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Lorico [155]
3 years ago
11

Determine how many ml of water you need to remove, by evaporation, if you have a 500 ml of 10.20 M HNO3 dilute solution and you

need to make it a 12 M HNO3 stock solution.
Chemistry
1 answer:
Ludmilka [50]3 years ago
7 0

The total volume of water that would be removed will be 75 mL

<h3>Dilution equation</h3>

Using the dilution equation:

M1V1 = M2V2

In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M

Substitute:

V2 = 500 x 10.20/12

         = 425 mL

The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:

500 - 425 = 75 mL

More on dilution can be found here: brainly.com/question/7208939

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Consider the following reaction and situations 1 through 10. In the spaces provided, clearly indicate the best response to each
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2. A. increase

3. E. shift to the right

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<em>According to Le Chaterlier principle, when a reaction is in equilibrium and one of the constraints that affect reactions is applied, the equilibrium will shift so as annul the effects of the constraints.</em>

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1. If the pressure of the system is increased, there would be no change to the system because there are equal number of moles of products and reactants.

2. If H2 concentration is decreased, the equilibrium will shift to the right and more products will be formed. Hence, the concentration of CO will increase.

3. If H2 concentration is decreased, the equilibrium will shift to the right to annul the effects of the decrease in the concentration of a product.

4. If the concentration of H2 is increased, the equilibrium will shift to the left to annul the effects of increased concentration of a product. Hence, more H2O would be formed.

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6. If the amount of C (a reactant) is increased, the equilibrium will shift to the right. Hence, more H2 will be formed.

7. The reaction is endothermic, hence an increase in temperature will ordinarily shift the equilibrium to the right. However, the addition of H2 (a product) is supposed to shift the equilibrium to the left. Hence, the effects of simultaneous addition of the two actions become indeterminate.

8. Since the reaction is endothermic, increase in the temperature of the system will shift the equilibrium to the right. Hence, more CO will be formed.

9. If the concentration of H2O (a reactant) is decreased and that of CO (a product) is increased, both actions lead to the equilibrium being shifted to the left.

10. Addition of catalyst to the system will only speed up the rate at which the system reach the equilibrium.

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