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Fantom [35]
3 years ago
14

Can anyone tell me about sulfuric acid in shampoo/soap?

Chemistry
1 answer:
yanalaym [24]3 years ago
4 0
The term sulfate is used in chemistry to denote a salt of sulfuric acid. ... This is the sulfate type which can be found in many cleaning and hygiene products including shampoos. The main reasons why it is added to shampoos are because this sulfate produces foam and is a powerful detergent.
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Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of
harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g)   (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

n = \frac{m}{M}

Where:    

m: is the mass

M: is the molar mass

  • For <em>hydrogen </em>we have:

n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles

  • And for <em>nitrogen</em>:

n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles

Since we have <u>3.47 moles of hydrogen</u> and we need <u>7.50 moles</u> to react with all the mass of nitrogen, the <em>limiting reactant</em> is <em>hydrogen</em>.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

   

I hope it helps you!                        

6 0
3 years ago
How many grams of copper (II) oxide will react with 10 liters of hydrogen gas?​
Svetradugi [14.3K]

Answer:

maybe 200g

I don't know thats just a guess

HOPE ITS TURE

8 0
2 years ago
If a “sample” of pennies contained 75 heads and 25 tails, how many half‐lives would have passed since the “sample” formed. Expla
finlep [7]
..............................
8 0
3 years ago
Glucose, C 6 H 12 O 6 , is used as an energy source by the human body. The overall reaction in the body is described by the equa
MA_775_DIABLO [31]

Answer:

m_{O_2}=61.87gO_2

m_{CO_2}=85.07gCO_2

Explanation:

Hello,

Considering the given reaction's stoichiometry, grams of oxygen result:

m_{O_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molO_2}{1molC_6H_{12}O_6}*\frac{32gO_2}{1molO_2}\\m_{O_2}=61.87gO_2

Moreover, the mass of produced carbon dioxide turns out:

m_{CO_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molCO_2}{1molC_6H_{12}O_6}*\frac{44gCO_2}{1molCO_2}\\m_{O_2}=85.07gCO_2

Best regards.

6 0
3 years ago
Using the diagram , what is the intersection of plan ABCD and GC?
Oduvanchick [21]

Answer:

C

Explanation:

When a line segment and a plane intersect, they intersect at a point. They both coincide on point C

7 0
3 years ago
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