江苏省泰州第二中学化学组

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阿伏加德罗定律. 江苏省泰州第二中学化学组. 戴 连 久. 阿伏加德罗定律. 在相同的温度和压强下, 相同体积的任何气体 都含有相同数目的分子. 二、阿伏加德罗定律的推论. 同温同压下: V 1 =V 2 即 N 1 =N 2. n 1 =n 2. 推论 1. 同温同压下, 任何气体的体积与物质的量、 分子数成正比. 即: V 1 ∶V 2 = n 1 ∶ n 2 = N 1 : N 2. 练习:. 1:1. 在标准状况下, 22.4LCO 和 17gNH 3 的体积之比为 - PowerPoint PPT Presentation

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  • V1=V2 N1=N2 V1V2 = n1 n2 = N1 : N2n1=n2122.4LCO17gNH3 :2gH2 3.01x1023CO20.8molHCl1.12LCl2 1:11:11:2

  • m1m2=M 1M 22V1=V2 N1=N2 n1=n2=m1M1m2M2=m1M1m2M2n1=n212.580g/mol

  • V1V2=M 2M 1m1=m2 3n1xM1=n2xM2V1VmV2Vm xM1= x M2V1V2=M2M1 CO2H2Cl2HClSO2

  • 4 12=M 1M 2V1V2n1n2==m1m2m1m212M1M2=12M1M229 CO2H2Cl2HClN2

  • 5 p1p2=n1n2

    67V1 V2 =T1 T2V1 V2 =P2P1

  • mMMXXNANAVVmXVmX n N

  • 118.5g1717gmol

    V(NH3)=22.4Lmol0.5mol=11.2L

    8.5g11.2Lm(NH3)M(NH3)8.5g17g/moln(NH3)= = =0.5mol

  • 1.40gN21.60gO24.00gArn(N2) =M(N2) m(N2) ==1.40g28g/mol0.0500moln(O2) =M(O2) m(O2) ==1.60g32 g/mol0.0500moln(Ar)+M(Ar) m(Ar) == 4.00g40g/mol0.100moln()=n(Ar) =n(O2)+n(Ar) =0.0500mol0.0500mol0.100mol++=0.200molV=n( Vm=0.200molx22.4L/mol=4.48L

  • 20.2L 0.25g 2 M=m/nn=V/Vm28

  • 1.92g672mln=672ml=0.672L0.672L22.4L/mol=0.03molM=m/n=1.92g 0.03mol=64g/mol64g/mol

  • H=1C=12O=164gH222.4LCO2___1.5molH2CO2 _____g100mL0.179gH2 CO2 664040g/mol

  • 3 23.36LHClZnZnxHCly

    0.15molZn0.30molHCl

  • 1 [ ]A 22.4L/molB22.4L 6.021023C0101kPa5.6LNH3 6.021023D100101kPa 22.4L18gB C

  • 22XY2g 1Y2g2 [ ] AX3Y BX3Y2 CX2Y3 DXY3322.4L [ ] A1molH2Sg B0101kPa28gCO C1molH2O D64gSO2DB

  • 516.5gX 12gO2 X [ ] A44 B16 C28.25 D8840.2mol N2ONO2 [ ] A B C DCDA

  • 6VL H2NH2 [ ] AVN/22.4 B22.4N/V C22.4VN D22.4V/N

    BC79.6gSO2O2 4.48L SO2O2 [ ] A21 B12 C11 D

  • 9 [ ] A B5L H2S5L NH3 34 CN2CO D 8AH2BNH3 [ ] A31 B12 C13 D21

    DAD

  • 10W 18W [ ] A63 B36 C126 D2521154g CO260.6g CO [ ] A60.6g B58.2g C56.2g D48.4gAB

  • 12NA,[ ] A18g8NA B22.4L 2NA CNA D12g 0.1NA 13 X20.17g 100mL ________ _____ O2____B3838g/mol1.7g/L

  • 15agb NA cg____ 141.293g/L ____ 21 25 ____16agb ____ 29g/molC4H1022.4bcaNALa4bmol-1

  • 1730gCOCO2 17.6L

  • 1730gCOCO2 17.6L CO8gCO222g CO6.4LCO211.2LMCO =28g/mol MCO2=44g/molCOxgCO230-xgx=8g mCO2=30g-8g=22gVCO2=17.6L-6.4L=11.2L