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在273K、1.01×105Pa条件下,将1.40g氮气、1.60g氧气和4.00g氩气混合.该混合气体的体积是()A.3.36LB.6.72LC.8.96LD.4.48L

题目详情
在273K、1.01×105Pa条件下,将1.40g氮气、1.60g氧气和4.00g氩气混合.该混合气体的体积是(  )
A. 3.36L
B. 6.72 L
C. 8.96 L
D. 4.48 L
5



▼优质解答
答案和解析
n(N22)=
1.40g
28g/mol
=0.05mol;
n(O2)=
1.60g
32g/mol
=0.05mol;
n(Ar)=
4.00g
40g/mol
=0.1mol,
气体的总物质的量为0.2mol,则总体积为0.2mol×22.4L/mol=4.48L.
故选:D.
1.40g
28g/mol
1.40g1.40g1.40g28g/mol28g/mol28g/mol=0.05mol;
n(O22)=
1.60g
32g/mol
=0.05mol;
n(Ar)=
4.00g
40g/mol
=0.1mol,
气体的总物质的量为0.2mol,则总体积为0.2mol×22.4L/mol=4.48L.
故选:D.
1.60g
32g/mol
1.60g1.60g1.60g32g/mol32g/mol32g/mol=0.05mol;
n(Ar)=
4.00g
40g/mol
=0.1mol,
气体的总物质的量为0.2mol,则总体积为0.2mol×22.4L/mol=4.48L.
故选:D.
4.00g
40g/mol
4.00g4.00g4.00g40g/mol40g/mol40g/mol=0.1mol,
气体的总物质的量为0.2mol,则总体积为0.2mol×22.4L/mol=4.48L.
故选:D.