaGymaiaaiIdacaWG5baaaa!52EE! \xrightarrow{{BTKL}}16,32 + 40.(x + 2y) = \,18,78 + {m_X} + 18y\)(2)

Khi cho X tác dụng với ancol thì: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaaiaacaGaaeqabaWaaeaaeaaakeaacaWGUbWaaSbaaS % qaaiaadIeadaWgaaadbaGaaGOmaaqabaaaleqaaOGaeyypa0JaaGim % aiaacYcacaaI1aGaamOBamaaBaaaleaacaWGHbGaamOBaiaadogaca % WGVbGaamiBaaqabaGccqGH9aqpcaaIWaGaaiilaiaaiwdacaWG4baa % aa!46AE! {n_{{H_2}}} = 0,5{n_{ancol}} = 0,5x\) và m_b.tăng = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaaiaacaGaaeqabaWaaeaaeaaakeaacaWGTbWaaSbaaS % qaaiaadIfaaeqaaOGaeyOeI0IaamyBamaaBaaaleaacaWGibWaaSba % aWqaaiaaikdaaeqaaaWcbeaakiabgkDiElaad2gadaWgaaWcbaGaam % iwaaqabaGccqGHsislcaWG4bGaeyypa0JaaG4maiaacYcacaaI4aGa % aG4maiaaysW7caGGOaGaaG4maiaacMcaaaa!4A56! {m_X} - {m_{{H_2}}} \Rightarrow {m_X} - x = 3,83\;(3)\)

Từ (1), (2), (3) suy ra: x = 0,05; y = 0,07 → n_NaOH = 0,19 mol → V_NaOH = 190 ml

(a) Sai, Điện phân dung dịch NaCl thì trên catot xảy ra sự khử H₂O.

(d) Sai, Kim loại Fe khử được H₂O ở nhiệt độ cao.

(g) Sai, Trong một chu kì, bán kính nguyên tử kim loại lớn hơn bán kính nguyên tử phi kim.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadoeacaWGpbGaaiOoaiaadIhaaeaacaWGdbGaam4tamaaBaaa % leaacaaIYaaabeaakiaacQdacaWG5baabaGaamisamaaBaaaleaaca % aIYaaabeaakiaacQdacaGG4bGaey4kaSIaaGOmaiaacMhaaaGaay5E % aaGaeyO0H49aaiqaaqaabeqaaiaadIhacqGHRaWkcaWG5bGaey4kaS % YaaeWaaeaacaWG4bGaey4kaSIaaGOmaiaadMhaaiaawIcacaGLPaaa % cqGH9aqpcaaIWaGaaiilaiaaiEdaaeaacaWGcbGaamivaiaaykW7ca % WGLbGaaiOoaiaaikdacaWG4bGaey4kaSIaaGOmaiaac6cadaqadaqa % aiaadIhacqGHRaWkcaaIYaGaamyEaaGaayjkaiaawMcaaiabg2da9i % aaicdacaGGSaGaaGinaiaac6cacaaIZaaaaiaawUhaaiabgkDiEpaa % ceaaeaqabeaacaWG4bGaeyypa0JaaGimaiaacYcacaaIYaaabaGaam % yEaiabg2da9iaaicdacaGGSaGaaGymaaaacaGL7baacqGHshI3caGG % LaGaamOBamaaBaaaleaacaWGibWaaSbaaWqaaiaaikdaaeqaaaWcbe % aakiabg2da9iaaiwdacaaI3aGaaiilaiaaigdacaaI0aGaaiyjaaaa % !7E0A! \left\{ \begin{gathered} CO:x \hfill \\ C{O_2}:y \hfill \\ {H_2}:x + 2y \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} x + y + \left( {x + 2y} \right) = 0,7 \hfill \\ BT\,e:2x + 2.\left( {x + 2y} \right) = 0,4.3 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} x = 0,2 \hfill \\ y = 0,1 \hfill \\ \end{gathered} \right. \Rightarrow \% {n_{{H_2}}} = 57,14\%\)

