Calculate the amount of Substance in each of the following
(a)120g of Trioxocarbonate(IV)ion CO3.
(b) 31.5g of trioxocarbonate(V) acid HNO3.
(C)123g of copper(II) trioxocarbonate(IV).
Relative atomic mass
C=12
O=16
N=17
Cu=63
4 answers
I really need the answer urgently please
CO3: C = 12 O3 = 48
so C = 12/(12+48) * 120g = 24g
and so on
so C = 12/(12+48) * 120g = 24g
and so on
Thanks oobleck
I need the answer to B and C too
I need the answer to B and C too
Ronald, I don't get it.
First, the IUPAC names are carbonate ion for A, carbonic acid for B (which btw, is H2CO3 and not HNO3) and C is copper(II) carbonate.
If I read the problem right, there are 123 g CuCO3 in 123 g of C. There are 31.5 g H2CO3 in 31.5 g H2CO3 in B and there are 120 g carbonate ion in A.
Surely that isn't the question.
Second, oobleck showed you how to solve one of the three. The others are done the same way and I'm sure your calculator works the same way as his, assuming that he has interpreted the problem correctly.
First, the IUPAC names are carbonate ion for A, carbonic acid for B (which btw, is H2CO3 and not HNO3) and C is copper(II) carbonate.
If I read the problem right, there are 123 g CuCO3 in 123 g of C. There are 31.5 g H2CO3 in 31.5 g H2CO3 in B and there are 120 g carbonate ion in A.
Surely that isn't the question.
Second, oobleck showed you how to solve one of the three. The others are done the same way and I'm sure your calculator works the same way as his, assuming that he has interpreted the problem correctly.
0 answers