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ira [324]
3 years ago
12

The reduction of nitrogen monoxide is described by the following chemical equation: 2H2 (g) +2NO (g) 2H20 ()N2 (g Suppose a two-

step mechanism is proposed for this reaction, beginning with this elementary reaction: H2 g+2NO(g)- N20 (g)+H20(g) Suppose also that the second step of the mechanism should be bimolecular Suggest a reasonable second step. That is, write the balanced chemical equation of a bimolecular elementary reaction that would complete the proposed mechanism
Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

Reasonable Second step- N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)

Explanation:

The given single step chemical reaction is as follows.

2H_{2}(g)+2NO \rightarrow 2H_{2}O(g)+N_{2}(g)

Suppose a two-step mechanism is proposed for this reaction,

The reaction occured in two steps they are as follows.

Step -1:H_{2}(g)+2NO \rightarrow N_{2}O(g)+H_{2}O(g)

Step-2:N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)

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Arbon dioxide is dissolved in blood (ph 7.5) to form a mixture of carbonic acid and bicarbonate. Part a neglecting free co2, wha
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Answer : The fraction of carbonic acid present in the blood is 5.95%

Explanation :

The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.

The pH of a buffer is calculated using Henderson equation which is given below.

pH = pKa + log \frac{[Base]}{[Acid]}

We have been given,

pH = 7.5

pKa of carbonic acid = 6.3

Let us plug in the values in Henderson equation to find the ratio Base/Acid.

7.5 = 6.3 + log \frac{[base]}{[acid]}

1.2 = log \frac{[base]}{[acid]}

\frac{[Base]}{[Acid]} = 10^{1.2}

\frac{[Base]}{[Acid]} = 15.8

[Base] = 15.8 \times [Acid]

The total of mole fraction of acid and base is 1. Therefore we have,

[Acid] + [Base] = 1

But Base = 15.8 x [Acid]. Let us plug in this value in above equation.

[Acid] + 15.8 \times [Acid] = 1

16.8 [Acid] = 1

[Acid] = \frac{1}{16.8}

[Acid] = 0.0595

[Acid] = 0.0595 x 100 = 5.95 %

The fraction of carbonic acid present in the blood is 5.95%

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