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wolverine [178]
3 years ago
13

The ____ are new substances that are created following a chemical reaction. *

Chemistry
1 answer:
KonstantinChe [14]3 years ago
8 0

Answer:

b

Explanation:

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The monoprotic acid from among the following is
oksian1 [2.3K]

Monoprotic acid are acids having only one hydrogen atoms after dissociation into ions from its compound. The monoprotic acid from among the following is HCl. The answer is letter D. HCl → H+ + Cl-. Note that there is only one H+ ion upon dissociation.

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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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3 years ago
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An element bonds chemically to fill it's outer valence electron shell, when that occurs it becomes stable.
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saveliy_v [14]

A displacement reaction will occur from the system given above. The chlorine molecules will displace the bromide ions in the solution of sodium bromide. The reaction will yield to sodium chloride and bromine. The reaction will be:

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