Đề thi cuối kì 2 Hóa 11 THPT Tân Kỳ 3 (Nghệ An) năm 2023-2024 có đáp án

Đáp án: 18,9

Để cung cấp Q kJ nhiệt lượng cho đun nấu:

+ Nếu dùng biogas thì nCH$_{4}$ = Q/890

nCO$_{2}$ phát thải = nCH$_{4}$ = Q/890

+ Nếu dùng gas thì nC$_{3}$H$_{8}$ = 2x và nC$_{4}$H$_{10}$ = 3x

⟶ Q = 2220.2x + 2850.3x ⟶ x = Q/12990

nCO$_{2}$ phát thải = 3.2x + 4.3x = 3Q/2165 > Q/890 nên với cùng 1 nhiệt lượng cung cấp ra thì dùng biogas sẽ phát thải ít CO$_{2}$ hơn gas.

Lượng CO$_{2}$ giảm = (3Q/2165 – Q/890) / (3Q/2165) = 18,9%

Đáp án: 99,1

Công thức phân tử của vanillin: C$_{8}$H$_{8}$O$_{3}$.

${n}_{NaOH}=7{,}82.1{0}^{-4}\ \text{(}\text{m}\text{o}\text{l}\text{)}$

HOC$_{6}$H$_{3}$(OCH$_{3}$)(CHO) + NaOH $\overset{}{\rightarrow }$ NaOC$_{6}$H$_{3}$(OCH$_{3}$)(CHO) + H$_{2}$O

$\overset{PTHH}{\rightarrow }\ {\text{n}}_{Va}\ \text{=}\ {\text{n}}_{NaOH}\ \text{=}\ \text{7}\text{,}\text{8}\text{2}\text{.}\text{1}{\text{0}}^{-4}\ \text{m}\text{o}\text{l}\ \overset{}{\rightarrow }\ \text{\%}{\text{m}}_{Va}\ \text{=}\ \frac{7{,}82.1{0}^{-4}.152}{0{,}12}.100\%\ \text{=}\ \text{9}\text{9}\text{,}\text{0}\text{5}\text{\%}$≈ 99,1%

Đáp án: 33,8

$A〈{C}_{6}{H}_{5}OH\text{;}\ {\text{C}}_{2}{H}_{5}OH〉\{\begin{matrix}\overset{+Na}{\rightarrow }\ {\text{H}}_{2}\ \text{(}\text{0}\text{,}\text{1}\text{5}\text{)} \\ \overset{+B{r}_{2}}{\rightarrow }\ {\text{C}}_{6}{H}_{2}B{r}_{3}OH\ \text{(}\text{0}\text{,}\text{0}\text{6}\text{)}\end{matrix}$

$\overset{A+B{r}_{2}}{\rightarrow }\ {\text{n}}_{{C}_{6}{H}_{5}OH}\ \text{=}\ {\text{n}}_{↓}\ \text{=}\ \text{0}\text{,}\text{0}\text{6}\ \text{m}\text{o}\text{l}$

$\overset{A+Na}{\rightarrow }\ {\text{n}}_{{C}_{6}{H}_{5}OH}\ \text{+}\ {\text{n}}_{{C}_{2}{H}_{5}OH}\ \text{=}\ \text{2}{\text{n}}_{{H}_{2}}\ \Rightarrow \ {\text{n}}_{{C}_{2}{H}_{5}OH}\ \text{=}\ \text{0}\text{,}\text{2}\text{4}\ \text{m}\text{o}\text{l}\ \overset{}{\rightarrow }\ \text{\%}{\text{m}}_{{C}_{6}{H}_{5}OH(A)}\ \text{=}\ \text{3}\text{3}\text{,}\text{8}\text{\%}$

Đáp án: 0,2

Phương trình hóa học:

$\begin{matrix}HOOC-{C}_{6}{H}_{4}-OH+C{H}_{3}OH\overset{{\text{H}}_{\text{2}}\text{S}{\text{O}}_{\text{4}}\ \text{®}\text{Æ}\text{c}\text{,}\ {\text{t}}^{\text{o}}\ }{⇌}HO-{C}_{6}{H}_{4}-COOC{H}_{3}+{H}_{2}O \\ salicylic\ acid\ \ \ \ \ \ \ methyl\ salicylate\end{matrix}$

$\begin{matrix}HO-{C}_{6}{H}_{4}-COOC{H}_{3}+2NaOH\overset{}{\rightarrow }NaOOC-{C}_{6}{H}_{4}-ONa+C{H}_{3}OH+{H}_{2}O \\ \ 0{,}1\ \rightarrow \ \ 0{,}2\end{matrix}$ mol

Đáp án: 100

V $_{ethanol (mL) có trong xăng}$ = $\frac{2000\ \ .\ \ 5}{100}$ = 100 ml

Đáp án: 0,5

${\overline{V}}_{NaOH}=\frac{20+19{,}9+20}{3}\approx 20{,}0\ \text{m}\text{l}$

$6.{C}_{M(C{H}_{3}\text{C}\text{O}\text{O}H)}=20.0{,}15\rightarrow {C}_{M(C{H}_{3}\text{C}\text{O}\text{O}H)}=0{,}5M$