题目内容
(1)0.3molNH3中所含质子数与______gH2O分子中所含质子数相等.
(2)100mLAl2(SO4)3溶液中c(SO42-)=0.9mol/L,则n[Al2(SO4)3]=______.
(3)10mL6mol/L稀硫酸加水稀释到200mL,稀释后c(H+)=______.
(4)已知16gA和20gB恰好完全反应生成0.04molC和31.76gD,则C的摩尔质量为______.
(2)100mLAl2(SO4)3溶液中c(SO42-)=0.9mol/L,则n[Al2(SO4)3]=______.
(3)10mL6mol/L稀硫酸加水稀释到200mL,稀释后c(H+)=______.
(4)已知16gA和20gB恰好完全反应生成0.04molC和31.76gD,则C的摩尔质量为______.
(1)0.3mol NH3含有的质子数的物质的量为0.3mol×(7+3)=3mol,而1molH2O含有(2×1+8)mol=10mol质子,则应有0.3molH2O,质量为m(H2O)=0.3mol×18g/mol=5.4g,
故答案为:5.4;
(2)100mL Al2(SO4)3溶液中c(SO42-)=0.9mol/L,则n(SO42-)=0.1L×0.9mol/L=0.09mol,
n[Al2(SO4)3]=
n(SO42-)=0.03mol,
故答案为:0.03mol;
(3)10mL 6mol/L 稀硫酸加水稀释到200mL,稀释后c(H+)=
×2=0.6mol/L,
故答案为:0.6mol/L;
(4)16g A和20g B 恰好完全反应生成0.04mol C 和31.76g D,由质量守恒可知m(C)=16g+20g-31.76g=4.24g,
M(C)=
=106g/mol,
故答案为:106g/mol.
故答案为:5.4;
(2)100mL Al2(SO4)3溶液中c(SO42-)=0.9mol/L,则n(SO42-)=0.1L×0.9mol/L=0.09mol,
n[Al2(SO4)3]=
1 |
3 |
故答案为:0.03mol;
(3)10mL 6mol/L 稀硫酸加水稀释到200mL,稀释后c(H+)=
0.01L×6mol/L |
0.2L |
故答案为:0.6mol/L;
(4)16g A和20g B 恰好完全反应生成0.04mol C 和31.76g D,由质量守恒可知m(C)=16g+20g-31.76g=4.24g,
M(C)=
4.24g |
0.04mol |
故答案为:106g/mol.
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