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Chapter 8: Problem 45

An \(11.1\) - \(\mathrm{L}\) sample? sample of gas is determined to contain \(0.50 \) \(\mathrm{mole}\) sample?of \(\mathrm{N}_{2} .\) At the same temperature and pressure, how many moles of gas would there be in a \(20 .\) - \(\mathrm{L}\) sample?

Short Answer

Expert verified
In the 20L container, there would be approximately \(0.899\) moles of gas under the same temperature and pressure conditions.

Step by step solution

01

Write down the Ideal Gas Law and the given information.

The ideal gas law is given by the equation: PV = nRT Where, P = Pressure, V = Volume, n = moles of gas, R = Ideal gas constant, and T = Temperature We don't have values for P and T, but since we know P and T remain constant for both samples, we can use the ratio form, given as: \( \frac{n_1}{V_1} = \frac{n_2}{V_2} \) Given: \(V_1 = 11.1 L\), \(n_1 = 0.50 mol\), \(V_2 = 20 L\) We need to find the moles of gas in the 20L container: \(n_2\)
02

Calculate the number of moles in the 20L container using the ratio form of the Ideal Gas Law.

We have the ratio form equation as: \( \frac{n_1}{V_1} = \frac{n_2}{V_2} \) We can solve for \(n_2\) by multiplying both sides by \(V_2\): \( n_2 = \frac{n_1}{V_1} \cdot V_2 \) Now we can substitute the given values into the equation to find the moles of gas in the 20L container: \( n_2 = \frac{0.50 mol}{11.1 L} \cdot 20 L \)
03

Solve for the number of moles in the 20L container.

Calculating the number of moles using the given data: \( n_2 = \frac{0.50 mol}{11.1 L} \cdot 20 L =0.04495495495 mol \cdot 20 L = 0.899099099 mol\) So, in the 20L container, there would be approximately 0.899 moles of gas under the same temperature and pressure conditions.

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Most popular questions from this chapter

Atmospheric scientists often use mixing ratios to express the concentrations of trace compounds in air. Mixing ratios are often expressed as ppmv (parts per million volume): $$\text { ppmv of } X=\frac{\text { vol of } X \text { at } \mathrm{STP}}{\text { total vol of air at } \mathrm{STP}} \times 10^{6}$$ On a certain November day, the concentration of carbon monoxide in the air in downtown Denver, Colorado, reached \(3.0 \times 10^{2}\) ppmv. The atmospheric pressure at that time was \(628\) torr and the temperature was \(0^{\circ} \mathrm{C}\) a. What was the partial pressure of \(\mathrm{CO}\)? b. What was the concentration of \(\mathrm{CO}\) in molecules per cubic meter? c. What was the concentration of \(\mathrm{CO}\) in molecules per cubic centimeter?

A student adds \(4.00 \mathrm{g}\) of dry ice (solid \(\mathrm{CO}_{2}\) ) to an empty balloon. What will be the volume of the balloon at STP after all the dry ice sublimes (converts to gaseous \(\mathrm{CO}_{2}\) )?

A piece of solid carbon dioxide, with a mass of \(7.8 \mathrm{g},\) is placed in a \(4.0\)-\(\mathrm{L}\) otherwise empty container at \(27^{\circ} \mathrm{C}\). What is the pressure in the container after all the carbon dioxide vaporizes? If \(7.8 \mathrm{g},\) solid carbon dioxide were placed in the same container but it already contained air at \(740\) torr, what would be the partial pressure of carbon dioxide and the total pressure in the container after the carbon dioxide vaporizes?

Which of the following statements is(are) true? a. If the number of moles of a gas is doubled, the volume will double, assuming the pressure and temperature of the gas remain constant. b. If the temperature of a gas increases from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) the volume of the gas would double, assuming that the pressure and the number of moles of gas remain constant. c. The device that measures atmospheric pressure is called a barometer. d. If the volume of a gas decreases by one half, then the pressure would double, assuming that the number of moles and the temperature of the gas remain constant.

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\). This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus 1894 p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows:If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.
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