E có công thức chung là C₅₇H_yO₆ → \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xb9qqFj0d % Xdbba91qpepeI8k8fiI+fsY-rqaqpepae9pg0FirpepesP0xe9Fve9 % Fve9qapdbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGfbaabeaakiabg2da9maalaaabaGaamOBamaaBaaaleaa % caWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaaakeaacaaI1a % GaaG4naaaacqGH9aqpcaaIWaGaaiilaiaaicdacaaIZaGaaGjbVlaa % d2gacaWGVbGaamiBaaaa!4987! {n_E} = \frac{{{n_{C{O_2}}}}}{{57}} = 0,03\;mol\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xb9qqFj0d % Xdbba91qpepeI8k8fiI+fsY-rqaqpepae9pg0FirpepesP0xe9Fve9 % Fve9qapdbaqaaeaaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaamaaoqcaleaacaWGcbGaamivaiaacQdacaaMe8Uaam4taaqabOGa % ayPKHaGaamOBamaaBaaaleaacaWGibWaaSbaaWqaaiaaikdaaeqaaS % Gaam4taaqabaGccqGH9aqpcaaIXaGaaiilaiaaiAdacaaMe8UaamyB % aiaad+gacaWGSbaabaWaa4ajaSqaaiaadkeacaWGubGaam4saiaadY % eaaeqakiaawkziaiaad2gadaWgaaWcbaGaamyraaqabaGccqGH9aqp % caaIYaGaaGOnaiaacYcacaaI2aGaam4zaaaacaGL7baaaaa!576E! \left\{ \begin{gathered} \xrightarrow{{BT:\;O}}{n_{{H_2}O}} = 1,6\;mol \hfill \\ \xrightarrow{{BTKL}}{m_E} = 26,6g \hfill \\ \end{gathered} \right.\) mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xb9qqFj0d % Xdbba91qpepeI8k8fiI+fsY-rqaqpepae9pg0FirpepesP0xe9Fve9 % Fve9qapdbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % sislcaWGUbWaaSbaaSqaaiaadIeadaWgaaadbaGaaGOmaaqabaWcca % WGpbaabeaakiabg2da9iaad6gadaWgaaWcbaGaamOqaiaadkhadaWg % aaadbaGaaGOmaaqabaaaleqaaOGaey4kaSIaaGOmaiaad6gadaWgaa % WcbaGaamiwaaqabaGccqGHshI3caWGUbWaaSbaaSqaaiaadkeacaWG % YbWaaSbaaWqaaiaaikdaaeqaaaWcbeaakiabg2da9iaaicdacaGGSa % GaaGimaiaaiwdacaaMe8UaamyBaiaad+gacaWGSbGaeyO0H49aaSaa % aeaacaWGTbaabaGaamyyaaaacqGH9aqpcaaI1aGaaG4maiaaikdaaa % a!5FEB! {n_{C{O_2}}} - {n_{{H_2}O}} = {n_{B{r_2}}} + 2{n_X} \Rightarrow {n_{B{r_2}}} = 0,05\;mol \Rightarrow \frac{m}{a} = 532\)

Phát biểu: (b) Sai, Xenlulozơ có PTK lớn hơn tinh bột.

(b) Sai, Phản ứng xảy ra trong bước 1 là phản ứng một chiều.

(c) Sai, Dung dịch H₂SO₄ đặc ngăn cản quá trình thủy phân của xenlulozơ.

(f) Sai, Thí nghiệm trên dùng để chứng minh xenlulozơ có phản ứng thủy phân tạo glucozơ.

Hỗn hợp E gồm X: HCOOH (a mol); Y: RCOOH (b mol); T (c mol)

