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太原市2024年高三年级模拟试题（二)
理综化学部分参考答案和评分建议
一、
选择题（每个小题6分，共7个小题，共42分）
题号
7
8
9
10
11
12
13
答案
B
B
0
0
B
c
D
三、非选择题
27.（共11分，除标注外，每空2分)
(1)浓氨水(1分)
(2)6TiO2+8NH,
800C6TiN+12H,0+N,
(3)吸收多余的氨气与水蒸气
(4)②④⑤③
(5)产生氨气的速率较快，反应时间较短（或其他合理答案）
4×62
(6)
0.423×107N
28.(共16分，除标注外，每空2分)
(1)d(1分)15
(2)3SiCl4 +4LiCo02
500℃
4LiCl+4C0Cl2+3SiO2+02
(3)焰色试验(1分)
(4)9.39
(5)①SiCl4+60H=SiO32+4C1+3H20(或其他合理答案)
②Si-C1键极性更强，易断裂（或Si的原子半径更大，Si-C1键键能更小，易断裂：
或Sⅰ有更多的价层轨道易接受O的孤电子对形成配位键等合理答案)
(6)⑥⑩③⑤
SiCl4
29.(共16分，除标注外，每空2分)
(1)-233
(2)BC(有错为零分，不全扣1分)
(3)随着反应温度T升高，1000
减小，lnk变大，则k变大(1分)
T
(4)①b(1分)根据C0和CO2的选择性定义，两者之和为1，图中曲线a与c此消彼
长，则表示氢气产率的曲线不可能是a或c(或其他合理解释)
②CaO与CO2反应生成CaCO3,使CO2浓度减小，平衡I正向移动，平衡IⅡ逆向移动
(5)①号P
768
②0.05 mol-L-.min
p
30.(共15分，除标注外，每空2分)
(1)羧基、酮羰基
(2)
140℃，重二甲苯
2HCI
(3
取代反应(1分)
CHCOOH
(4)17
CHO
(5)共4分
以上试题其他合理答案或说法也可给分。{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
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{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}太原市 2023年高三年级模拟考试（二）
物理参考答案及评分建议
二、选择题:本题共 8小题，每小题 6分，共 48分。在每小题给出的四个选项中，第 14～
18题只有一项符合题目要求，第 19～21题有多项符合题目要求。全部选对的得 6分，选对
但不全的得 3分，有选错的得 0分。
题号 14 15 16 17 18 19 20 21
选项 C D C D C CD CD AB
三、非选择题：共 62分。
22.（8分）
（1）A （2分）
4 sin 2 cos cos sin + sin （ ） （ 分） 2 1 （2分） 1 2 （2分）
sin 1 cos + 2 cos
其他正确答案均可得分
23.（10分）
（1）C（1分） D（1分） E（1分）
1
（2）b（1分） b 1 11（2分） 或 （2分） 2 21 1+ 2
（3） V + 2（2分）（其他合理说法均可得分）
24.（10分）
解：（1）
对 a束光
sin 60
na 3 ···········································································（1分）sin
sin 1 ，α=30°············································································（1分）
2
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对 b束光
sin 60 6
nb ··········································································（1分）sin 2
sin 2 ，β=45°·········································································（1分）
2
由光路图，出射 a光束平行 PO，出射 b光束与 PO成 30°角，
a、b两光束成 30°············································································（1分）
（2）由 v c 得
n
v ca n ························································································· (1分)a
v cb n ·························································································（1分）b
t 3R 3Ra ··············································································（1分）va c
t 2R 3Rb ············································································（1分）vb c
ta 3 ························································································（1分）
tb 1
25．（14分）
（1）粒子在磁场做匀速圆周运动，在电场中做斜抛运动。
由受力分析，粒子带正电，粒子回到 O点时与 MN的夹角为 ·················· （3分）
（2）在磁场中，洛伦兹力提供向心力
2
qvB m v ·····················································································（1分）
r
r mv
qB
从边界 MN射出的位置记为 P点，OP的距离为 L
= 2 sin ·····················································································（1分）
在 MN上方的电场中运动时，粒子做匀变速曲线运动
竖直方向，粒子做匀变速直线运动
= ··························································································（1分）

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= 2 sin ····················································································（1分）

水平方向，粒子做匀速直线运动
= cos ·················································································· （1分）
= ························································································（1分）
cos
（3）当速度变为 2 后，粒子的运动轨迹如图所示，粒子在磁场中运动时
2π
=

= 2π = 2π cos
··········································································· （1分）
= 2π 2 1 ·····················································································（1分）2π
粒子在电场中运动的时间设为 2
= 2 2 sin 2 ···················································································（1分）
粒子从 O点射出后，又返回出发点 O所用的时间为 总 = 2 1 + 2··············（1分）
= 4 π cos + 4 sin 总 ································································· （1分）
26.（20分）
（1）小滑块的加速度为 1，圆弧槽和板的加速度为 2
1 = 1 ···············································································（1分）
3 2 = 1 24 ································································· （1分）
1 2 10 1 2 2 = ································································· （2分）2 2
圆弧槽和板的速度 2 = 2 =1 m/s·····················································（1分）
（2）P到达 B点时的速度 1 = 0 1 ···············································（1分）
1 + 3 2 =

1 + 3

2······························································（2分）
1 2 + 11 3 2
1
2 =
2 + 11 3
2
2 + 2 ·········································（2分）2 2 2 2

解得 1 = 4 m/s·············································································（1分）
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P到达 C点时的速度大小为 4 m/s，方向水平向左·································（1分）
（3）P C 4 从 点落到长木板的时间 = ·············································（1分）

P = + 相对板向左运动 1 2 ····················································（1分）
P第一次弹起后的水平速度 1 =
1····················································· （1分）
2

碰后板的速度 ，
板 1
1 + 3 2 = 1 + 3 ···························································（1分）板 1
P 相对板向左运动 1 = ( 1 +
) 2 ············································（1分）
板 1
P


= 1 + 3 = + 3 第二次弹起后的水平速度 2 ，4 1 2 2 板 2
P 相对板向左运动 2 = ( 2 + ) 2 ············································（1分）板 2
+ 1 + 1 + 2····································································（1分）
P被弹起两次后弹离木板···································································（1分）
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}

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