Calculate the mass, in kilograms, of 4.35 mol of gold (III) sulfate. (For the final answer, must do significant figures) *This is a grade 11 practice question please do not answer in a level above that because I won't understand.*
*Also please show me the solutions so that I know how you got the answer*

first one is the correct answer

answer7,570,82356 cubic centimeters;

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o