Answer:
cerium (iii) sulfate is less soluble
Answer:

Explanation:
First, find the mass of empirical formula, CH. 12.01 g/mol is for carbon, and 1.008 g/mol is for hydrogen. 12.01+1.008=13.018 G/mol CH. Divide 78.110 G/mol by 13.018 g/mol. You get approximately 6. Multiply that by the subscript of each element. 6(CH)=

Answer:
The answer to your question is: 65.9 g released of CO2
Explanation:
MW CO2 = 44 g
MW CuCO3 = 123.5 g
CO2 released = ?
CuCO3 = 185 g
CuCO3 ⇒ CO2 + CuO
123.5 ----------- 44g
185 g ----------- x
x = (185 x 44) / 123.5
x = 65.9 g released of CO2
The answer & explanation for this question is given in the attachment below.