資源簡介

太原市2024年高三年級模擬試題（二)
理綜化學部分參考答案和評分建議
一、
選擇題（每個小題6分，共7個小題，共42分）
題號
7
8
9
10
11
12
13
答案
B
B
0
0
B
c
D
三、非選擇題
27.（共11分，除標注外，每空2分)
(1)濃氨水(1分)
(2)6TiO2+8NH,
800C6TiN+12H,0+N,
(3)吸收多余的氨氣與水蒸氣
(4)②④⑤③
(5)產生氨氣的速率較快，反應時間較短（或其他合理答案）
4×62
(6)
0.423×107N
28.(共16分，除標注外，每空2分)
(1)d(1分)15
(2)3SiCl4 +4LiCo02
500℃
4LiCl+4C0Cl2+3SiO2+02
(3)焰色試驗(1分)
(4)9.39
(5)①SiCl4+60H=SiO32+4C1+3H20(或其他合理答案)
②Si-C1鍵極性更強，易斷裂（或Si的原子半徑更大，Si-C1鍵鍵能更小，易斷裂：
或Sⅰ有更多的價層軌道易接受O的孤電子對形成配位鍵等合理答案)
(6)⑥⑩③⑤
SiCl4
29.(共16分，除標注外，每空2分)
(1)-233
(2)BC(有錯為零分，不全扣1分)
(3)隨著反應溫度T升高，1000
減小，lnk變大，則k變大(1分)
T
(4)①b(1分)根據C0和CO2的選擇性定義，兩者之和為1，圖中曲線a與c此消彼
長，則表示氫氣產率的曲線不可能是a或c(或其他合理解釋)
②CaO與CO2反應生成CaCO3,使CO2濃度減小，平衡I正向移動，平衡IⅡ逆向移動
(5)①號P
768
②0.05 mol-L-.min
p
30.(共15分，除標注外，每空2分)
(1)羧基、酮羰基
(2)
140℃，重二甲苯
2HCI
(3
取代反應(1分)
CHCOOH
(4)17
CHO
(5)共4分
以上試題其他合理答案或說法也可給分。{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}太原市 2023年高三年級模擬考試（二）
物理參考答案及評分建議
二、選擇題:本題共 8小題，每小題 6分，共 48分。在每小題給出的四個選項中，第 14～
18題只有一項符合題目要求，第 19～21題有多項符合題目要求。全部選對的得 6分，選對
但不全的得 3分，有選錯的得 0分。
題號 14 15 16 17 18 19 20 21
選項 C D C D C CD CD AB
三、非選擇題：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
其他正確答案均可得分
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）（其他合理說法均可得分）
24.（10分）
解：（1）
對 a束光
sin 60
na 3 ···········································································（1分）sin
sin 1 ，α=30°············································································（1分）
2
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
對 b束光
sin 60 6
nb ··········································································（1分）sin 2
sin 2 ，β=45°·········································································（1分）
2
由光路圖，出射 a光束平行 PO，出射 b光束與 PO成 30°角，
a、b兩光束成 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）粒子在磁場做勻速圓周運動，在電場中做斜拋運動。
由受力分析，粒子帶正電，粒子回到 O點時與 MN的夾角為 ·················· （3分）
（2）在磁場中，洛倫茲力提供向心力
2
qvB m v ·····················································································（1分）
r
r mv
qB
從邊界 MN射出的位置記為 P點，OP的距離為 L
= 2 sin ·····················································································（1分）
在 MN上方的電場中運動時，粒子做勻變速曲線運動
豎直方向，粒子做勻變速直線運動
= ··························································································（1分）

{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
= 2 sin ····················································································（1分）

水平方向，粒子做勻速直線運動
= cos ·················································································· （1分）
= ························································································（1分）
cos
（3）當速度變為 2 后，粒子的運動軌跡如圖所示，粒子在磁場中運動時
2π
=

= 2π = 2π cos
··········································································· （1分）
= 2π 2 1 ·····················································································（1分）2π
粒子在電場中運動的時間設為 2
= 2 2 sin 2 ···················································································（1分）
粒子從 O點射出后，又返回出發點 O所用的時間為 總 = 2 1 + 2··············（1分）
= 4 π cos + 4 sin 總 ································································· （1分）
26.（20分）
（1）小滑塊的加速度為 1，圓弧槽和板的加速度為 2
1 = 1 ···············································································（1分）
3 2 = 1 24 ································································· （1分）
1 2 10 1 2 2 = ································································· （2分）2 2
圓弧槽和板的速度 2 = 2 =1 m/s·····················································（1分）
（2）P到達 B點時的速度 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）
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}
P到達 C點時的速度大小為 4 m/s，方向水平向左·································（1分）
（3）P C 4 從 點落到長木板的時間 = ·············································（1分）

P = + 相對板向左運動 1 2 ····················································（1分）
P第一次彈起后的水平速度 1 =
1····················································· （1分）
2

碰后板的速度 ，
板 1
1 + 3 2 = 1 + 3 ···························································（1分）板 1
P 相對板向左運動 1 = ( 1 +
) 2 ············································（1分）
板 1
P


= 1 + 3 = + 3 第二次彈起后的水平速度 2 ，4 1 2 2 板 2
P 相對板向左運動 2 = ( 2 + ) 2 ············································（1分）板 2
+ 1 + 1 + 2····································································（1分）
P被彈起兩次后彈離木板···································································（1分）
{#{QQABLYAQoggIAJIAABgCQQXACEGQkBAAACoGBFAEIAAByRNABAA=}#}

展開更多......

收起↑