Benchmarks

17. (1) 原子 石墨 (2) B (3) BD AC $\backslash text\{ClO\}\_2$

解析: (3) 17号元素是氯元素, 8号元素是氧元素, 设D与A形成化合物的化学式为 $\backslash text\{Cl\}\_x\backslash text\{O\}\_y$ , 则 $35.5x:\; 16y\; =\; 71:\; 64,\; x:y\; =\; 1:2$ , M的化学式为 $\backslash text\{ClO\}\_2$ 。

18. (1) 属于 (2) a $2\backslash text\{H\}\_2\backslash text\{O\}\; \backslash xrightarrow\{\backslash text\{通电\}\}\; 2\backslash text\{H\}\_2\backslash uparrow\; +\; \backslash text\{O\}\_2\backslash uparrow$

(3) ① 2 ② 活性炭 ③ $\backslash text\{AgCl\}$ ④ $\backslash text\{N\}\_2$

19. (1) $\backslash text\{CaCO\}\_3\; +\; 2\backslash text\{HCl\}\; =\; \backslash text\{CaCl\}\_2\; +\; \backslash text\{H\}\_2\backslash text\{O\}\; +\; \backslash text\{CO\}\_2\backslash uparrow$ (2) $\backslash text\{S\}\; +\; \backslash text\{O\}\_2\; \backslash xrightarrow\{\backslash text\{点燃\}\}\; \backslash text\{SO\}\_2$

(3) $2\backslash text\{KClO\}\_3\; \backslash xrightarrow[\backslash Delta]\{\backslash text\{MnO\}\_2\}\; 2\backslash text\{KCl\}\; +\; 3\backslash text\{O\}\_2\backslash uparrow$

20. (1) B (2) A (3) ① 20 ② 丙 ③ 1:7 ④ 降温结晶 ⑤ C

解析: (3) ③ $t\_2^\backslash circ\backslash text\{C\}$ 时, 从50 g甲的饱和溶液中取出10 g, 加10 g水, 该温度下, 此时所得溶液的溶质与溶液质量比为

$(10\; \backslash text\{\; g\}\; \backslash times\; \backslash frac\{40\; \backslash text\{\; g\}\}\{40\; \backslash text\{\; g\}\; +\; 100\; \backslash text\{\; g\}\}\; \backslash times\; 100\backslash \%)\; :\; 20\; \backslash text\{\; g\}\; =\; 1\; :\; 7$ ; ⑤ C. 将 $t\_2^\backslash circ\backslash text\{C\}$ 时等质

量的甲、乙、丙三种物质的饱和溶液分别降温到 $t\_1^\backslash circ\backslash text\{C\}$ , 降温后, 甲、乙的溶解度减小, 有溶质析出, 且甲的溶解度受温度影响比乙大, 甲析出的晶体多, 丙的溶解度增加, 变为不饱和溶液, 无晶体析出, 因此溶液质量: 丙>乙>甲; D. 降温后, 甲、乙的溶解度减小, 还是饱和溶液, 丙变为不饱和溶液, 降温后, 丙的溶质质量分数不变。 $t\_1^\backslash circ\backslash text\{C\}$ 时, 三种溶液中溶质的质量从多到少的顺序为乙、甲、丙, 则所得溶液的溶质质量分数: 乙>甲>丙。

21. (1) 延展性 金属 (2) $\backslash text\{Fe\}\_2\backslash text\{O\}\_3\; +\; 3\backslash text\{CO\}\; \backslash xrightarrow\{\backslash text\{高温\}\}\; 2\backslash text\{Fe\}\; +\; 3\backslash text\{CO\}\_2$

(3) ① $\backslash text\{Fe\}^\{2+\}$ ② $\backslash text\{Fe\}$ 、 $\backslash text\{Cu\}$ (4) 0.15

解析: (3) ① 滤液中加入一枚洁净的铁钉, 发现铁钉表面无明显现象, 说明硫酸铜和硫酸已经完全反应, 则滤液中一定含硫酸亚铁, 一定大量存在的阳离子是 $\backslash text\{Fe\}^\{2+\}$ ; ② 向滤渣中再滴入稀硫酸产生气泡, 说明铁过量, 则滤渣中一定含 $\backslash text\{Fe\}$ 以及铁与硫酸铜反应生成的 $\backslash text\{Cu\}$ ; (4) 铝、锌、铁与稀硫酸反应的化学方程式可知, 硫酸中的硫酸根离子完全转化到反应后的盐中, 则硫酸根离子的质量为 $11\; \backslash text\{\; g\}\; -\; 3.8\; \backslash text\{\; g\}\; =\; 7.2\; \backslash text\{\; g\}$ , 硫酸中的氢元素完全转化到了氢气中, 设生成氢气的质量为 $x$ , 则 $\backslash frac\{96\}\{7.2\; \backslash text\{\; g\}\}\; =\; \backslash frac\{2\}\{x\}$ , $x\; =\; 0.15\; \backslash text\{\; g\}$ 。

22. (1) 锥形瓶 (2) $2\backslash text\{H\}\_2\backslash text\{O\}\_2\; \backslash xrightarrow\{\backslash text\{MnO\}\_2\}\; 2\backslash text\{H\}\_2\backslash text\{O\}\; +\; \backslash text\{O\}\_2\backslash uparrow$ E 浓硫酸

(3) $\backslash text\{Ca\}(\backslash text\{OH\})\_2$ 黑色粉末逐渐变为红色

23. (1) $\backslash text\{CO\}\_2$ (2) $\backslash text\{BaCO\}\_3\; +\; 2\backslash text\{HNO\}\_3\; =\; \backslash text\{Ba\}(\backslash text\{NO\}\_3)\_2\; +\; \backslash text\{CO\}\_2\backslash uparrow\; +\; \backslash text\{H\}\_2\backslash text\{O\}$

