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Yuri [45]
3 years ago
7

How many grams are in 9.97 moles of Be(NO3)2? Use two digits past the decimal for all values.

Chemistry
1 answer:
Simora [160]3 years ago
4 0

Answer:

1,869.97 grams of Be(NO3)2

Explanation:

Be(NO3)2 = Be  N2  O6

Be=9.012182g/mole

N2=28.0134g/mole

O6=96g/mole

therefore Be(NO3)2 gives you 187.56g in one mole

so 9.97 moles means there is 9.97 times more

9.97mole Be(NO3)2 * 187.56g Be(NO3)2/1mole Be(NO3)2 = 1,869.97g of Be(NO3)2

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What is the mass, in grams, of 7.20×1020 molecules of caffeine, c8h10n4o2?
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3 0
3 years ago
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If 143.56 mL of 0.6653 M ammonium carbonate reacts with 175.37 mL of 0.8732 M chromium(III) sulfate in a double replacement reac
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Answer:

<h3>The answer is 0.75 g/mL</h3>

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density =  \frac{mass}{volume}  \\

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Hope this helps you

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