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Chapter 5: Problem 48

Calculate the density of hydrogen bromide (HBr) gas in grams per liter at \(733 \mathrm{mmHg}\) and \(46^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The density of HBr gas under the given conditions is calculated to be approximately 2.40 g/L.

Step by step solution

01

Determine the Molar Mass of Hydrogen Bromide

Hydrogen bromide is made up of one atom of hydrogen (H) and one atom of bromine (Br). The molar mass of hydrogen is approximately 1 g/mol, and that of bromine is approximately 79.9 g/mol. Apply these values to calculate the molar mass of HBr. \( M_{HBr} = 1g/mol + 79.9g/mol = 80.9 g/mol \)
02

Convert Temperature from Celsius to Kelvin

The ideal gas law requires temperature to be in Kelvin. You can convert from Celsius to Kelvin using the formula \( K = C + 273.15 \), where \( K \) is temperature in Kelvin and \( C \) is temperature in Celsius. Therefore, the temperature in Kelvin is \( 46°C + 273.15 = 319.15 K \).
03

Convert Pressure from mmHg to atm

We also need to convert the pressure from millimeters of mercury (mmHg) to atmospheres (atm) because the ideal gas law commonly uses options for the pressure in atm. To convert mmHg to atm, use the conversion rate 1 atm = 760 mmHg. Therefore, the pressure in atm is \( 733 mmHg / 760 mmHg/atm = 0.964 atm \).
04

Calculate the Density of HBr Gas

The general formula for gas density is \( d = \frac{MP}{RT} \), where \( d \) is the density, \( M \) is the molar mass, \( P \) is the pressure, \( R \) is the ideal gas constant (which equals to 0.0821 L∙atm/mol∙K), and \( T \) is the temperature. Substituting the known values into the formula will yield \( d = \frac{(80.9 g/mol) × (0.964 atm)}{(0.0821 L∙atm/mol∙K) × (319.15 K)} \).

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Most popular questions from this chapter

A sample of zinc metal is allowed to react completely with an excess of hydrochloric acid: $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ The hydrogen gas produced is collected over water at \(25.0^{\circ} \mathrm{C}\) using an arrangement similar to that shown in Figure \(5.14 .\) The volume of the gas is \(7.80 \mathrm{~L},\) and the atmospheric pressure is \(0.980 \mathrm{~atm} .\) Calculate the amount of zinc metal in grams consumed in the reaction. (Vapor pressure of water at \(25^{\circ} \mathrm{C}=\) \(23.8 \mathrm{mmHg} .)\)

What is the difference between a gas and a vapor? At \(25^{\circ} \mathrm{C},\) which of the following substances in the gas phase should be properly called a gas and which should be called a vapor: molecular nitrogen \(\left(\mathrm{N}_{2}\right),\) mercury?

A \(2.5-\mathrm{L}\) flask at \(15^{\circ} \mathrm{C}\) contains a mixture of three gases, \(\mathrm{N}_{2},\) He, and \(\mathrm{Ne},\) at partial pressures of \(0.32 \mathrm{~atm}\) for \(\mathrm{N}_{2}, 0.15 \mathrm{~atm}\) for \(\mathrm{He},\) and \(0.42 \mathrm{~atm}\) for Ne. (a) Calculate the total pressure of the mixture. (b) Calculate the volume in liters at STP occupied by He and Ne if the \(\mathrm{N}_{2}\) is removed selectively.

Estimate the distance (in nanometers) between molecules of water vapor at \(100^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\). Assume ideal behavior. Repeat the calculation for liquid water at \(100^{\circ} \mathrm{C}\), given that the density of water is \(0.96 \mathrm{~g} / \mathrm{cm}^{3}\) at that temperature. Comment on your results. (Assume water molecule to be a sphere with a diameter of \(0.3 \mathrm{nm} .\) )

A piece of sodium metal undergoes complete reaction with water as follows $$2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(aq)+\mathrm{H}_{2}(g)$$ The hydrogen gas generated is collected over water at \(25.0^{\circ} \mathrm{C}\). The volume of the gas is \(246 \mathrm{~mL}\) measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at \(\left.25^{\circ} \mathrm{C}=0.0313 \mathrm{~atm} .\right)\)
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