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2020-2021ѧѧǦɽһкѧ߶θİ
ѧԾ
: time90
ʱprobablyõԭH-1C-12N-14O-16Na-23Mg-24Al-27Fe-56Cu-64Zn-65 S-16Cl-35.5Ag-108Br-80I-127Ca-40K-39
һchoice⣨16С⣬ÿС3ֹ48֣ÿСֻһѡ⣩
1Chinaͼʾװƻ¶ˮȡϡλ¼šмȡ¶ˮķΪС굣ʵһһʹ㣬ת֮йҺ֮װõʵǣ
A B Cleg D
2֪ȻѧʽSO2(g)1/2O2(g) SO3(g) H98.32kJmol14 molSO2μӦų314.3 kJʱSO2תӽ( )
A40% B50% C70% D80%
3˵ȷǣ
AȷӦԷӦ
BS(g)O2(g)===SO2(g) H1 S(s)O2(g)===SO2(g)H2H1<H2
Cstay½еķӦȷӦ
Dͬͬѹ£H2(g)Cl2(g)===2HCl(g)stayպȼµĦHͬ
4зӦӷʽвȷǣ
AH218OͶNa2O2壺2Na2O2+2H218O=4Na++4OH+18O2
Bstay£NH+4NO3NH+4+2O2+2OHNO3+3H2O
CtakeKMnO4ҺH2C2O4Һϣ2MnO4+5H2C2O4+6H+2Mn2++10CO2+8H2O
DʵBa(OH)2ˮУ3Ba2++6OH+2Al3++3SO243BaSO4+2Al(OH)3
5ResearchͨѭϵͳͼsureҵеSO2ƱһҪĻraw materialA ͬʱԴ˵ȷǣ
AAΪ
BI2stayѭϵͳ߻
CϷH2HIķֽ
DȼֵߡԴḻ޶Ⱦ桢㣬Դ
6NAΪ٤ֵйȷ( )
A0.1mol KHSO4кH+Ϊ0.1NA
B׼״£5.6L SO3кеĵΪ10NA
C33.6g FeˮȳַӦתƵΪ1.6NA
D5.6g FeͶ100mL3.5molL1УַӦתƵΪ0.3NA
7.ѧӦʵ˵ȷǣ
ٺʱѹǿѧӦһfast
䣬¶ԽߣѧӦԽfast
ʹô߻ɸı䷴ӦʣӶı÷Ӧų
3 molL1s1ķӦһ1 molL1s1ķӦʴ
¶ʹѧӦmainreason ˷Ӧлӵİٷ
μĻѧӦѹǿСӦӻӵİٷ
ӶӦ
ӦŨȣӵİٷӶʹλtimeЧײ࣬
߻뷴ӦܽͻӵİٷӶӦ
A ڢݢ B ڢޢ C ڢۢݢߢ D ٢ڢܢݢ
8б仯һcauseƽƶ
ٷӦʵı仯Ũȵı仯۸ְٷֺı仯ƽԷı仯
ɫı仯޻ܶȵı仯תʵı仯¶ȵı仯
A٢ڢޢߢ Bڢܢݢޢ Cڢܢݢޢ Dۢܢߢ
9ӦmAsnBg pCg H0stayһ¶£ƽʱBѹǿ仯Ĺϵͼʾȷǣ
m+n

¶ȣ÷Ӧƽⳣ
A٢ڢ Bڢ Cۢ D٢ۢ
10ϩֱˮϷӦƱҴC2H4(g)+H2O(g) C2H5OH(g)ϩƽת¶ȡѹǿı仯ϵ£ʼʱn(H2O)n(C2H4)1molΪ1Lзȷǣ
AϩֱˮϷӦH0
BIn the pictureѹǿĴСϵΪp1p2p3
CIn the pictureaӦƽⳣK
Dﵽƽ״̬abҪtimeab
11stayFe2(SO4)3ҺУa gͭȫܽټb gַӦõc g壬a>c˵ȷǣ
AõҺһCu2+ Bͭ
Ctake뵽ϡH2SO4Уݲ DõҺprobablyFe3+
12ķͼʾexpressϢӦǣ

