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irakobra [83]
3 years ago
8

How many electrons are there in second principal energy level(n=2) of a phosphorus atom?

Chemistry
1 answer:
salantis [7]3 years ago
6 0

Answer:

8 electrons

Explanation:

Phosphorous is the element of group 15 and third period. The atomic number of chlorine is 15 and the symbol of the element is P.

The electronic configuration of the element phosphorus is:-

1s^22s^22p^63s^23p^3

Thus, from the electronic configuration shown above,

In n = 2, the total number of electrons are 8 which are present in 2s^22p^6

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The question is incomplete, complete question is :

The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia:

Step 1 : 4NH_3(g)+5O_2(g)\rightleftharpoons 4NO(g)+6H_2O(g)

Step 2 :  2NO(g) +O_2(g) \rightleftharpoons 2NO_2 (g)

The net reaction is:

4NH_3(g)+7O_2(g)\rightleftharpoons 4NO_2(g)+6H_2O(g)

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols

Answer:

Equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 \& K_2:

K=K_1\time (K_2)^2

Explanation:

Step 1 : 4NH_3(g)+5O_2(g)\rightleftharpoons 4NO(g)+6H_2O(g)

Expression of an equilibrium constant can be written as:

K_1=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}

Step 2 :  2NO(g) +O_2(g) \rightleftharpoons 2NO_2 (g)

Expression of an equilibrium constant can be written as:

K_2=\frac{[NO_2]^2}{[NO]^2[O_2]}

The net reaction is:

4NH_3(g)+7O_2(g)\rightleftharpoons 4NO_2(g)+6H_2O(g)

Expression of an equilibrium constant can be written as:

K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}

Multiply and divide [NO]^4;

K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO]^4}{[NO]^4}

K=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO_2]^4}{[NO]^4}

K=K_1\times \frac{[NO_2]^4}{[O_2]^2[NO]^4}

K=K_1\times (\frac{[NO_2]^2}{[O_2]^1[NO]^2})^2

K=K_1\time (K_2)^2

So , the equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 \& K_2:

K=K_1\time (K_2)^2

4 0
3 years ago
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