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nikitadnepr [17]
3 years ago
11

Enter the atomic symbol, including mass number and atomic number, for iodine-130.

Chemistry
1 answer:
WINSTONCH [101]3 years ago
6 0

<u>Answer:</u> The atomic symbol of the given element is _{53}^{130}\textrm{I}

<u>Explanation:</u>

The general isotopic representation of an element is given as:  _Z^A\textrm{X}

where,

Z represents the atomic number of the element

A represents the mass number of the element

X represents the symbol of an element

For the given isotope: 130-iodine

Mass number = 130

Atomic number = 53

Hence, the atomic symbol of the given element is _{53}^{130}\textrm{I}

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<u>Answer:</u> The value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

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We are given:

Initial moles of nitrogen gas = 1.30 moles

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Equilibrium moles of ammonia = 0.100 moles

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For the given chemical equation:

                N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

<u>Initial:</u>            1.30       1.65

<u>At eqllm:</u>       1.30-x    1.65-3x             2x

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The expression of K_c for above equation follows:

K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}

Equilibrium moles of nitrogen gas = (1.30-x)=(1.30-0.05)=1.25mol

Equilibrium moles of hydrogen gas = (1.65-x)=(1.65-0.05)=1.60mol

Putting values in above expression, we get:

K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}

Calculating the K_c' for the given chemical equation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2

Hence, the value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

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