Asked by Anthony
If you are given 718 grams of lead (iv) oxalate, how many atoms of oxygen do you have?
Answers
Answered by
Devron
Pb(C2O4)2
718g of Pb(C2O4)2 *(1 mole of Pb(C2O4)2/383.2g)= moles of Pb(C2O4)2
moles of Pb(C2O4)2 *(8 moles of O/1 mole of Pb(C2O4)2)= moles of O
Moles of O*(6.02 x 10^23 atoms/1 mole of O)= atoms of O
718g of Pb(C2O4)2 *(1 mole of Pb(C2O4)2/383.2g)= moles of Pb(C2O4)2
moles of Pb(C2O4)2 *(8 moles of O/1 mole of Pb(C2O4)2)= moles of O
Moles of O*(6.02 x 10^23 atoms/1 mole of O)= atoms of O
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