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Chapter 5: Problem 30

A \(2.00-g\) sample of \(S X_{6}(g)\) has a volume of \(329.5 \mathrm{~cm}^{3}\) at \(1.00 \mathrm{~atm}\) and \(20^{\circ} \mathrm{C}\). Identify the element \(\mathrm{X}\). Name the compound.

Short Answer

Expert verified
Answer: The element X is chlorine (Cl), and the compound is sulfur hexachloride (SCL₆).

Step by step solution

01

Convert the given values to appropriate units

First, we need to convert the given values to appropriate units. Volume: Convert the volume from cm³ to liters: 1 L = 1000 cm³ 329.5 cm³ = 329.5 / 1000 L = 0.3295 L Temperature: Convert the temperature from Celsius to Kelvin: T(K) = T(°C) + 273.15 T(K) = 20 °C + 273.15 = 293.15 K Now, we have: P = 1.00 atm V = 0.3295 L T = 293.15 K
02

Use the Ideal Gas Law to find the number of moles of the given sample

We will use the Ideal Gas Law equation PV = nRT to find the number of moles, n, of the given gaseous sample. We can rearrange the equation to solve for moles as follows: n = PV/(RT) Now, plug in the values: n = (1.00 atm)(0.3295 L) / (0.0821 L atm / K mol)(293.15 K) n = 0.0148 moles
03

Calculate the molar mass of the compound

We know the mass of the sample and the number of moles, so we can calculate the molar mass of the compound: Molar mass (M) = mass (g) / moles (n) M = 2.00 g / 0.0148 moles M = 135.14 g/mol
04

Divide the molar mass of the compound by the molar mass of sulfur to get the molar mass of the element X

The chemical formula of the compound is \(S X_{6}\). So, to find the molar mass of element X, divide the total molar mass of the compound by the molar mass of sulfur. Molar mass of sulfur, \(S= 32.07 g/mol\) Molar mass of \(6X = 135.14 g/mol - 32.07 g/mol = 103.07 g/mol\) Molar mass of element X = 103.07 g/mol ÷ 6 Molar mass of element X = 17.18 g/mol
05

Identify the element X and name the compound

The element with a molar mass of approximately 17.18 g/mol is chlorine (Cl). Thus, the compound is sulfur hexachloride (SCL₆). Compound name: Sulfur hexachloride.

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Most popular questions from this chapter

A tire is inflated to a gauge pressure of \(28.0\) psi at \(71^{\circ} \mathrm{F}\). Gauge pressure is the pressure above atmospheric pressure, which is \(14.7\) psi. After several hours of driving, the air in the tire has a temperature of \(115^{\circ} \mathrm{F}\). What is the gauge pressure of the air in the tire? What is the actual pressure of the air in the tire? Assume that the tire volume changes are negligible.

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