2024届四川省乐山市高三上学期第一次调研考试 理科综合答案

2023-12-25 · U1 上传 · 7页 · 850.2 K

乐山市高中2024届第一次调查研究考试理科综合参考答案化学一、选择题7.A8.C9.B10.C11.D12.B13.D三、非选择题26.(14分)(1)粉碎、研磨等(1分)(2)SO2(2分)++(3)CuSO4(1分)Bi2O3+2H=2BiO+H2O(2分)SiO2和GeO2(2分)(4)Na2GeO3+6HCl=GeCl4+2NaCl+3H2O(2分)抑制GeCl4水解(2分)(5)蒸馏(2分)27.(14分)(1)稀HCl(1分)排尽装置的空气防止FeCO3被氧化(2分)(2)饱和碳酸氢钠溶液(1分)HCl将与丙中的NaHCO3(生成的FeCO3)反应,降低FeCO3的产率(2分)2+-(3)Fe+2HCO3=FeCO3↓+CO2↑+H2O(2分)(4)玻璃棒和漏斗(2分)4FeCO3∙2H2O+O2=4FeOOH+4CO2↑+6H2O(2分)-2+(5)Na2CO3溶液碱性更强,水解产生的OH易与Fe反应生成Fe(OH)2(2分)28.(15分)퐾2(1)-747(2分)2(2分)퐾1(2)3(2分)N2O2+CO=N2O+CO2(2分)(3)①a(1分);等温过程增大压强平衡正向移动,α(NO)增大与曲线a的变化相符(2分)1②25(2分)4.4×10-3/4×10-3/4.44×10-3/(2分)22535.(15分)(1)3d104s2(1分)ds(1分)(2)O>N>C>H(2分)>(1分)(3)sp2、sp3杂化(2分)分子晶体(1分)(4)三角锥形(1分)NH3在形成配离子时,孤电子对转化为成键电子对,对N-H键的排斥作用减弱,H-N-H键角增大(2分)2×(65+16)2×(65+16)(5)4(2分)3/3(2分)√푎2푏푁×10−30√(푎×10−10)2(푏×10−10))푁2퐴2퐴{#{QQABCYCEggggAAAAARhCEQVYCkKQkBAAAIoGwAAMoAIBwRFABAA=}#}36.(15分)(1)(2分)2-氨基丙酸/α-氨基丙酸/丙氨酸(2分)(2)羟基和羧基(2分)消去反应(1分)(3)CO2(2分)(4)(2分)(5)16(2分)(2分){#{QQABCYCEggggAAAAARhCEQVYCkKQkBAAAIoGwAAMoAIBwRFABAA=}#}乐山市2024届“一调”考试(生物)参考答案及评分标准第I卷第1~6题,每题6分,共36分。题号123456答案BBCDDA第Ⅱ卷29.(除注明外,每空1分,共9分)(1)磷脂双分子层0(2)二甲双胍抑制线粒体的功能,细胞中ATP供给不足(缺乏)(1分),直接影响无活型和激活型RagC进出细胞核(1分),最终抑制细胞生长(共2分)(3)RNA或mRNA或rRNA或tRNADNA聚合酶或RNA聚合酶(4)AB(5)抑制(2分)30.(除注明外,每空2分,共10分)(1)叶绿体(1分)光反应(1分)(2)①光合膜系统的数量;光反应中心蛋白质的含量②光合膜系统正常(或数量多,或发达);蛋白质含量正常③低温处理④光合膜系统少;蛋白质含量低31.(每空2分,共8分)(1)神经中枢(2)A、B(一点1分,共2分)(3)局部麻醉剂作用于神经纤维细胞膜Na+的载体,导致Na+内流受阻(1分),感受器(神经)不能产生兴奋,神经也不能传导兴奋到大脑皮层(1分)(4)内啡肽作用于突触后膜上的受体,相应的神经元兴奋(1分)并传导到大脑皮层的相应中枢,产生愉悦的感觉(1分)32.(除注明外,每空2分,共12分)(1)相对性状多(品种多);相对性状差异显著;繁殖周期短;繁殖能力强(后代数量多);雌雄异花。(合理即可。一点1分,共2分)(2)Aa、aa(一点1分,共2分)(3)花粉鉴定法(待开花后,取花粉以碘液染色,显微镜观察并统计记录花粉的形状和颜色(1分),花粉均长形且都被染成蓝色的即为BBCC植株(1分),共2分)(4)生殖隔离(5)选择基因型AaCc为亲本自交(1分),统计记录后代各表现型的数量(1分)。若后代4种表现型之比为9:3:3:1,则Aa、Cc两对等位基因分别位于两对不同的同源染色体上(1分);若后代的表现型之比不是9:3:3:1,则Aa、Cc两对等位基因不是分别位于两对不同的同源染色体上(1分)。(共4分)或选择基因型AaCc和aacc为亲本测交(1分),统计记录后代各表现型的数量(1分)。若后代4种表现型之比为1:1:1:1,则Aa、Cc两对等位基因分别位于两{#{QQABCYCEggggAAAAARhCEQVYCkKQkBAAAIoGwAAMoAIBwRFABAA=}#}对不同的同源染色体上(1分);若后代的表现型之比不是1:1:1:1,则Aa、Cc两对等位基因不是分别位于两对不同的同源染色体上(1分)。(共4分)37.(除注明外,每空2分,共15分)(1)食用油、超临界CO2(正确一点1分,正确两点3分,共3分)水不溶性、水不溶性(正确一点1分,正确两点3分,共3分)(2)具有惊人的溶解能力;没有毒性;萃取剂容易从产品中完全除去;萃取剂不污染环境;不影响产品质量(正确一点1分,正确两点3分,共3分)(3)温度不能太高、时间不能太长(每点1分,共2分)被破坏萃取效率38.(除注明外,每空2分,共15分)(1)逆转录酶、Taq酶(每点1分,共2分)界定要扩增的目的基因;是逆转录酶、Taq酶的结合部位;是子链延伸的起点(一点1分,共4分)(2)5%CO2、95%空气(每点1分,共2分)血清(3)S基因、GS基因(正确一点1分,两点3分,共3分)(4)限制腺病毒DNA的复制,提高安全性将E1基因导入人胚胎肾细胞{#{QQABCYCEggggAAAAARhCEQVYCkKQkBAAAIoGwAAMoAIBwRFABAA=}#}2024届第一次调研考试(物理)答案选择题.1415161718192021BCCCABDBCAD22.(每空两分,共6分,选择题漏选得1分.)(1)BD(2)①BC②23.