Aͼ1expressH2O2Ӧе仯H2ȼΪ241.8kJmol1
Bͼ2expressijȷӦֱstayС޴߻·Ӧе仯
Cͼ3expressһH2Cl2HClķӦ;޹أH1=H2+H3
Dͼ4expressѹǿԿ淴Ӧ2A(g)+2B(g) 3C(g)+D(s)Ӱ죬ҵǿ
13Ԫȷǣ
ԪصԭֻлԭԣֻԣԪstayۣԭһõʣԪزprobablystayУԪ̬Ϊ̬һԭ
A٢ Bڢ C٢ܢ Dڢۢ
14ʵװãּгװȥӦʵ飬ܴﵽʵĵǣ

A.װIеձI2Fe Bװâϳɰ鰱
CװâƱ Dװâȡ

15ѧϳһµĻṹʽͼʾWXYZMǶгԪأԭ֪MԪԭӺYԪԭӺ˵ȷǣ
A. XYԪsureɶֻ
B. WZԪλڱͬһ
C. Ԫصķǿ˳ΪZYX
D. YMԪص̬⻯ߵΪYM
16ijҺprobablySiO23OHCO23SO24AlO2HCO3Na+Fe3+Mg2+Al3+ӡҺμһʵŨȵҺʱfind ɳʵҺ仯ͼʾ˵ȷǣ
AԭҺһ
BӦγɵҺеֻ
CԭҺкCO23AlO2ʵ֮Ϊ12
DԭҺһеǣOHCO23SiO23AlO2
choice⣨5С⹲52)
1712֣50mL1.0 mol/L50mL1.1mol/LNaOHҺstayͼʾװýкͷӦͨⶨӦųкlegȡanswerproblem
1ʵװIn the pictureȱٵһֲ____________
2ձӲֽ壬粻Ӳֽ壬кlegֵӰ______(ƫߡƫ͡Ӱ족)
3if60 mL 1.0 molL150 mL 1.1 molL1NaOHҺзӦʵȣ_______(ӡ١򡰲䡱)reason______________кlegֵ________(ӡ١򡰲䡱)reason______________
4ͬŨȺİˮNaOHҺʵ飬õкlegȵֵ________(ƫ󡱡ƫСӰ족)
5 ܡ򡰲ܡBa(OH)2Һreason


18(10)(1)ӦFe(s)CO2(g) FeO(s)CO(g)H1ƽⳣΪK1
ӦFe(s)H2O(g) FeO(s)H2(g)H2ƽⳣΪK2
700 900
K1 1.47 2.15
K2 2.38 1.67
stayͬ¶ʱK1K2ֵ±ӦCO2(g)H2(g) CO(g)H2O(g)HƽⳣΪKH________(æH1ͦH2express)K________(K1K2express)֪ӦCO2(g)H2(g) CO(g)H2O(g) ______Ӧ(ȡ򡰷ȡ)
(2)һ¶£ijܱм۲һCO2壬ӦFe(s)CO2(g) FeO(s)CO(g)H0CO2ŨtimeĹϵͼʾ
700 900
K1 1.47 2.15
K2 2.38 1.67

