/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 96 Nickel carbonyl, \(\mathrm{Ni}(\... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 96

Nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4}\), is one of the most toxic substances known. The present maximum allowable concentration in laboratory air during an 8 -hr workday is 1 part in \(10^{9}\) parts 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 \(1.00 \mathrm{~atm}\) pressure. What mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is allowable in a laboratory that is \(54 \mathrm{~m}^{2}\) in area, with a ceiling height of \(3.1 \mathrm{~m}\) ?

Short Answer

Expert verified
The allowable mass of Ni(CO)鈧� in a laboratory that is 54 m虏 in area, with a ceiling height of 3.1 m at 24 掳C and 1.00 atm pressure is approximately 0.0017 g.

Step by step solution

01

Determine the volume of the laboratory

The volume of the laboratory can be found by multiplying its length, width, and height. Given the area and height, we can calculate the volume (V) as follows: \[V = 54 m^2 \times 3.1 m = 167.4 m^3\]
02

Convert the volume to liters

To make calculations easier, we'll convert the volume from cubic meters to liters, using the conversion factor 1 m鲁 = 1000 L: \[167.4 m^3 \times 1000 L/m^3 = 167400 L\]
03

Calculate the moles of gas in the laboratory

We'll use the ideal gas law, PV = nRT, to find the moles of gas (n) in the laboratory, where P is the pressure, V is volume, R is the ideal gas constant, and T is the temperature in Kelvin. First, convert the temperature in Celsius to Kelvin: \[T = 24 ^\circ C + 273.15 = 297.15 K\] Using the known pressure (1 atm) and volume (167400 L) in the ideal gas law and R = 0.0821 L atm / (mol K): \[n = \dfrac{PV}{RT} = \dfrac{(1 \mathrm{~atm})(167400 \mathrm{~L})}{(0.0821 \mathrm{~L~atm / (mol~K)})(297.15 \mathrm{~K})} \approx 68164.7 \text{ moles}\]
04

Find the moles of Ni(CO)鈧�

Given the allowable concentration of Ni(CO)鈧� is 1 in 10鈦� parts by volume, we can find the moles of Ni(CO)鈧� by dividing the total moles of gas by 10鈦�: \[n_{Ni(CO)鈧剗 = \dfrac{68164.7 \text{ moles}}{10^9} \approx 6.81647 \times 10^{-5} \text{ moles}\]
05

Convert moles of Ni(CO)鈧� to mass

Finally, we'll find the mass of Ni(CO)鈧� by multiplying its moles with its molar mass (Ni = 58.69 g/mol, C = 12.01 g/mol, O = 16.00 g/mol): \[m_{Ni(CO)鈧剗 = (6.81647 \times 10^{-5} \text{ moles}) \times [(1\times58.69) + (4\times12.01) + (4\times16)]\,\text{g/mol}\approx 0.0017\, g\] Thus, the allowable mass of Ni(CO)鈧� in the laboratory is approximately 0.0017 g.

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