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"empirical formulas" p 889(153e)Asked by rose
empirical formulas"
p 889(153e)
p 889(153e)
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Answered by
DrBob222
This doesn't tell me much. We don't know what course you're taking, much less the book being used.
Answered by
rose
chemistry, my textbook is modern chemistry
i found the actual problem:
2.8% hydrogen, 9.8% nitrogen,20.5% nickel, 44.5% oxygen, and 22.4% sulfur
determine empirical formula for compounds
i found the actual problem:
2.8% hydrogen, 9.8% nitrogen,20.5% nickel, 44.5% oxygen, and 22.4% sulfur
determine empirical formula for compounds
Answered by
DrBob222
You do this the same way I've show you in the last few days to do them.
Take 100 g sample which gives you
2.8 g H.
9.8 g N
20.5g Ni
44.5 O
22.4 S.
Convert each to mols by mols = grams/atomic mass.
Then find the ratio of the elements to each other. The easy way is to divide the smallest number by itself then divide all of the other numbers by the same small number. Post your work if you get stuck.
Take 100 g sample which gives you
2.8 g H.
9.8 g N
20.5g Ni
44.5 O
22.4 S.
Convert each to mols by mols = grams/atomic mass.
Then find the ratio of the elements to each other. The easy way is to divide the smallest number by itself then divide all of the other numbers by the same small number. Post your work if you get stuck.
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