【文档说明】广东省佛山市南海区2024-2025学年高三上学期开学摸底测试 化学 PDF版含答案.pdf,共(10)页,10.209 MB,由小赞的店铺上传
转载请保留链接:https://www.doc5u.com/view-9ef439860a6103a2ee1c1747c86eb09d.html
以下为本文档部分文字说明:
{#{QQABRQaEogAoQJAAARgCAwV6CAKQkAACCYgOxBAIsAAAABNABAA=}#}{#{QQABSYIk5gg40JTACZ4KAwE+CQmQkJEhLUgEBRADOAwqQI
NIBIA=}#}{#{QQABRQaEogAoQJAAARgCAwV6CAKQkAACCYgOxBAIsAAAABNABAA=}#}{#{QQABSYIk5gg40JTACZ4KAwE+CQmQkJEhLUgEBRADO
AwqQINIBIA=}#}{#{QQABRQaEogAoQJAAARgCAwV6CAKQkAACCYgOxBAIsAAAABNABAA=}#}{#{QQABSYIk5gg40JTACZ4KAwE+CQmQkJEhLUgEBR
ADOAwqQINIBIA=}#}{#{QQABRQaEogAoQJAAARgCAwV6CAKQkAACCYgOxBAIsAAAABNABAA=}#}{#{QQABSYIk5gg40JTACZ4KAwE+CQmQ
kJEhLUgEBRADOAwqQINIBIA=}#}{#{QQABRQaEogAoQJAAARgCAwV6CAKQkAACCYgOxBAIsAAAABNABAA=}#}{#{QQABSYIk5gg40JTACZ4KAwE+CQmQkJEhLUgEBRADOAwqQINIBI
A=}#}{#{QQABRQaEogAoQJAAARgCAwV6CAKQkAACCYgOxBAIsAAAABNABAA=}#}{#{QQABSYIk5gg40JTACZ4KAwE+CQmQkJEhLUgEBRADOAwqQINIBIA=}#}{#{QQABRQaEogAoQJAAAR
gCAwV6CAKQkAACCYgOxBAIsAAAABNABAA=}#}{#{QQABSYIk5gg40JTACZ4KAwE+CQmQkJEhLUgEBRADOAwqQINIBIA=}#}{#{QQABRQ
aEogAoQJAAARgCAwV6CAKQkAACCYgOxBAIsAAAABNABAA=}#}{#{QQABSYIk5gg40JTACZ4KAwE+CQmQkJEhLUgEBRADOAwqQINIBIA=}#}南海区2025届高三摸底测
试答案化学第1页共2页南海区2025届高三摸底测试化学答案一、选择题:本题共16小题,共44分。第1~10小题,每小题2分;第11~16小题,每小题4分。在每小题给出的四个选项中,只有一项是符合题目要求的。题号12345678910选项ACDADBBACD题号111213141516选项BCCDC
B二、非选择题:本题共4小题,共56分。17.(14分)(1)N(1分)NaOH(1分)(2)抑制Fe3+水解(1分)(3)K3[Fe(CN)6]或铁氰化钾溶液(1分)0.30(1分)Fe3++Ag⇌Fe2++
Ag+(1分)(4)②i.BD(2分)ii.96.00%(2分)③i.(2分)ii.NaCl溶液(1分)(5)Fe(NO3)3溶银不产生污染大气的气体(1分)18.(14分)(1)+6(1分)(2)Fe(VO3)3+6
H+=Fe3++3VO+2+3H2O(2分)(3)B(1分)(4)V(1分)适当调高pH(1分,或适当升高温度也合理)(5)0.8(2分)(6)MgSiO3(1分)(7)6(1分)棱心、体心(2分)A321a1033N×(2分)c(A
g+)/mol·L-10.10.064{#{QQABRQaEogAoQJAAARgCAwV6CAKQkAACCYgOxBAIsAAAABNABAA=}#}{#{QQABSYIk5gg40JTACZ4KAwE+CQmQkJEhLUgEBRAD
OAwqQINIBIA=}#}南海区2025届高三摸底测试答案化学第2页共2页19.(14分)(1)2s22p4(1分)(2)Ca(OH)2+2HCOOH=(HCOO)2Ca+2H2O(2分)(3)①ΔH1—Δ
H2—ΔH3(1分)②BD(2分)(4)①<(1分)②>(1分)C2H5-是推电子基,使羧基中的O-H极性减小,酸性减弱(2分)③(4分)由图可知,Ka(HCOOH)=10—3.75,Ka(R1COOH)=10—5.75可
求反应HCOOH+R1COO−=R1COOH+HCOO−的平衡常数为:100)COOHR()HCOOH()COOR()HCOOH()HCOO()COOHR(1aa11==⋅⋅=−−KKccccK甲酸和丙酸钠等物质的量混合时,设两者的初始浓度均为c0,变化浓度为x,则H
COOH+R1COO−=R1COOH+HCOO−初始浓度c0c000变化浓度xxxx平衡浓度c0—xc0—xxx则K=������������·������������(������������0—����������
��)·(������������0—������������)=100,求得x=0.909c0,因此平衡时甲酸的转化率为90.9%。20.(14分)(1)C9H15N(1分)(2)醛基或羟基(1分)CH
3COOC2H5或CH3CH2COOCH3(2分)(3)CCCH3HCHOHn(1分)新制Cu(OH)2溶液、加热(1分)氧化反应(1分)(4)H2N(或CH2=CHCH2CH2CH2NH2)(2分)(5)AC(2分)(6)(1分)BrBr+2NaO
H乙乙乙乙+2NaBr+2H2O(2分){#{QQABRQaEogAoQJAAARgCAwV6CAKQkAACCYgOxBAIsAAAABNABAA=}#}{#{QQABSYIk5gg40JTACZ4KAwE+CQmQk
JEhLUgEBRADOAwqQINIBIA=}#}