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Andreyy89
3 years ago
13

The equilibrium fraction of lattice sites that are vacant in silver (ag) at 500°c is 0.5 x 10-6. calculate the number of vacanci

es (per meter cubed
Chemistry
1 answer:
Sedbober [7]3 years ago
6 0
Missing question: <span>Assume a density of 10.35 g/cm3 for Ag, A(Ag) = 107.87 g/mol.
N(Ag) = Na </span>· d(Ag) ÷ A(Ag).
N(Ag) = 6,023·10²³ atoms/mol · 10,35 g/cm³ · 10⁶ cm³/m³ ÷ 107,87 g/mol.
N(Ag) = 5,78·10²⁸ atoms/mol.
Nv = 5,78·10²⁸ atoms/mol · 5·10⁻⁵.
Nv = 2,89·10²².
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3 years ago
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Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange
SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)

A

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate  = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

=\frac{5.75 g}{10.0 g}\times 100=57.5\%

57.5% was the percent yield.

C

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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<h3><u>Explanation</u>:</h3>

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