Ta có:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbw5usTq % vATv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwz % YbItLDharqqtubsr4rNCHbGeaGqiVGI8Ffeu0dXdh9vqqj-hEeeu0x % Xdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9q8qqaq-dir-f % 0-yqaiVgFr0xfr-xfr-xb9adbaGaaiaadaWaamaaceGaaqaacaqbaa % GcbaWaaiqaaqaabeqaaiaaikdacaWGHbGaey4kaSIaaGOmaiaadoga % cqGH9aqpcaWGUbWaaSbaaSqaaiaadgeacaWGNbaabeaakiabg2da9i % aaicdacaGGSaGaaGymaiaaiAdaaeaadaGdKaWcbaGaamOqaiaadsfa % caGG6aGaaGjbVlaad+eaaeqakiaawkziaiaaikdacaWGHbGaey4kaS % IaaGOmaiaadkgacqGHRaWkcaaI0aGaam4yaiabg2da9iaaicdacaGG % SaGaaG4maiaaiIdaaeaacaGGOaGaam4AaiabgkHiTiaaigdacaGGPa % GaaiOyaiabgUcaRiaadUgacaWGJbGaeyypa0JaamOBamaaBaaaleaa % caWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGHsislca % WGUbWaaSbaaSqaaiaadIeadaWgaaadbaGaaGOmaaqabaWccaWGpbaa % beaakiabg2da9iaaicdacaGGSaGaaGymaiaaisdaaaGaay5EaaWaa4 % ajaSqaaiaadUgacqGH9aqpcaaIYaaabeGccaGLsgcadaGabaabaeqa % baGaamyyaiabg2da9iaaicdacaGGSaGaaGimaiaaiwdaaeaacaWGIb % Gaeyypa0JaaGimaiaacYcacaaIWaGaaGioaaqaaiaadogacqGH9aqp % caaIWaGaaiilaiaaicdacaaIZaaaaiaawUhaaiabgkDiElaacwcaca % WGTbWaaSbaaSqaaiaadIfaaeqaaOGaeyypa0JaaGymaiaaiIdacaGG % SaGaaGynaiaaiIdacaGGLaaaaa!8AF9! \left\{ \begin{gathered} 2a + 2c = {n_{Ag}} = 0,16 \hfill \\ \xrightarrow{{BT:\;O}}2a + 2b + 4c = 0,38 \hfill \\ (k - 1)b + kc = {n_{C{O_2}}} - {n_{{H_2}O}} = 0,14 \hfill \\ \end{gathered} \right.\xrightarrow{{k = 2}}\left\{ \begin{gathered} a = 0,05 \hfill \\ b = 0,08 \hfill \\ c = 0,03 \hfill \\ \end{gathered} \right. \Rightarrow \% {m_X} = 18,58\% \)→ (d) Sai.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2Caerbw5usTq % vATv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwz % YbItLDharqqtubsr4rNCHbGeaGqiVGI8Ffeu0dXdh9vqqj-hEeeu0x % Xdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9q8qqaq-dir-f % 0-yqaiVgFr0xfr-xfr-xb9adbaGaaiaadaWaamaaceGaaqaacaqbaa % GcbaWaa4ajaSqaaiaadkeacaWGubGaaiOoaiaaysW7caWGdbaabeGc % caGLsgcacaaIWaGaaiilaiaaicdacaaI1aGaaiOlaiaaigdacqGHRa % WkcaaIWaGaaiilaiaaicdacaaI4aGaaiOlaiaadoeadaWgaaWcbaGa % amywaaqabaGccqGHRaWkcaaIWaGaaiilaiaaicdacaaIZaGaaiOlai % aacIcacaaIXaGaey4kaSIaam4qamaaBaaaleaacaWGzbaabeaakiab % gUcaRiaadoeadaWgaaWcbaGaamyyaiaad6gacaWGJbGaam4BaiaadY % gaaeqaaOGaaiykaiabg2da9iaaicdacaGGSaGaaGinaiaaiEdacqGH % shI3daGabaabaeqabaGaam4qamaaBaaaleaacaWGzbaabeaakiabg2 % da9iaaiodacaGG6aGaai4qaiaacIeadaWgaaWcbaGaaGOmaaqabaGc % cqGH9aqpcaWGdbGaamisaiabgkHiTiaadoeacaWGpbGaam4taiaadI % eaaeaacaWGdbWaaSbaaSqaaiaadggacaWGUbGaam4yaiaad+gacaWG % Sbaabeaakiabg2da9iaaikdacaGG6aGaam4qamaaBaaaleaacaaIYa % aabeaakiaadIeadaWgaaWcbaGaaGinaaqabaGccaGGOaGaam4taiaa % dIeacaGGPaWaaSbaaSqaaiaaikdaaeqaaaaakiaawUhaaaaa!7F22! \xrightarrow{{BT:\;C}}0,05.1 + 0,08.{C_Y} + 0,03.(1 + {C_Y} + {C_{ancol}}) = 0,47 \Rightarrow \left\{ \begin{gathered} {C_Y} = 3:C{H_2} = CH - COOH \hfill \\ {C_{ancol}} = 2:{C_2}{H_4}{(OH)_2} \hfill \\ \end{gathered} \right.\)

(e) Sai, E làm mất màu dung dịch Br₂.

