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3 years ago

What is an empirical formula for a compound that is 82.2% nitrogen and 17.8% hydrogen

1 answer:

8 0

Percent to mass
Mass to mole
Divide by small
Multiply 'til whole

Assume 100 g of substance, then 82.2 g N and 17.8 g H.

82.2 g N * (1 mol/14.01 g N) = 5.86 mol N
17.8 g H * (1 mol/1.01 g H) = 17.6 mol H

Divide by the smallest mole
17.6 H / 17.6 = 1 H
5.86 N / 17.6 = 1/3 N

Multiply to make whole number ( x3 in this case)
3 x 1 H = 3H
3 x 1/3 N = 1 N

NH3

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When glucose is fermented, ethanol is formed together with ………………….

levacccp [35]

Answer:

carbon dioxide and oxygen

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3 years ago

For the following reaction, 4.07 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 10.4 grams of

torisob [31]

Answer:

Theoretical yield = 13.7 g

% yield =76 %

Explanation:

For $Al_2O_3$

Mass of $Al_2O_3$ = 4.07 g

Molar mass of $Al_2O_3$ = 101.96 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{4.07\ g}{101.96\ g/mol}$

$Moles\ of\ Al_2O_3= 0.0399\ mol$

According to the reaction:

$Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O$

1 mole of $Al_2O_3$ on reaction produces 1 mole of $Al_2(SO_4)_3$

So,

0.0399 mole of $Al_2O_3$ on reaction produces 0.0399 mole of $Al_2(SO_4)_3$

Moles of $Al_2(SO_4)_3$ obtained = 0.0399 mole

Molar mass of $Al_2(SO_4)_3$ = 342.2 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0399= \frac{Mass}{342.2\ g/mol}$

$Mass= 13.7\ g$

<u>Theoretical yield = 13.7 g</u>

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

$\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100$

Given , Values from the question:-

Theoretical yield = 13.7 g

Experimental yield = 10.4 g

Applying the values in the above expression as:-

$\%\ yield =\frac{10.4}{13.7}\times 100$

<u>% yield =76 %</u>

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Answer:

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Explanation:

hope this helps have a good rest of your day :) ❤

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Alex17521 [72]

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3 years ago

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