普化无机试卷（气体一）答案.docx

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1、普化无机试卷（气体一）答案一、选择题1010 0101 t z(x7 Xl/ 7 7 x)/ JZ l JZ x)/ 7 7 KJZ JZ VUZ )z JZ )z 7A ”“Cbccbbcbbccdcddbdc CBB A A B z( /k z( /( /( l z( /V /Ix z( l z( z( l l z( l z( z(x z(x z(x Zlx Z(N z(x z(x03Xlz JZ Xlz J )x 1 7 7 J JZ 7 1 45683422233 0000221111 (0(0(0(0(0(0(0Ooooo z(x z( Zfx z( Ztxlz J lz )z I7

2、 VIZ Klz l lz )z fz J- HeNeCO2;(2)分子平均能量都相等。74.(0163)1.004三、计算题(共3题155分)75.(0109)pV96.9150IO3x,二-0.00250(mol)RT8.31(427+273)2MX2(g):=2MX(g)+X2(g)1.120g0.720g0.400gX2的摩尔质量为:0.400/0.00250=160(gmol-1)X的相对原子质量为80“MX=2X0.00250=5.00IO3(mol)MX的摩尔质量为:0.720/(5.00IO3)=144(gmol,)M的相对原子质量为6476. (OIlO)n1.OIO9P=R

3、T=X8.31Xl200=1.710,1(kPa)V6.0XlO2377. (Olll)pV=nRT当p,V一定时,随7变，pV1In=X()R7(冬)7(夏)1.00l0611=103()=I0.5103(mol)8.31248314MCHII=I5X16.0=168(kg)78. (0112)2H2OCaC2=Ca(OH)2+C2H2t设反应后得到了nmol的C2H2,即含2molH2O,21.0103所以n=l.00102=8.621O-4(mol)8.3129328.62104l8.0H2O的质量分数=X100%=2.06%1.50879. (0113)TlP2273207“2=d=1

4、.43,=2.76(gdm3)T2Pl290IOI80. (0114)pp2V2(1)膨胀前为状态1,膨胀后为状态2,=T1T2202V1014V=T2=600(K)300T2(2) Ph,=总XH,二:+Ne25.075.0I2.5(mol)Ne=3.71(mol)2.020.212.5=0.77212.5+3.71Ph,=2020.772=156(kPa)81. (0115)PIVZI=P2L1011.01011.0=50.5V2=2.0(dm3)50.5ViV21.0V373(2) =V2=1.0=1.1(dm3)TlT233037333082. (0116)d)pv=pr(不考虑水)(

5、101.3-17.3)1.0=(202.6-17.3)V,V=0.45(dm3)pVpV,(2) =(不考虑水)TT(101.3-17.3)1.0(101.3-1.2)V,=V=0.72(dm3)33028383. (0117)设球的体积为V,当两球都在沸水中时101(2V)=RX373(1)若一球在沸水中，另一球在冰水中,两球平衡时压力为p,在沸水中有/nmolN2,而在冰水中有2molN2贝(:加+2=n1373=/12273p(2V)=mRX373+n2R213=RX373+iRX373=2mRX373(2)(1) 101w1W21373=(1)=(H)=1.18(2) p2m2川227

6、3P=85.6kPa84. (0118)(I)冷却前为状态1,冷却后为状态2,冷却前后空气质量不变。pPiV2TyT2(10l.3-293)10.0(101.3-6.66)V229337329372.05.98 (m3)冷却后气体体积=10.037394.6pV29.3X10.0X1()3(2)冷却前乙醇含量m=94.5(mol)RT8.313736.665.98103冷却后乙醇含量2=16.4(mol)8.31293乙醇凝聚量W=94.516.4=78.1(mol)85. (0119)01194.62初态时:(Cb)=0.0651(mol)35.524.161(SO2)=0.0650(mol

7、)64.0终态时，设有ymolS2起反应则：SO2+Cl2=SO2CI2初态ZZZmol0.06500.06510终态mol0.0650-y0.0651-jy在终态气体总量(0.130y)mol用pV=nRT式求y202X2.00=(0.130一5)义8.31X463202X2.00y=0.130=0.02508.314630.0650-0.0250XSO2=0.3810.130-0.0250Pso,=202X0.381=77.0(kPa)0.0651-0.0250XCI2=0.3810.130-0.0250Pcb=77.0kPa0.0250xSO2Cl20.2380.130-0.0250Ps

8、25=2020238=48.1(kPa)86. (0120)设混合前氢压力为P=nN2+nH2n=pVRT400XV总500X%p%=+RX400RX500R300V总kPh?”=+1 1300按定义力=VNJyH2Ph2%vN2=vH2=300.*.Ph,=300kPa87. (0121)燃烧反应方程式:2C2H6(g)+7O2(g)=6H2O+4CO2(g)pV50.0Ho,=99.2X=2.00(mol)RT8.31298几c,网=6.00/30=0.200mol7C2K完全燃烧时需O?为：X0.200=0.700(mol)2剩余O2为:2.00-0.70=1.30(mol)生成的水：1

9、1h2o=0.600mol生成的CO2:11c2=0.400mol(I)T=573K,H2O为气体,nlRTP=Po2+Ph2o+Pco2=V(0.600+0.400+130)8.31573=219(kPa)50.0(2)T=363K时,如H2O为气体，wH2ORTPh2o=36.2(kPa)比90时饱和水蒸气压要小HzO全部气化。V(0.6+0.4+1.3)8.3IX363P=Ph2O+Pco2+Po2=139(kPa)50.088. (0122)0.475,2c6hjb=3.03IO3(mol)157101X10.0Wair=0.415(mol)8.312933.03IO3p=IOl=0.

10、732(kPa)3.03X10-3+041589. (0129)RT8.31313按理想气体方程式计算：P理=2.17103(kPa)V1.20按VanderWaalS方程式计算：3.65XlO2(p+)(1.20-0.0427尸8.31313p=l.99lO3(kPa)(1.20)2按理想气体状态方程计算和实验值偏离较大,按范德华方程式计算和实验值很接近。90. (0138)经干燥后山的体积为V,按Boyle定律，结合Dalton分压定律：97.2XV=(97.2-2.64)X850V=827cm3在标准状态下的体积Vstp,按理想气体状态方程，结合道尔顿的分压(97.2-2.64)X850101.3XVstp定律：=Vstp=734cm329527391. (0139)先用pV=nRT求总压力PX5.00=(0.850+0.100)8.3l293p=463kPa再用PI=P总Xi求分