Asked by Heather
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?
Answers
Answered by
DrBob222
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.
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.
Answered by
Lucas
wewwo
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