CO2ʼŨΪ2.0 molL1ƽʱCO2ŨΪ________ molL1
дʩʹƽʱC(CO)C(CO2)ֵ________(ĸ)
A¶ Bѹǿ CһCO2 Dټһ
1910֣һҪĻraw materialͨúϳɰϳIJ
(1)֪ϳɰȻѧʽΪN2g+ 3H2g 2NH3gH=92kJ/molʵ֮13ıֱstay200桢400桢600½зӦNH3ʵѹǿı仯ͼʾ
cӦ¶Ϊ___________ 档
˵ȷ___________
AŨȣɴٽƽƶ ߵת
BQӦķӦ£ʵĴ߻NH3stayеʵ
CMNƽⳣСϵΪMN
DMNQӦʵĴСϵΪNMQ
2[CO(NH2)2]һveryҪĸЧʣҵNH3CO2Ϊraw materialأ÷ӦʵΪӦ
һ2NH3(g)CO2(g)=H2NCOONH4(s) H=272kJmol1
H2NCOONH4(s)=CO(NH2)2(s)H2O(g) H=138kJmol1
дҵNH3CO2Ϊraw materialϳصȻѧʽ
3automobileͨΪΪ޶ȾijʣдNOӦĻѧʽ 1molȫӦʱתƵӵĿΪ___________

20(10)COںϳɼ״ѧʽΪ
CO(g)2H2(g) CH3OH(g)
(1)ͼ1ǷӦCO(g)2H2(g) CH3OH(g)stayͬ¶COתtime仯ߡ
ٸ÷Ӧʱ䦤H________(><򡰣)0
stayT1¶£Ϊ1 LܱУ1 mol CO2 mol H2COCH3OH(g)Ũtime仯ͼ2ʾ÷ӦƽⳣΪ______________
ݻ䣬дʩCOתʵ________(ĸ)
a¶ btakeCH3OH(g)ϵз
cʹúʵĴ߻ dHeʹϵѹǿ
(2)stayݻΪ1 LĺУֱResearchstay230 桢250 270 ¶ºϳɼ״Ĺɡͼ3¶H2COʼɱ(ʼʱCOʵΪ1 mol)COƽתʵĹϵzӦ¶________棻abcӦĻѧƽⳣֱΪK1K2K3K1K2K3ĴСϵΪ____________________

2110֣ͻԭstaylifeй㷺ʹá
(1)غҺsure·Ӧ2KMnO4+16HBr=2KBr+2MnBr2+8H2O+5Br2
amongԭʵ֮Ϊ
0.1molĻԭʵΪ_________תƵӵĿ____
(2)ԪFe2+Fe3+stayijƬк΢ĿϸСĻԭۣthisЩstayθeffectתΣӦӷʽΪ Fe2+stayºeasilyед÷Ӧӷʽ
(3)(H3PO2)һ־ϸproductһԪǿᣬ뷽ʽǣH3PO2 H++H2PO2haveǿԭԣanswerproblem
H3PO2УԪصĻϼΪ_________
H3PO2NaH2PO2takeҺеӻԭΪʣӶڻѧ(H3PO2)лѧӦУ뻹ԭʵ֮Ϊ4U1Ϊ_________
APH3 BH3PO3 CP DH3PO4
NaH2PO2 λʽΣ








2020-2021ѧѧǦɽһкѧ߶θİ
ѧο

1-4 B D B A 5-8 D C A D
9-12 B B A C 13-16 D C B D

1712֣ע⣬ÿ1֣

2ƫ͡
3ӣ 0.055molH2Oǰֻ0.050molH2O2֣ 䣻
кlegָкͷӦ1molH2Oʱų޹ء2֣
4ƫС
5 becauseBa(OH)2ҺӦBaSO4ȻӰ췴ӦķӦȣ2֣

1810֣ÿ2֣
(1 )H1H2 K1K2
(2) 0.67(2/3) A

1910֣ÿ2֣
1 600 ACD
22NH3(g)CO2(g)=CO(NH2)2(s)H2O(g) H=134kJmol1
32CO(NH2)2 + 6NO = 2CO2 + 5N2 + 4H2O 6NA

2010֣ÿ2֣
(1) < 12 b
(2) 270 K1K2
21. 10֣ע⣬ÿ1֣

151 0.5mol 0.5NA

22H++Fe=Fe2++H22֣ 4Fe2++O2+4H+=4Fe3++2H2O2֣

3 +1 D 

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