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Chapter 13: Problem 34

How many moles of He are in \(13 \mathrm{g}\) of He?

Short Answer

Expert verified
Answer: There are 3.25 moles of helium in a 13 gram sample.

Step by step solution

01

Identify given information

We are given the mass of helium (He), which is \(13 \mathrm{g}\), and the molar mass of helium, which is \(4 \mathrm{g/mol}\).
02

Write the formula to find the number of moles

Using the formula moles = (mass of substance) / (molar mass of substance), we can find the number of moles of He.
03

Plug the given values into the formula

We will now plug the given values into the formula: moles = (13 g) / (4 g/mol).
04

Calculate the number of moles

On solving the equation, we get moles = \(\dfrac{13 \mathrm{g}}{4 \mathrm{g/mol}}= 3.25 \mathrm{mol}\). Thus, there are 3.25 moles of He in 13 g of He.

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Most popular questions from this chapter

A hydrogen balloon at Earth's surface has a volume of \(5.0 \mathrm{m}^{3}\) on a day when the temperature is \(27^{\circ} \mathrm{C}\) and the pressure is \(1.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2} .\) The balloon rises and expands as the pressure drops. What would the volume of the same number of moles of hydrogen be at an altitude of \(40 \mathrm{km}\) where the pressure is \(0.33 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}\) and the temperature is \(-13^{\circ} \mathrm{C} ?\)
A smoke particle has a mass of \(1.38 \times 10^{-17} \mathrm{kg}\) and it is randomly moving about in thermal equilibrium with room temperature air at \(27^{\circ} \mathrm{C} .\) What is the rms speed of the particle?
If 2.0 mol of nitrogen gas \(\left(\mathrm{N}_{2}\right)\) are placed in a cubic box, \(25 \mathrm{cm}\) on each side, at 1.6 atm of pressure, what is the rms speed of the nitrogen molecules?
Show that, in two gases at the same temperature, the rms speeds are inversely proportional to the square root of the molecular masses: $$ \frac{\left(v_{\mathrm{rms}}\right)_{1}}{\left(v_{\mathrm{rms}}\right)_{2}}=\sqrt{\frac{m_{2}}{m_{1}}} $$
What is the mass density of air at \(P=1.0\) atm and $T=(a)-10^{\circ} \mathrm{C}\( and \)(\mathrm{b}) 30^{\circ} \mathrm{C} ?$ The average molecular mass of air is approximately 29 u.
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