H2CO3 - 2 И 3 АТОМЫ 1) Mr вещества; 2) найти массовую долю химических элементов в веществе; 3) решить задачу: найти массу 0.4 моль H2CO3.
1)
Mr = 1*2+12+16*3 = 62
2)
w(H2) = 2/62 =3%
w(C) = 12/62 = 19,3%
w(O3) = 48/62 = 77,7%
3)
n = m/M
M=Mr
m = nM = 0.4 * 62 = 24.8
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Answers & Comments
1)
Mr = 1*2+12+16*3 = 62
2)
w(H2) = 2/62 =3%
w(C) = 12/62 = 19,3%
w(O3) = 48/62 = 77,7%
3)
n = m/M
M=Mr
m = nM = 0.4 * 62 = 24.8