XX年扬州市中考化学试题(带答案和解释).pdf

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1、c XXxxH1c12o16g24S32c135.5ca40cu64 A.B.c.D.AABcDAo194%cuSo4o cuSo4 A.B.c.D.cuSo4CA.B.c.D.AABcDAoA.B.c.D.cABcDc o1821 A.B.c.D.DDoA.H2B.N2c.o2D.So2AAH2BN2co2D So2AoA.B.c.D.BXX21 A.AB.Bc.cD.D BABcDBoA.N2B.co2c.So2D.H2o cPH5.6AN2Bco2pH5.6 cSo2pH5.6DpH=7c o0.A.cuSo4Na2So4B.clNacl c.caco3Na2co3D.NH4No3 Na

2、No3 BAB cDBo1.A.2g+o22goB.cu+2AgNo3=cu2+2Ag c.cucl2cu+cl2 T D.2NaoH+H2So4=Na2So4+2H2o DA2g+o22goBcu+2AgNo3=cu2+2Agccucl2cu+cl2 T,D2NaoH+H2So4=Na2So4+2H2oDo3cl2+6NaoH5Nacl+X+3H2oX A.o2B.Nacloc.HclD.Naclo3 D3cl2+6NaoH5Nacl+X+3H2o 66665530X113XNaclo3Do3.A.o3B.4:3c.9D.24%BA03B.43cD.=X 100%=31.6%Bo A.B.9

3、c.D.DAB44cDDoA.B.pHpHc.D.cAco2HclNaoHBpHpHpH pH.pHCDC oA.B.c.D.ADABc DADbNo3clA.No3clB.tlclt2c.Acl29%D.No3No3DABt1clt2cAcl29g=x 100%=22.5%DNo3No3Do=x 100%oA.B.c.D.AAB CDAopHA.B.AHclNacl c.D.BcABApH7HclNaclcDBe0.cuocu2o14.4gcuo B4.4gA.BH2oco2B.21coc.cu2oD.ccuoBAAcuo cocu+co2 cu2o co2cu+co2Bco2B21cocB

4、4.4g 4.4g xx 100%x=1.6g=21cu2o cucuocu2ocucuoDcBcuoB1.co2：X cH4 cH48gcH432go2agco2 18gH2oU a=_ cH4NH3.N2.caco3+2Hcl cacl2+H2o+co2 T.22.co2+2NH3co2+H2o X N2caco3+2Hcl cacl2+H2o+co2 f;cH48gcH432go2agco218gH2o,U a=8g+32g 18g=22g NH3co2+2NH3co2+H2o 2.I.100g/g/g/g/gc/g6.70.9787136 0.05 0.60.328370.411 3.

5、13.56120900.11 A.B.c.A.B.c.cc6H8o6cHoccII.2019 A.B.c.3.c.ca.A.176.9 112.A .CcaAcc6H8o6=12 X 6+1 x 8+16X 6=176cHo=9112cc A33.I.no4o2 _，II.13240333：12379.779.679.988.489.087.990.089.889.3 79.788.489.7 123100%.2no42no4+no2+o2 T.3Fe+2o2Fe3o4.co2.Fe+2Hcl Fecl2+H2 T 2no42no4+no2+o2 f;1223 100%3Fe+2o2Fe3o4

6、 co2Fe+2HclFecl2+H2 I.6ZnFe2o46Zno+4Fe3o4+o2 T 3Zno+2Fe3o4+H2o3ZnFe2o4+HZnFe2o4FeH2o2n.g2cugcu 2g2cu+3H23gH2+gcu2g2cugH2H21.2gg2cuH2g.+3.H2O.1 8.gH2+2H2o=g2 J+2H2T.0.7g ZnFe2o4+22Fex+2x+X 4=0 x=+3gcug2cu gH2H2,gH2+2H2o=g2j+2H2f;1.2gg2cugcu211.2gg2cuxgcu2yg2cu+3H23gH2+gcu2 2478152 1.2gx x=3.9g y=7.6g

7、 3.9gagH2+2H2o=g2j+2H2f 4 9ga a=0.6g 6ggcu2 7.6g xx 100%=1.2g 1.2gbg+2Hcl gcl2+H2 f 2 2gb b=0.1g 0.6g+0.1g=0.7g cacl2NH4clXRDXRDNH3cacl2CNH4clcacl2NH4clFe2o3coa.14.7000gcacl2.b.Na2co3c.lO.OOOOg x bx.ca2cao+H2o.85.5.Fecl3.Fe2o3+3co2Fe+3co2.2.cacl2XRDca2cao+H2o85.5 Ccacl285.5 CNH4cl cacl2NH4clFecl3 Fe2o3+3co2Fe+3co2 10.00gycacl2+Na2co3 caco3 J+2Nacl 11100 y10g y=11.1g=14.7000g 11.1g=3.6g 14.7g xx 100%=3.6gx=2 x