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3 years ago

How many grams of NH3 are needed to provide the same number of molecules as in 0.55 grams of SF6?

Chemistry

1 answer:

xenn [34]3 years ago

4 0

Above it says the molecular weights are

NH3- 17g/mol and SF6-146 g/mol

Well 1 mole of SF6 is 146.048 grams (i added hte atomic masses of each element). So then the number of moles in 0.85 grams would be 0.00582000438 moles.

= 1mole / 146.048g * 0.85g

so we would need 0.00582000438 moles of NH3 to have the same number of molecules.

One mole of NH3 is 17.030519999989988 grams (i added each atoms mass). so 0.00582000438 moles of NH3 would be:

= 17.030519999989988 g / mole * 0.00582000438moles

that equals 0.09911770099 grams.

so 0.09911770099 grams is the answer if you rounded, you get about 0.1 grams

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A compound containing chromium, Cr; chlorine, Cl; and oxygen, O, is analyzed and found to be 33.6% chromium, 45.8% chlorine, and

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Answer

The empirical formula is CrO₂Cl₂

Explanation:

Empirical formula is the simplest whole number ratio of an atom present in a compound.

The compound contain, Chromium=33.6%

Chlorine=45.8%

Oxygen=20.6%

And the molar mass of Chromium(Cr)=51.996 g mol.

Chlorine containing molar mass (Cl)= 35.45 g mol.

Oxygen containing molar mass (O)=15.999 g mol.

Step-1

Then,we will get,

Cr= $\frac{1}{51.996} \times33.6=0.64$ mol

Cl= $\frac{1}{35.45} \times45.8=1.29$ mol.

O= $\frac{1}{15.99} \times=1.28$ mol.

Step-2

Divide the mole value with the smallest number of mole, we will get,

Cr= $\frac{0.64}{0.64} =1$

Cl= $\frac{1.29}{0.64} =2$

O= $\frac{1.28}{0.64} =2$

Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)

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