Khi cho F tác dụng với AgNO₃ thì:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xb9qqFj0d % Xdbba91qpepeI8k8fiI+fsY-rqaqpepae9pg0FirpepesP0xe9Fve9 % Fve9qapdbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGibGaam4qaiaad+eacaWGpbGaamOtaiaadggaaeqaaOGa % eyypa0ZaaSaaaeaacaWGUbWaaSbaaSqaaiaadgeacaWGNbaabeaaaO % qaaiaaikdaaaGaeyypa0JaaGimaiaacYcacaaIWaGaaGinaiaaykW7 % caWGTbGaam4BaiaadYgacaaMc8UaeyO0H4TaamOBamaaBaaaleaaca % WGdbGaamisamaaBaaameaacaaIZaaabeaaliaadoeacaWGpbGaam4t % aiaad6eacaWGHbaabeaakiabg2da9iaaicdacaGGSaGaaGimaiaaik % dacaaMc8UaamyBaiaad+gacaWGSbaaaa!601B! {n_{HCOONa}} = \frac{{{n_{Ag}}}}{2} = 0,04\,mol\, \Rightarrow {n_{C{H_3}COONa}} = 0,02\,mol\)

mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xb9qqFj0d % Xdbba91qpepeI8k8fiI+fsY-rqaqpepae9pg0FirpepesP0xe9Fve9 % Fve9qapdbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGfbaabeaakiabg2da9iaad6gadaWgaaWcbaGaam4qaiaa % dIeadaWgaaadbaGaaG4maaqabaWccaWGdbGaam4taiaad+eacaWGob % GaamyyaaqabaGccqGHRaWkcaWGUbWaaSbaaSqaaiaadIeacaWGdbGa % am4taiaad+eacaWGobGaamyyaaqabaGccqGH9aqpcaaIWaGaaiilai % aaicdacaaI2aGaeyO0H4TaamytamaaBaaaleaacaWGfbaabeaakiab % g2da9iaaiIdacaaI2aGaaGPaVlaacIcacaWGdbWaaSbaaSqaaiaais % daaeqaaOGaamisamaaBaaaleaacaaI2aaabeaakiaad+eadaWgaaWc % baGaaGOmaaqabaGccaGGPaaaaa!5D99! {n_E} = {n_{C{H_3}COONa}} + {n_{HCOONa}} = 0,06 \Rightarrow {M_E} = 86\,({C_4}{H_6}{O_2})\) có các đồng phân là:

HCOOCH=CH-CH₃ ; HCOOC(CH₃)=CH₂ ; CH₃COOCH=CH₂

Cho M tác dụng với AgNO₃ thì: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xb9qqFj0d % Xdbba91qpepeI8k8fiI+fsY-rqaqpepae9pg0FirpepesP0xe9Fve9 % Fve9qapdbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaad6gadaWgaaWcbaGaam4qaiaadIeadaWgaaadbaGaaG4maaqa % baWccaWGdbGaamisaiaad+eaaeqaaOGaeyypa0JaaGimaiaacYcaca % aI1aGaaiOBamaaBaaaleaacaWGbbGaam4zaaqabaGccqGH9aqpcaaI % WaGaaiilaiaaicdacaaIZaaabaGaamOBamaaBaaaleaacaWGdbGaam % isamaaBaaameaacaaIZaaabeaaliaadoeacaWGibGaam4taaqabaGc % cqGH9aqpcaWGUbWaaSbaaSqaaiaadoeacaWGibWaaSbaaWqaaiaaio % daaeqaaSGaam4qaiaad+eacaWGpbGaam4qaiaadIeacqGH9aqpcaWG % dbGaamisamaaBaaameaacaaIYaaabeaaaSqabaGccqGH9aqpcaaIWa % GaaiilaiaaicdacaaIYaGaaGjbVlaad2gacaWGVbGaamiBaaaacaGL % 7baacqGHshI3caWGUbWaaSbaaSqaaiaadIeacaWGdbGaam4taiaad+ % eacaWGdbGaamisaiabg2da9iaadoeacaWGibGaeyOeI0Iaam4qaiaa % dIeadaWgaaadbaGaaG4maaqabaaaleqaaOGaeyypa0JaaGimaiaacY % cacaaIWaGaaGymaiaaysW7caWGTbGaam4BaiaadYgaaaa!7B8C! \left\{ \begin{gathered} {n_{C{H_3}CHO}} = 0,5{n_{Ag}} = 0,03 \hfill \\ {n_{C{H_3}CHO}} = {n_{C{H_3}COOCH = C{H_2}}} = 0,02\;mol \hfill \\ \end{gathered} \right. \Rightarrow {n_{HCOOCH = CH - C{H_3}}} = 0,01\;mol\)

