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adell [148]
3 years ago
15

For the reactionN2(g)+O2(g)?2NO(g)classify each of the following actions by whether it causes a leftward shift, a rightward shif

t, or no shift in the direction of the reaction.a) half oxygenb) double oxygenc) double nitrogend) half nitrogene) double nitrogen monoxidef) half nitrogen monoxide
Chemistry
1 answer:
Soloha48 [4]3 years ago
6 0

Answer: Le Chatleir Principle

Explanation:  The answer of this question can be easily explanied through the Le Chatleir Principle which states that In order to maintain the equilibrium of the reaction, the stress applied to that particular direction will move the equilibrium to the opposite direction in which the stress has been applied.

The given reaction is -

      N_{2}(g)+O_{2}(g)\rightleftharpoons 2NO(g)

If the amount of the oxygen is reduced to half and is being doubled, then the reaction will move in froward direction that is a rightward shift. Same situation will arise in option c and d.

If the amount of nitrogen monoxide is reduced to half or being doubled, the reaction will move towards backward direction that is a leftward reaction.

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When energy is added to solid manganese(ii) sulfate heptahydrate crystals they break down to form liquid water and solid mangane
RoseWind [281]
Chemical reaction: MnSO₄·7H₂O → MnSO₄·H₂O + 6H₂O.
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7 0
3 years ago
0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was
nikitadnepr [17]

Answer: The value of k_{a} for the weak acid is 3.07 \times 10^{-4}.

Explanation:

First, we will calculate the molarity of HA as follows.

     [HA] = \frac{\text{no. of moles}}{\text{volume}}

             = \frac{0.03 mol}{2 L}

             = 0.015

At equilibrium,

              HA + H_{2}O \rightleftharpoons H_{3}O^{+} + A^{-}

Initial:   0.015                  0          0

Change:  -x                     +x         +x

Equilibm: 0.015 - x          x           x

It is given that, at equilibrium

          [HA] = 0.015 - x = 0.013

             - x = 0.013 - 0.015

                  = 0.002

Now, expression for k_{a} of this reaction is as follows.

          k_{a} = \frac{[A^{-}][H_{3}O^{+}]}{[HA]}

                      = \frac{x^{2}}{[HA]}

                      = \frac{(0.002)^{2}}{0.013}

                      = 3.07 \times 10^{-4}

Thus, we can conclude that the value of k_{a} for the weak acid is 3.07 \times 10^{-4}.

4 0
3 years ago
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