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telo118 [61]
2 years ago
13

How many molecules are in a 6.82 gram sample of Hydrogen Sulfide (H2S)

Chemistry
1 answer:
Alinara [238K]2 years ago
5 0

Answer:

1.204×10^(23)

Explanation:

Molecular weight of H2S,  M = 1*2+32 = 34

weight of sample, m = 6.82 gram

now number of moles, n= m/M

n= 6.82/34 = 0.200 moles

1 mole has 6.023×10^(23) molecules of H2S.

therefore 0.2 moles of H2S will have  0.2×6.023×10^(23) = 1.204×10^(23) number of H2S molecules.

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Another way to look at this problem is to use Newton's second law of motion. The first law states that F = m\times a, where a is the accelerationF is the net force and m is the mass of the object.

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Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty
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Answer :

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(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)

The equilibrium expression :

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CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)

The equilibrium expression :

K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

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HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)

The equilibrium expression of HC_2H_3O_2 will be:

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(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)

The equilibrium expression of Co(H_2O)_6^{3+} will be:

K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}

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CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)

The equilibrium expression of CH_3NH_3^+ will be:

K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}

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