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Gennadij [26K]
3 years ago
7

A beaker containing a mixture of N2(g) and H2(g) is cooled by placing it in a tray of cold water. In response to the cooling, th

e likely number of collisions between the molecules of N2(g) and H2(g) will . The rate of the reaction will .
Chemistry
2 answers:
sashaice [31]3 years ago
5 0

Answer:

A beaker containing a mixture of N2(g) and H2(g) is cooled by placing it in a tray of cold water. In response to the cooling, the likely number of collisions between the molecules of N2(g) and H2(g) will  <em><u>decrease</u></em> . The rate of the reaction will  <em><u>decrease</u></em> .

Explanation:

#platofam

Nataliya [291]3 years ago
3 0

The frequency of collisions between the N₂ and H₂ molecules will decrease and the rate of reaction will also decrease.

Since the water is cooler than the gas mixture, heat will flow from the gas to the water.

The gas will cool down, so the average kinetic energy of the gas molecules will decrease.

The molecules will be moving more slowly, so there will be <em>fewer collisions</em> and <em>fewer of these collisions will have enough energy to react</em>.

The rate of reaction between H₂ and N₂ molecules at room temperature is exceedingly slow, <em>but cooling the gas mixture will make the reaction even slower</em>.

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ValentinkaMS [17]

Answer : The concentration of OH^- ion, pH and pOH of solution is, 1.12\times 10^{-13}M, 1.05 and 12.95 respectively.

Explanation : Given,

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pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

pH=-\log [H^+]

First we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.090)

pH=1.05

The pH of the solution is, 1.05

Now we have to calculate the pOH.

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The pOH of the solution is, 12.95

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

12.95=-\log [OH^-]

[OH^-]=1.12\times 10^{-13}M

The OH^- concentration is, 1.12\times 10^{-13}M

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