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navik [9.2K]
2 years ago
13

Consider the chemical equation in equilibrium. A(g) 2B(g) Double headed arrow. C(g) D(g) heat What will happen to the equilibriu

m of this reaction if the temperature is increased? It will shift to the left because heat is released. It will shift to the left because heat is absorbed. It will shift to the right because heat is released. It will shift to the right because heat is absorbed.
Chemistry
1 answer:
Mazyrski [523]2 years ago
6 0

If the temperature is increased then reaction will shift to the left because heat is absorbed.

<h3>What is equilibrium state?</h3>

Equilibrium of any reaction is that state in which concentration of reactant and concentration of product will be constant.

Given chemical reaction is:

A(g) + 2B(g) ⇄ C(g) + D(g)

From the equilibrium state reaction will move only that side which will contribute to maintain the stable state. In the forward reaction heat is released as mention in the question. So, when the temperature of reaction is increased then it shifts towards the left side by absorbing the heat and maintain the stability.

Hence, option (2) is correct, i.e. It will shift to the left because heat is absorbed.

To know more about equilibrium, visit the below link:

brainly.com/question/14297698

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6 0
3 years ago
A formula unit represents the simplest ratio of elements in a______of an ionic compound.
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A formula unit represents the simplest ratio of elements in a mole of an ionic compound.

Explanation:

The formula unit, also known as the empirical formula, gives the simplest whole number ratio of the components of the elements in a mole of an ionic compound.

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For example: Glucose with C6H12O6 will have a formula unit of CH2O. Thus, the simplest ratio of carbon, hydrogen and oxygen in glucose is 1:2:1

3 0
2 years ago
Calculate the pH after 10 mL of 1.0 M sodium hydroxide is added to 60 mL of 0.5M acetic acid.
baherus [9]

Explanation:

It is given that the total volume is (10 mL + 60 mL) = 70 mL.

Also, it is known that M_{1}V_{1} = M_{2}V_{2}

Where,    V_{1} = total volume

               V_{2} = initial volume

Therefore, new concentration of CH_{3}COOH = \frac{M_{2}V_{2}}{V_{1}}

                                        = \frac{60 \times 0.5}{70}

                                        = 0.43 M

New concentration of NaOH = \frac{M_{2}V_{2}}{V_{1}}

                                               = \frac{10 \times 1.0}{70}

                                               = 0.14 M

So, the given reaction will be as follows.

              CH_{3}COOH + OH^{-} \rightarrow CH_{3}COO^{-} + H_{2}O

Initial:             0.43          0.14                     0

Change:          -0.14        -0.14                    0.14

Equilibrium:    0.29          0                       0.14

As it is known that value of pK_{a} = 4.74

Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.

           pH = pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

                 = 4.74 + log \frac{0.14}{0.29}

                 = 4.74 + (-0.316)

                 = 4.42

Therefore, we can conclude that the pH of given reaction is 4.42.

6 0
2 years ago
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