Đề thi cuối kì 2 Hóa học 10 THPT Nguyễn Trãi (Đồng Nai) năm 2023-2024 có đáp án

a. Cl$_{2}$ + 2Al $\overset{}{\rightarrow }$ 2AlCl$_{3}$ ( 0.5 điểm)

b. 3Cl$_{2}$ + 6KOH $\overset{100{\ }^{o}C}{\rightarrow }$5KCl + KClO$_{3}$ + 3H$_{2}$O ( 0.5điểm)

$\overline{v}=-\frac{1}{2}.\frac{Δ{C}_{{N}_{2}{O}_{5}}}{Δt}\ \text{=}\ \frac{1}{4}.\frac{Δ{C}_{N{O}_{2}}}{Δt}\ \text{=}\ \frac{1}{1}.\frac{Δ{C}_{{O}_{2}}}{Δt}$

$\overline{v}=-\frac{1}{2}.\frac{(0{,}0388-0{,}0196)}{360}\ \text{=}\ \frac{1}{4}.\frac{(0{,}0699-0{,}0315)}{360}\ \text{=}\ \frac{1}{2}.\frac{(0{,}0175-0{,}0079)}{360}\ \text{=}\ \text{2}\text{,}\text{6}\text{7}\text{.}\text{1}{\text{0}}^{-5}\ \text{(}\text{M}\text{/}\text{s}\text{)}$

2 Na + X$_{2}$$\overset{}{\rightarrow }$ 2 NaX

Số mol Na = 2,3: 23 = 0,1 $\overset{}{\rightarrow }$số mol NaX = 0,1

Vậy M$_{NaX}$= 5,85 : 0,1 = 58,5 $\overset{}{\rightarrow }$23 + M$_{X}$ = 58,5 $\overset{}{\rightarrow }$ M$_{X}$ = 35,5 (chlorine )

Viết đúng phương trinh: 0,125

Tinh được số mol Na suy ra số mol NaX : 0,125

Tính đúng M của halogen : 0,125

Xác định được tên Halogen: Chlorine : 0,125

${n}_{{K}_{2}C{r}_{2}{O}_{7}}\ \text{=}\ \text{2}\text{.}\text{1}{\text{0}}^{-4}\ \text{m}\text{o}\text{l}$

a) $3\overset{-2}{C}{}_{2}{\text{H}}_{5}\text{O}\text{H}\ \text{+}\ {\text{K}}_{2}\overset{+6}{\text{C}\text{r}}{}_{2}{\text{O}}_{7}\ \text{+}\ \text{8}{H}^{+}\ \overset{}{\rightarrow }\ \text{3}\overset{-1}{\text{C}}{}_{2}{\text{H}}_{4}\text{O}\ \text{+}\ \text{2}\overset{+3}{Cr}\ \text{+}\ \text{2}{\text{K}}^{+}\ \text{+}\ \text{7}{\text{H}}_{2}O$

$\begin{matrix}C{r}^{+6}\ \ \ \ \text{+}\ \ \ \ \ \ \ \ \ \ \ \ \text{3}\text{e}\ \overset{}{\rightarrow }\ \text{C}{\text{r}}^{+3} \\ \text{2}\text{.}\text{2}\text{.}\text{1}{\text{0}}^{-4}\ \rightarrow \ \ \text{1}\text{,}\text{2}\text{.}\text{1}{\text{0}}^{-3}\end{matrix}||\begin{matrix}\ {C}_{2}^{-2}\ \overset{}{\rightarrow }\ {C}_{2}^{-1}\ \ \ \ \text{+}\ \ \ \text{2}\text{.}\text{1}\text{e} \\ \ \text{6}\text{.}\text{1}{\text{0}}^{-4}\ \ \ \ \overset{}{\leftarrow }\ \ \ \ \ \text{1}\text{,}\text{2}\text{.}\text{1}{\text{0}}^{-3}\end{matrix}$

${m}_{{C}_{2}{H}_{5}OH}\ \text{=}\ \text{6}\text{.}\text{1}{\text{0}}^{-4}.46\ \text{=}\ \text{0}\text{,}\text{0}\text{2}\text{7}\text{6}\ \text{g}\text{a}\text{m}\ \overset{}{\rightarrow }\ \text{\%}{m}_{{C}_{2}{H}_{5}OH(25\ \text{g}\text{a}\text{m})}\ \text{=}\ \text{0}\text{,}\text{1}\text{1}\text{\%}\ \text{>}\ \text{0}\text{,}\text{0}\text{2}\text{\%}$.

Vậy người lái xe vi phạm luật giao thông.