2013年长沙市初中毕业学业水平考试试卷 .doc

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1精品文档,欢迎下载! 2013年长沙市初中毕业学业水平考试试卷

理科综合参考答案及评分标准

一、选择题(本大题包括27个小题,每小题3分,共81分。每小题只有1个选项符合题意。)

题号 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

答案 A B D C B C B A D D A A B C C

题号 16 17 18 19 20 21 22 23 24 25 26 27

答案 C A B A D A B C C D B B

二、(本大题共5小题,每空2分,共22分)

28.凝华;放出 29.1.68×105;168 30.1.25;不变

31.240;96 32.100;250;80%

三、(本大题共4小题,第33题4分,第34题、35题、36题每题6分,共22分)

33.(1)10 ················································································· (2分)

(2)靠近 ·············································································· (2分)

34.(1)晶体 ·············································································· (2分)

(2)固液共存 ········································································ (2分)

(3)小 ················································································· (2分)

35.(1)匀速直线 ········································································ (2分)

(2)接触面越粗糙 ·································································· (2分)

(3)接触面不同(没有控制变量/变量不唯一,只要合理即可得分) ····· (2分)

36.(1)连线正确 ········································································ (2分)

(2)闭合开关前滑动变阻器应调到阻值最大处/滑动变阻器变化范围太小

(其他答案只要合理均可给分) ········ (2分)

(3)0.625 ············································································· (1分)

1.444 ············································································· (1分)

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2精品文档,欢迎下载! 四、(本大题共3小题,第37、38题每题6分,第39题8分,共20分)

37.(1)①受重力作用 ②一阵急促的助跑动作

说明:每空2分,其他答案只要合理,均可给分。

(2)问题:欢呼声响彻湘江沿岸说明什么?答案:空气可以传声

问题:风呼呼的从脸庞刮过,他为什么会倍感凉爽?答案:蒸发吸热

问题:保持一定高度滑翔时,他和三角翼受什么力作用?答案:重力和升力

说明:问题和答案各1分,共2分。其他答案只要合理,均可给分。

38.解:(1)根据s=vt可得

st=v ··········································································· (1分)

300m200s1.5m/s==(或3.33min或0.056h) ··························· (1分)

(2)pghr=海水 ································································ (1分)

3351.010kg/m10N/kg10.5m1.0510=创?? Pa ············· (1分)

(3)航母处于漂浮状态,浮力等于总的重力,当飞机起飞后,航母总重力减小,

其减小的重力与减小的浮力相等。

F=V浮G飞机=m飞机g=352010kg10N/kg210N创=? ··········· (1分)

根据阿基米德原理 FgVr=浮液排 得

5333210N20m1.010kg/m10N/kgFVgr´===创VV浮排海水 ···················· (1分)

39.解:(1)LLLPIU= ··································································· (1分)

3W0.25A12V== ······················································ (1分)

(2)闭合S,断开S1、S2,此时R1与L串联

由灯泡正常工作和串联电路中电流处处相等知UL=12V,I=IL=0.25A

················································································· (1分)

根据欧姆定律得

1172V12V2400.25AUURI--===W ·································· (1分)

(3)S、S1、S2均闭合时,R1与R2并联

2min2maxUPR= ······························································ (1分)

2(72V)25.92W200==W ··············································· (1分)

1172V0.3A240UIR===W

A2允许通过的最大电流为3A

则允许通过滑动变阻器的最大电流2max3A0.3A=2.7AI=- ···· (1分)

max2max72V2.7A194.4WPUI==? ·································· (1分)

说明:计算题其他解法只要合理均可给分。 本人提供的文档均由本人编辑如成,如对你有帮助,请下载支持!

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五、(本大题共4小题,化学方程式每个3分,其余每空2分,共20分)

40.(1)2P

(2)CH4

41.(1)t1℃时,甲、乙两物质的溶解度都为15g

(2)大于

42.CaCO3CaO+CO2­

CaO+H2O=Ca(OH)2

Ca(OH)2+CO2=CaCO3¯+H2O(任写两个,只要正确均计分)

43.(1)①可燃

②导电

(2)拨打119火警电话 迅速离开现场(其他合理答案亦可)

六、(本大题共2小题,化学方程式每个3分,其余每空2分,共11分)

44.(1)元素

(2)镉等重金属造成了水污染或滥用化肥(其它合理答案亦可)

(3)防治污水,合理使用化肥(其它合理答案亦可)

45.(1)H2SO4

(2)ZnSO4+BaCl2=ZnCl2+BaSO4¯(其他合理答案亦可)

七、(本大题共2小题,每空2分,共18分)

46.(1)试管

(2)e

(3)a、b, g

47.(1)盐

(2)探究Na+是否能使酚酞试液变红

(3)酚酞试液中含有水分子(或NaCl溶液中含有水分子)

(4)白色沉淀

(5)CO32-与H2O发生反应产生OH-(其它合理合理答案亦可)

八、(本大题共1小题,共6分)

48.(1)2.5 ················································································ (1分)

(2)解:设H2O2的质量为x ····················································· (1分)

2H2O2 2H2O+O2­ ·············································································· (1分)

68 32

x

1.6g

6832=1.6xg ········································································ (1分)

x=3.4g ·········································································· (1分)

过氧化氢溶液中溶质质量分数为:3.4100gg´100%=3.4% ············· (1分) 煅烧 本人提供的文档均由本人编辑如成,如对你有帮助,请下载支持!

4精品文档,欢迎下载! 答:原过氧化氢溶液中溶质质量分数为3.4%。