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adelina 88 [10]
3 years ago
11

Predict, using Boyle’s Law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa (normal atmospheric

pressure is 101.3 kPa).
Chemistry
1 answer:
kaheart [24]3 years ago
7 0

Explanation:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.

According to Boyle's law: The pressure of the gas  is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).

pressure\propto \frac{1}{volume} (At constant temperature)

The pressure beneath the sea is 202 kPa and the atmospheric pressure  is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).

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Answer:

pH = 10.11

Explanation:

Hello there!

In this case, since it is possible to realize that this base is able to acquire one hydrogen atom from the water:

C_{18}H_{21}NO_4+H_2O\rightleftharpoons C_{18}H_{21}NO_4H^+OH^-

We can therefore set up the corresponding equilibrium expression:

Kb=\frac{[C_{18}H_{21}NO_4H^+][OH^-]}{[C_{18}H_{21}NO_4]}

Which can be written in terms of the reaction extent, x:

Kb=\frac{x^2}{0.00500M-x}=3.39x10^{-6}

Thus, by solving for x we obtain:

x_1=-0.000132M\\\\x_2=0.0001285M

However, since negative solutions are now allowed, we infer the correct x is 0.0001285 M; thus, the pOH can be computed:

pOH=-log(x)=-log(0.0001285)=3.89

And finally the pH:

pH=14-pOH=14-3.89\\\\pH=10.11

Best regards!

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