yunginjay817 yunginjay817
  • 16-10-2020
  • Chemistry
contestada

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

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whited7188 whited7188
  • 16-10-2020

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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