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3 years ago

Consider the following reaction in chemical equilibrium:

Chemistry

1 answer:

Eddi Din [679]3 years ago

6 0

Answer:

1- Adding Br₂ increases the concentration of Br₂, causing the reaction progress to shift to the left (away from the NO).

2- Adding BrNO increases the concentration of BrNO, causing the reaction progress to shift to the right.

Explanation:

To solve this question, we should mention Le Chatelier's principle states that if we affect on an equilibrium by an external factor, the reactions shifts to the direction that can suppress the effect of the external factor.
1- the effect of adding additional Br₂(g) to the reaction mixture:

Adding Br₂ increases the concentration of Br₂, causing the reaction progress to shift to the left (away from the NO).

Adding Br₂, increases the concentration of Br₂, so the reaction shifts to the reactants side (left) to cancel the increase in Br₂ concentration.

2- The effect of adding additional BrNO:

Adding BrNO increases the concentration of BrNO, causing the reaction progress to shift to the right.

Adding BrNO, increases the concentration of BrNO, so the reaction shifts to the products side (right) to cancel the increase in BrNO concentration.

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In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53 × x10−

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4 years ago

Calculate the Molarity when 14.6 g Ethanol (C2H5OH) is dissolved in a 2.10 L solution.

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3 years ago

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3 years ago

A chemist must prepare 200.0 mL of hydrochloric acid solution with a pH of 0.60 at 25 °C. He will do this in three steps: • Fill

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Answer:

In the second step, the chemist must measure 8.3 ml of concentrated acid

Explanation:

The concentration of the final solution can be obtained using the pH value:

pH = -log[H] = 0.60

[H] = 10^(-0.60) = 0.25 M

Then, the final concentration of HCl will be 0.25 M because HCl is a monoprotic acid, which means that HCl only has one hydrolyzable proton. Therefore: [HCl] = [H].

The number of moles of HCl in the final solution will be equal to the number of moles present in the volume taken from the stock solution:

n° of moles in the volume taken from stock solution = n° moles in the final solution.

The number of moles can be calculated as follows:

number of moles = concentration * volume

Then:

Ci * Vi = Cf * Vf

where

Ci = concentration of the stock solution

Vi = volume taken from the stock solution

Cf = concentration of the final solution

Vf = volume of the final solution

Replacing with the data, we can obtain Vi:

6.0 M * Vi = 0.25 M * 200.0 ml

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3 years ago