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fiasKO [112]
3 years ago
5

Two samples of a compound containing elements a and b are decomposed. the first sample produces 15 g of a and 35 g of

Chemistry
1 answer:
Alborosie3 years ago
3 0

According to law of definite proportion, for a compound, elements always combine in fixed ratio by mass.

The formula of compound remains the same, let it be a_{x}b_{y} where, a and b are two different elements.

Since, the ratio of mass remains the same , calculate the ratio of masses of element a and b in both cases

\frac{a}{b}=\frac{15}{35}=\frac{10}{y}

rearranging,

y=\frac{10\times 35}{15}=23.3

Thus, mass of b produced will be 23.3 g.

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What really is the Ebola virus?
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3 years ago
What are the chemical symbols for the two elements found in iron oxide
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For Iron: Fe

For Oxygen: O

These are the two chemical symbols for the two elements found in Iron Oxide.
7 0
3 years ago
Read 2 more answers
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.

8 0
3 years ago
Assume that the pressure inside of the 1 liter balloon is 200,000 Pa, what is the explosive energy in the inside the balloon? Ex
AysviL [449]

Answer:

200 Joules is the explosive energy in the inside the balloon. And that is  9.523\times 10^{-5}1 lb of TNT.

Explanation:

Pressure=\frac{force}{Area}=\frac{Force\times distance}{Area \times distance}=\frac{Energy}{Volume}

Volume of the balloon = V = 1 L = 0.001 m^3

Pressure inside the balloon ,P= 200,000 Pa =200,000 N/m^2

Explosive energy in the inside the balloon be E.

E = Pressure × Volume

E=200,000 N/m^2\times 0.001 m^3=200 Joules

1 lb of TNT = 2.1\times 10^6 J

200 Joules = 200\times \frac{1}{2.1\times 10^6 }1 lb of TNT

= 9.523\times 10^{-5}1 lb of TNT

5 0
3 years ago
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