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mrs_skeptik [129]
3 years ago
8

A 50.0 mL sample of 12.0 M HCl is diluted to 200 mL. What is true about the diluted solution?

Chemistry
1 answer:
kobusy [5.1K]3 years ago
7 0

The concentration of the solution reduces and the number of moles of solute isn't affected.

Data;

  • V1 = 50mL
  • C1 = 12.0M
  • V2 = 200mL
  • C2 = ?
<h3>Facts about the diluted solution</h3>

1. When the solution is diluted, the concentration changes and this time, the concentration reduces.

Using dilution formula

c_1 v_1 = c_2 v_2\\12 * 50 = c_2 * 200\\c_2 = \frac{600}{200} \\c_2 = 3M

The concentration of the solution reduces.

2. The number of moles remains the same.

When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.

Learn more on concentration of a solution here;

brainly.com/question/2201903

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the mass by mass percentage of a solution is 20%. find the mass of the solute dissolved in 500g of solution​
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Answer:

m_{solute}=100g

Explanation:

Hello,

In this case, by considering the mathematical definition of by mass percentage of a solute as shown below:

\%m/m=\frac{m_{solute}}{m_{solution}}*100\%

We are able to compute the mass of the solute in a 20% solution having 500 g of solution as follows:

m_{solute}=\frac{m_{solution}*\%m/m}{100\%} =\frac{500g*20\%}{100\%}\\ \\m_{solute}=100g

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Part B When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 22.8 g of carbon were burned in the pres
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<u>Answer:</u> The amount of carbon dioxide gas produced in the reaction is 83.6 grams

<u>Explanation:</u>

As, some amount of oxygen gas is left after the reaction is completed. So, it is present in excess and is considered as an excess reagent.

Thus, carbon is considered as a limiting reagent because it limits the formation of product.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of carbon = 22.8 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon}=\frac{22.8g}{12g/mol}=1.9mol

The chemical equation for the reaction of carbon and oxygen gas follows:

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By Stoichiometry of the reaction:

1 mole of carbon produces 1 mole of carbon dioxide gas

So, 1.9 moles of carbon will produce = \frac{1}{1}\times 1.9=1.9moles of carbon dioxide gas

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 1.9 moles

Putting values in equation 1, we get:

1.9mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(1.9mol\times 44g/mol)=83.6g

Hence, the amount of carbon dioxide gas produced in the reaction is 83.6 grams

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