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xb9qqFj0d % Xdbba91qpepeI8k8fiI+fsY-rqaqpepae9pg0FirpepesP0xe9Fve9 % Fve9qapdbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taai % yjaiaad2gadaWgaaWcbaGaamisaiaadoeacaWGpbGaam4taiaadoea % caWGibGaeyypa0Jaam4qaiaadIeacqGHsislcaWGdbGaamisamaaBa % aameaacaaIZaaabeaaaSqabaGccqGH9aqpcaaIXaGaaGOnaiaacYca % caaI2aGaaG4naiaacwcaaaa!4D66! \Rightarrow \% {m_{HCOOCH = CH - C{H_3}}} = 16,67\% \) → (a) Đúng.

(b) Sai, Số mol Y trong 5,16 gam E là 0,02 mol.

(c) Sai.

(d) Đúng.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xb9qqFj0d % Xdbba91qpepeI8k8fiI+fsY-rqaqpepae9pg0FirpepesP0xe9Fve9 % Fve9qapdbaqaaeaaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIfacaaMe8Uaaiikaiaad6eacaWGibWaaSbaaSqaaiaaisda % aeqaaOGaai4taiaac+eacaGGdbGaeyOeI0Iaai4qaiaac+eacaGGpb % GaaiOtaiaacIeadaWgaaWcbaGaaG4maaqabaGccaGGdbGaaiisamaa % BaaaleaacaaIZaaabeaakiaacMcacaGG6aGaaiyyaiaaysW7caGGTb % Gaai4BaiaacYgaaeaacaWGzbGaaiikaiaadEeacaWGSbGaamyEaiab % gkHiTiaadEeacaWGSbGaamyEaiabgkHiTiaadgeacaWGSbGaamyyai % aacMcacaGG6aGaaGjbVlaadkgacaaMe8UaamyBaiaad+gacaWGSbaa % aiaawUhaaiabgkziUoaaceaaeaqabeaacaaIXaGaaG4maiaaiIdaca % WGHbGaey4kaSIaaGOmaiaaicdacaaIZaGaamOyaiabg2da9iaaikda % caaI3aGaaiilaiaaikdaaeaacaaIYaGaamyyaiabg2da9iaaicdaca % GGSaGaaGymaaaacaGL7baacqGHshI3daGabaabaeqabaGaamyyaiab % g2da9iaaicdacaGGSaGaaGimaiaaiwdaaeaacaWGIbGaeyypa0JaaG % imaiaacYcacaaIXaaaaiaawUhaaaaa!8453! \left\{ \begin{gathered} X\;(N{H_4}OOC - COON{H_3}C{H_3}):a\;mol \hfill \\ Y(Gly - Gly - Ala):\;b\;mol \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} 138a + 203b = 27,2 \hfill \\ 2a = 0,1 \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} a = 0,05 \hfill \\ b = 0,1 \hfill \\ \end{gathered} \right.\)

Khi cho E tác dụng với HCl thu được các chất hữu cơ lần lượt là (COOH)₂: 0,05; CH₃NH₃Cl: 0,05;  GlyHCl: 0,2; AlaHCl: 0,1 → m = 42,725 (g)