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djyliett [7]
3 years ago
14

PLEASE HELP ME DUE TOMORROW IT REALLY HELPS IF YOU HELP ME THANK YOU SO MUCH IS YOU DO HELP ME!!

Chemistry
1 answer:
aliya0001 [1]3 years ago
3 0

concentration of the contact solution = 3 %

Explanation:

We have the following data:

10.7 g of hydrogen peroxide H₂O₂

355 g of solution

To calculate the concentration of the contact solution we use the following formula:

concentration = (mass of solute / mass of solution) × 100

concentration of the contact solution = (10.7 / 355) × 100

concentration of the contact solution = 3 %

Learn more about:

concentration of solutions

brainly.com/question/7053858

brainly.com/question/4293889

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Common brass is a copper and zinc alloy containing 37.0% zinc by mass and having a density of 8.48g/cm3. A fitting composed of c
inysia [295]
First, we calculate the mass of the sample:

mass = density x volume
mass = 8.48 x 112.5
mass = 954 grams

Now, we will calculate the mass of each component using its percentage mass, then divide it by its atomic mass to find the moles and finally multiply the number of moles by the number of particles in a mole, that is, 6.02 x 10²³.

Zinc mass = 0.37 x 954
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Copper mass = 0.63 x 954
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3 0
2 years ago
Bases in solution produce what type of ions?
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Bases produce hydroxide ions, while acids produce hydrogen ions. 

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How will the concentration of H+ and OH− ions change when a substance with a pH 3.2 is added to water? Both H+ and OH− will incr
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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
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Answer:

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% 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>

6 0
2 years ago
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