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2 years ago

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An empty water bottle is full of air at 15°C and standard pressure. The volume of the bottle is 0. 500 liter. How many moles of

air are in the bottle? The water bottle contains mole of air.

1 answer:

drek231 [11]2 years ago

5 0

The moles of gas in the bottle has been 0.021 mol.

The ideal gas has been given as the gas where there has been negligible amount of interatomic collisions. The ideal gas equation has been given as:

$PV=nRT$

<h3>Computation for the moles of gas</h3>

The gi<em>ve</em>n gas has standard pressure, $P=1\rm atm$

The volume of the gas has been, $V= 0. 500 \;\rm L$

The temperature of the gas has been, $T=15^\circ \text C\\
T=288\;\rm K$

Substituting the values for the moles of gas, <em>n:</em>

<em />

<em /> $\rm 1\;\times\; 0. 500 =\textit n\;\times\;0.08214\;atm.L/mol.K\;\times\;288\;K\\\\
\textit n=\dfrac{0. 500}{0.08214\;\times\;288} \;mol\\\\
\textit n=0.021\;mol$

The moles of gas in the bottle has been 0.021 mol.

Learn more about ideal gas, here:
brainly.com/question/8711877

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The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

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$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

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$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

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