(前三空每空2分,最有一空3分,共9分.)(1)不是2.66(2.63-2.69)5.52(5.38-5.67)0.316(0.310-0.324)24.(12分)解:(1)由受力分析可得,···········································(2分)···············································(1分)···········································(2分)(2)细绳断裂后,小球在竖直平面内做类平抛运动,可分解为:竖直方向的匀加速直线运动:·················································(2分)························································(1分)小球落地时与水平线的夹角为,···············································(1分)···············································(1分)答:略25.(20分)解:令水平向右为正方向。(1)由图像可知,图线与时间轴围成的图形面积为力F在0-4s内的冲量,动量定理可得:···············································(2分)···············································(1分)(2)AB发生弹性碰撞,系统动量守恒和能量守恒,·······································(1分)···································(1分),·····································(1分)由于小车足够长,离墙壁足够远,故小车与墙壁碰撞前已和滑块B共速,由动量守恒可得,·······································(1分)············································(1分)小车与墙壁完成第1次碰撞后,车速变为小车与墙壁第2次碰撞前与滑块B再次共速,由动量守恒可得,···········································(1分)由系统功能关系可得,小车与墙壁第1次碰撞后到与墙壁第2次碰撞前瞬间的过程中滑块与小车间由于摩擦产生的热量··················(1分){#{QQABCYCEggggAAAAARhCEQVYCkKQkBAAAIoGwAAMoAIBwRFABAA=}#}··············································(1分)(3)由小车与墙壁第1次碰后和第2次碰前小车与滑块共速时的速度大小可推断,从第1次碰撞结束,每次小车与滑块共速后立即与墙壁发生碰撞。·····························(1分)以小车为研究对象,由牛顿第二定律可得······································(1分)···········································(1分)第1次碰后到减速到零有:··············································(1分)从第1次碰后到第2次碰前小车运动的路程有:·························(1分)同理可得:从第2次碰后到第3次碰前小车运动的路程有:·································(1分)从第3次碰后到第4次碰前小车运动的路程有:·······························(1分)从第4次碰后到第5次碰前小车运动的路程有:····························(1分)综上可得:从第1次碰后到第5次碰前小车运动的路程有:··························(1分)(第(2)(3)小题还可使用图像法求解)AB发生弹性碰撞,系统动量守恒和能量守恒,·······································(1分)···································(1分),·····································(1分)由于小车足够长,离墙壁足够远,故小车与墙壁碰撞前已和滑块B共速,由动量守恒可得,·······································(1分)············································(1分)以小车第一次碰后为即时起点,根据牛顿第二定律有······································(1分)······································(1分)滑块与小车运动过程中加速度大小分别为,·····························(1分)小车与滑块速度随时间变化的图像如图所示··················································(1分)小车与墙壁第1次碰撞后到与墙壁第2次碰撞前瞬间的过程中滑块与小车的相对路程为(0,8)(0,-4)(4,4)三点围成的三角形面积································································(1分)由于摩擦产生的热量·····················································(1分)由图像易知,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