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Diano4ka-milaya [45]
3 years ago
8

Jen can take either of two separate roads to drive to work. The first is a toll road that is rarely congested. The second road i

s a road with no tolls, but it is often congested and has many potholes. In this instance, the toll road is rivalrous and excludable nonrivalrous and nonexcludable rivalrous and nonexcludable nonrivalrous and excludable
Chemistry
1 answer:
Alchen [17]3 years ago
3 0
<span>A good is rivalrous if its consumption by one consumer prevents or impedes its use by other consumers. Roads are rivalrous to the extent that they are subject to congestion, since congestion is caused by use of the road and impedes the use of the road by others. However, a road which is not subject to congestion, like the toll road in this case, is non-rivalrous. A good is excludable if it is possible to prevent those who have not paid for it from using it. So a toll road is by design an excludable good. Hence, the toll road is rivalrous and excludable.</span>
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At what temperature do the gas and liquid phases become<br> indistinguishable from each other?
sergiy2304 [10]

Answer:

100 C or 212 F

Explanation:

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The pH of 0.10 M solution of an acid is 6. What is the acid dissociation constant of the acid?
svlad2 [7]

Answer: Dissociation constant of the acid is 10^{-11}.

Explanation: Assuming the acid to be monoprotic, the reaction follows:

                         HA\rightleftharpoons H^++A^-

pH of the solution = 6

and we know that

pH=-log([H^+])

[H^+]=antilog(-pH)

[H^+]=antilog(-6)=10^{-6}M

As HA ionizes into its ions in 1 : 1 ratio, hence

[H^+]=[A^-]=10^{-6}M

As the reaction proceeds, the concentration of acid decreases as it ionizes into its ions, hence the decreases concentration of acid at equilibrium will be:

[HA]=[HA]-[H^+]

[HA]=0.1M-10^{-6}M

[HA]=0.09999M

Dissociation Constant of acid, K_a is given as:

K_a=\frac{[A^-][H^+]}{HA}

Putting values of [H^+],[A^-]\text{ and }[HA] in the above equation, we get

K_a=\frac{(10^{-6}M).(10^{-6}M)}{0.09999M}

K_a=1.0001\times 10^{-11}M

Rounding it of to one significant figure, we get

K_a=1.0\times 10^{-11}M\approx 10^{-11}M

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