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mrs_skeptik [129]
2 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]2 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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Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:
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Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of HNO_3 = \frac{15.5g}{63g/mol\times 9.5L}=0.026M

Equilibrium concentration of NO = \frac{16.6g}{30g/mol\times 9.5L}=0.058M

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Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c  

For the given chemical reaction:

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The expression for K_c is written as:

K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}

K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}

K_c=3.72

Thus  the value of the equilibrium constant Kc for this reaction is 3.72

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