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levacccp [35]
3 years ago
12

Rank these hydrocarbons in order from predicted highest boiling point to lowest boiling point: Rank from highest boiling point t

o lowest boiling point. To rank items as equivalent, overlap them.
A. CH3CH3
B. CH3CH2CH3
C. CH3CH2CH2CH3
D. CH3CH(CH3)2
Chemistry
2 answers:
dexar [7]3 years ago
4 0

Answer: Arrangements of the hydrocarbon from the highest boiling point to the lowest is shown below

C--> D--> B--> A

Explanation:

Hydrocarbons are compounds which contains carbon and hydrogen only. The options given are all saturated because all the four bonds of the carbon are single bonds. The bonds could be a carbon- hydrogen bond or a carbon- carbon bond.

The boiling point of Hydrocarbons can be affected by the following factors:

--> Number of carbon atoms

--> molecular weight

Boiling point depends on the length of the carbon chain, it increases as number of carbon increase. That is the main reason why butane (option C) has the highest boiling point than the rest of the hydrocarbons with ethane( option A) being the one with lowest boiling point.

Increase in molecular weight Increases the boiling point of hydrocarbons. That is the reason why 2-methylpropane (option D) has a higher boiling point than propane( option B) because it has higher molecular weight.

kolezko [41]3 years ago
4 0

Answer:

1-D

2-B

3-C

4-A

Explanation:

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Answer :

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

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K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

(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 :

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 :

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.

(A) The dissociation reaction of HC_2H_3O_2 will be:

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

The equilibrium expression of HC_2H_3O_2 will be:

K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

(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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