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Chapter 5: Problem 51

The steel reaction vessel of a bomb calorimeter, which has a volume of \(75.0 \mathrm{~mL}\), is charged with oxygen gas to a pressure of \(14.5\) atm at \(22^{\circ} \mathrm{C}\). Calculate the moles of oxygen in the reaction vessel.

Short Answer

Expert verified
There are approximately 0.0396 moles of oxygen in the reaction vessel.

Step by step solution

01

Convert the given values to the appropriate units

We need to convert the given values to the appropriate units to use in the Ideal Gas Law formula. The volume should be in liters, the pressure should be in atm, and the temperature should be in Kelvin. 1. Convert the volume from mL to L: \[75.0\: mL \times \frac{1\:L}{1000\:mL} = 0.075\: L\] 2. The pressure is already given in atm, so we don't need to convert it. 3. Convert the temperature from Celsius to Kelvin: \[22^{\circ}\:C + 273.15 = 295.15\:K\] Now, we have the volume as 0.075 L, the pressure as 14.5 atm, and the temperature as 295.15 K.
02

Solve for the number of moles using the Ideal Gas Law

Using the Ideal Gas Law formula \(PV = nRT\), we can solve for the number of moles (n) of oxygen gas in the container with the given values: We are given: \(P = 14.5\: atm, V = 0.075\: L, T = 295.15\:K\) and \(R = 0.0821\: \frac{L\:atm}{mol\:K}\). Now we can rearrange the formula to solve for n: \[n = \frac{PV}{RT}\] Substitute the given values into the equation: \[n = \frac{(14.5\:atm)(0.075\:L)}{(0.0821\: \frac{L\:atm}{mol\:K})(295.15\:K)}\]
03

Calculate the number of moles of oxygen

Now, we can do the calculations to find the number of moles of oxygen gas in the container: \[n = \frac{(14.5)(0.075)}{(0.0821)(295.15)} = 0.0396\:mol\] Hence, there are approximately 0.0396 moles of oxygen in the container.

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Most popular questions from this chapter

Which of the following statements is(are) true? For the false statements, correct them. a. At constant temperature, the lighter the gas molecules, the faster the average velocity of the gas molecules. b. At constant temperature, the heavier the gas molecules, the larger the average kinetic energy of the gas molecules. c. A real gas behaves most ideally when the container volume is relatively large and the gas molecules are moving relatively quickly. d. As temperature increases, the effect of interparticle interactions on gas behavior is increased. e. At constant \(V\) and \(T\), as gas molecules are added into a container, the number of collisions per unit area increases resulting in a higher pressure. f. The kinetic molecular theory predicts that pressure is inversely proportional to temperature at constant volume and moles of gas.

Solid thorium(IV) fluoride has a boiling point of \(1680^{\circ} \mathrm{C}\). What is the density of a sample of gaseous thorium(IV) fluoride at its boiling point under a pressure of \(2.5\) atm in a 1.7-L container? Which gas will effuse faster at \(1680^{\circ} \mathrm{C}\), thorium(IV) fluoride or uranium(III) fluoride? How much faster?

A gauge on a compressed gas cylinder reads \(2200 \mathrm{psi}\) (pounds per square inch; \(1 \mathrm{~atm}=14.7 \mathrm{psi}\) ). Express this pressure in each of the following units. a. standard atmospheres b. megapascals (MPa) c. torr

Silane, \(\mathrm{SiH}_{4}\), is the silicon analogue of methane, \(\mathrm{CH}_{4}\). It is prepared industrially according to the following equations: $$ \begin{aligned} \mathrm{Si}(s)+3 \mathrm{HCl}(g) & \longrightarrow \operatorname{HSiCl}_{3}(l)+\mathrm{H}_{2}(g) \\ 4 \mathrm{HSiCl}_{3}(l) & \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l) \end{aligned} $$ a. If \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}(d=1.34 \mathrm{~g} / \mathrm{mL})\) is isolated when \(15.0 \mathrm{~L}\) \(\mathrm{HCl}\) at \(10.0 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3}\) ? b. When \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at \(10.0\) atm and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is \(93.1 \%\) ?

A hot-air balloon is filled with air to a volume of \(4.00 \times\) \(10^{3} \mathrm{~m}^{3}\) at 745 torr and \(21^{\circ} \mathrm{C}\). The air in the balloon is then heated to \(62^{\circ} \mathrm{C}\), causing the balloon to expand to a volume of \(4.20 \times 10^{3} \mathrm{~m}^{3}\). What is the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.)
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