(3) 氢氧化钠、碳酸钠、硫酸钠

(4) 3.51

解析: (4) 设氯化钡和硫酸钠反应生成氯化钠的质量为 $x$ ,

$$\backslash text\{BaCl\}\_2\; +\; \backslash text\{Na\}\_2\backslash text\{SO\}\_4\; =\; \backslash text\{BaSO\}\_4\backslash downarrow\; +\; 2\backslash text\{NaCl\}$$

$$\backslash frac\{233\}\{4.66\; \backslash text\{\; g\}\}\; =\; \backslash frac\{117\}\{x\}\; \backslash quad\; x\; =\; 2.34\; \backslash text\{\; g\}$$

固体甲中碳酸钡的质量 $=\; 6.63\; \backslash text\{\; g\}\; -\; 4.66\; \backslash text\{\; g\}\; =\; 1.97\; \backslash text\{\; g\}$ ,

设氯化钡和碳酸钠反应生成氯化钠的质量为 $y$ ,

$$\backslash text\{BaCl\}\_2\; +\; \backslash text\{Na\}\_2\backslash text\{CO\}\_3\; =\; \backslash text\{BaCO\}\_3\backslash downarrow\; +\; 2\backslash text\{NaCl\}$$

$$\backslash frac\{197\}\{1.97\; \backslash text\{\; g\}\}\; =\; \backslash frac\{117\}\{y\}\; \backslash quad\; y\; =\; 1.17\; \backslash text\{\; g\}$$

滤液A中氯化钠的质量为 $1.17\; \backslash text\{\; g\}\; +\; 2.34\; \backslash text\{\; g\}\; =\; 3.51\; \backslash text\{\; g\}$ 。

24. (1) $\backslash text\{CH\}\_4\; +\; 2\backslash text\{O\}\_2\; \backslash xrightarrow\{\backslash text\{点燃\}\}\; \backslash text\{CO\}\_2\; +\; 2\backslash text\{H\}\_2\backslash text\{O\}$ 释放 (2) B

(3) ① 8:9 ② 吸收二氧化碳

解析: (3) ① 转化炉中的化学方程式为 $\backslash text\{CH\}\_4\; +\; \backslash text\{H\}\_2\backslash text\{O\}\; \backslash xrightarrow\{\backslash text\{一定条件\}\}\; \backslash text\{CO\}\; +\; 3\backslash text\{H\}\_2$ , 参加反应的甲烷与水的质量比为 $16:18=8:9$ 。

25. (1) 有机物 (2) 2:1 (3) 16.8 g

解析: (3) 32.4 g纤维素中所含碳元素质量 $32.4\; \backslash text\{\; g\}\; \backslash times$

$$\backslash frac\{72n\}\{(72+10+16\backslash times\; 5)n\}\; \backslash times\; 100\backslash \%\; =\; 14.4\; \backslash text\{\; g\}$$

含碳元素质量相同的聚丙烯的质量为 $14.4\; \backslash text\{\; g\}\; \backslash div$

$$\backslash left(\; \backslash frac\{36n\}\{36n+6n\}\; \backslash times\; 100\backslash \%\; \backslash right)\; =\; 16.8\; \backslash text\{\; g\}。$$

26. 解: 设该合金样品中锌的质量为 $x$ , 反应生成硫酸锌的质量为 $y$ ,

参加反应的稀硫酸的质量为 $z$

$$\backslash text\{Zn\}\; +\; \backslash text\{H\}\_2\backslash text\{SO\}\_4\; =\; \backslash text\{ZnSO\}\_4\; +\; \backslash text\{H\}\_2\backslash uparrow$$

$$\backslash frac\{65\}\{x\}\; =\; \backslash frac\{2\}\{0.2\; \backslash text\{\; g\}\}\; \backslash quad\; x\; =\; 6.5\; \backslash text\{\; g\}$$

$$\backslash frac\{161\}\{y\}\; =\; \backslash frac\{2\}\{0.2\; \backslash text\{\; g\}\}\; \backslash quad\; y\; =\; 16.1\; \backslash text\{\; g\}$$

$$\backslash frac\{98\}\{10\backslash \%z\}\; =\; \backslash frac\{2\}\{0.2\; \backslash text\{\; g\}\}\; \backslash quad\; z\; =\; 98\; \backslash text\{\; g\}$$

该合金样品中铜的质量分数为: $\backslash frac\{20\; \backslash text\{\; g\}\; -\; 6.5\; \backslash text\{\; g\}\}\{20\; \backslash text\{\; g\}\}\; \backslash times\; 100\backslash \%\; =\; 67.5\backslash \%$

反应后所得溶液溶质质量分数 $=\; \backslash frac\{16.1\; \backslash text\{\; g\}\}\{6.5\; \backslash text\{\; g\}\; +\; 98\; \backslash text\{\; g\}\; -\; 0.2\; \backslash text\{\; g\}\}\; \backslash times\; 100\backslash \%\; \backslash approx$

15.4%

答: (1) 合金样品中铜的质量分数为67.5%;

(2) 反应后所得溶液中溶质的质量分数为15.4%。

①④2022 南开区初中毕业生学业考试模拟(一)

1. B 本题考查六大基本营养素。豆浆、鲜奶是富含蛋白质的食物。

2. C 本题考查物质的变化。C. 钻木取火的过程中发生了燃烧, 产生

中汇化学[261]

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