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Schach [20]
3 years ago
8

Both isopropyl alcohol, C3H8OH, and ethylene glycol, C2H6O2, are used as antifreeze. When equal masses of each are added to wate

r, which, if either, will be more effective at preventing freezing? A. Isopropyl alcohol B. Ethylene glycol C. They will produce the same effect D. Neither, because they do not dissolve in water.
Chemistry
2 answers:
kiruha [24]3 years ago
6 0
I believe the correct answer from the choices listed above is option B. When equal masses of each are added to water, ethylene glycol would be more effective. Ethylene glycol is the most widely used automotive cooling-systemantifreeze, although methanol, ethanol, isopropyl alcohol, and propylene glycol are also used.
Hope this answers the question. Have a nice day.
blagie [28]3 years ago
5 0
Hello there.

<span>Both isopropyl alcohol, C3H8OH, and ethylene glycol, C2H6O2, are used as antifreeze. When equal masses of each are added to water, which, if either, will be more effective at preventing freezing? 

</span><span>B. Ethylene glycol </span>
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Answer : The value of K_c for the given reaction is, 0.36

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.

The given equilibrium reaction is,

Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)

The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

Now we have to calculate the value of K_c for the given reaction.

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

K_c=\frac{(0.6)^2}{(1)\times (1)}

K_c=0.36

Therefore, the value of K_c for the given reaction is, 0.36

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