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mojhsa [17]
3 years ago
6

What is the empirical formula for a compound that contains 1.18 mol na, 1.18 mol n, and 3.53 mol o?

Chemistry
1 answer:
nikitadnepr [17]3 years ago
6 0
Answer is: NaNO₃.
n(Na) = 1,18mol.
n(N) = 1,18mol.
n(O) = 3,53mol.
n(Na) : n(N) : n(O) = 1,18mol : 1,18mol : 3,53mol / ÷1,18
n(Na) : n(N) : n(O) = 1 : 1 : 3.
Empirical formula is NaNO₃ (sodium-nitrate).
Empirical formula<span> of </span>compound<span> is the simplest </span>ratio<span> of </span>atoms<span> present in a compound.</span>

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3 0
3 years ago
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1.428 moles

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If 0.0714 moles of N2 gas occupies 1.25 L space,

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8 0
3 years ago
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