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3 years ago

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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3 years ago

What are the chemical symbols for the two elements found in iron oxide

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These are the two chemical symbols for the two elements found in Iron Oxide.

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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:

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

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.

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.

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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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

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3 years ago