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Chapter 10: Problem 103

Nickel carbonyl, Ni(CO) \(_{4},\) is one of the most toxic substances known. The present maximum allowable concentration in laboratory air during an 8 -hr workday is 1 ppb (parts per billion) by volume, which means that there is one mole of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for every \(10^{9}\) moles of gas. Assume \(24^{\circ} \mathrm{C}\) and 1.00 atm pressure. What mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is allowable in a laboratory room that is 12 \(\mathrm{ft} \times 20 \mathrm{ft} \times 9 \mathrm{ft}\) ?

Short Answer

Expert verified
The mass of allowable Nickel carbonyl (Ni(CO)4) in the laboratory room with dimensions 12 ft x 20 ft x 9 ft, at 24°C and 1.00 atm pressure, is approximately 0.000424 g.

Step by step solution

01

Calculate the Volume of Room in Liters

First, we need to convert the dimensions from feet to meters and then calculate the volume in liters (L). The conversion factors are: 1 foot = 0.3048 meters 1 cubic meter = 1000 liters Room dimensions = 12 ft × 20 ft × 9 ft = 2160 cubic feet. Convert to cubic meters: 2160 cubic feet × (0.3048 m / foot)^3 = \( 61.1323 \) cubic meters Convert to liters: \( 61.1323 \) cubic meters × 1000 L / cubic meter = 61132.3 L
02

Calculate Moles of Gas in the Room using Ideal Gas Law

Use the ideal gas law equation: PV = nRT Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We are given P = 1.00 atm, V = 61132.3 L, R = 0.0821 (atm·L) / (K·mol), and T is given in Celsius, we need to convert it to Kelvin: T = 24 + 273.15 = 297.15 K Solve for n (the number of moles of gas in the room): n = PV / RT n = (1.00 atm × 61132.3 L) / (0.0821 (atm·L) / (K·mol) × 297.15 K) n ≈ 2486.681 moles of gas
03

Find Moles of Ni(CO)4 Based on Maximum Allowable Concentration

Given one mole of Ni(CO)4 for every 10^9 moles of gas. Calculate the number of moles of Ni(CO)4: moles of Ni(CO)4 = (moles of gas) / 10^9 moles of Ni(CO)4 = 2486.681 / 10^9 moles of Ni(CO)4 ≈ 2.4867 × 10^(-6) moles
04

Calculate the Mass of Allowable Ni(CO)4

Use the molar mass of Ni(CO)4 to find its mass: Molar Mass of Ni(CO)4 = 58.69 (Ni) + 4 (12.01 (C) + 16.00 (O)) = 170.49 g/mol Mass = moles × molar mass Mass = 2.4867 × 10^(-6) moles × 170.49 g/mol ≈ 0.000424 g So, the mass of allowable Nickel carbonyl (Ni(CO)4) in the laboratory room is approximately 0.000424 g.

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Most popular questions from this chapter

Assume that a single cylinder of an automobile engine has a volume of 524 \(\mathrm{cm}^{3} .\) (a) If the cylinder is full of air at \(74^{\circ} \mathrm{C}\) and 0.980 atm, how many moles of \(\mathrm{O}_{2}\) are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 . )(\mathbf{b})\) How many grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2},\) assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ?

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? \((\mathbf{a})\) Lifting up on the piston to double the volume while keeping the temperature constant; \((\mathbf{b})\) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) , while keeping the volume constant; \((\mathbf{c})\) Pushing down on the piston to halve the volume while keeping the temperature constant.

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 105 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 31 s for 1.0 \(\mathrm{L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{L} ;\) in other words, rate is the amount that diffuses over the time it takes to diffuse.)

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Indicate which of the following statements regarding the kinetic-molecular theory of gases are correct. (a) The average kinetic energy of a collection of gas molecules at a given temperature is proportional to \(m^{1 / 2}\) . (b) The gas molecules are assumed to exert no forces on each other. (c) All the molecules of a gas at a given temperature have the same kinetic energy. (d) The volume of the gas molecules is negligible in comparison to the total volume in which the gas is contained. (e) All gas molecules move with the same speed if they are at the same temperature.
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