Lets say you have 25 ml of a 3.25M copper(II) chlorate solution. How many moles of copper (II) chlorate would be in the solution and how many moles of the copper(II) ion would be in the solution?

Answer 1

Copper(II) chlorate has the formula of Cu(ClO3)2.

mols Cu(ClO3)2 = M x L = ?
mols Cu(II) ion = mols Cu(ClO3)2 since there is 1 mol Cu ions in 1 mol Cu(ClO3)2.

The problem doesn't ask but how many mols chlorate (ClO3)^- do you have. You will have twice as many as there are mols Cu(ClO3)2 since there are two chlorataes for each copper chlorate.

Answer 2

wewwo