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Chapter 8: Problem 43

While resting, the average \(70 \mathrm{~kg}\) human male consumes \(16.6281\) of oxygen per hour at \(27^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\). How many moles of oxygen are consumed by the \(70 \mathrm{~kg}\) man while resting for lhour? (a) \(0.67\) (b) \(66.7\) (c) \(666.7\) (d) \(67.5\)

Short Answer

Expert verified
Question: Based on the given mass of consumed oxygen (16.6281 kg/h), calculate the number of moles of oxygen consumed in an hour. Answer: Approximately 519.6 mol/h

Step by step solution

01

Convert mass to grams #

Since we have the mass of consumed oxygen in kg per hour, we need to convert it to grams per hour. To do this, we'll use the conversion factor between kg and g as follows: 1 kg = 1000 g Using this conversion factor, we can convert the given mass to grams: 16.6281 kg/h × 1000 g/kg = 16628.1 g/h
02

Calculate the number of moles #

Now that we have the mass of consumed oxygen in grams per hour, we can calculate the number of moles using the formula: Number of moles = mass / molar mass The molar mass of oxygen (O2) is 32 g/mol. Plugging this into the formula, we get: Number of moles = 16628.1 g/h / 32 g/mol = 519.6281 mol/h
03

Round to the nearest decimal #

We need to round the number of moles to the nearest decimal and find which option matches our result. Rounding 519.6281 to the nearest decimal, we get: Approximately 519.6 mol/h This number is close to option (c), which is 666.7, but it's considerably different than any of the other options. Please check the given data and/or options for possible errors.

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Most popular questions from this chapter

A quantity of \(2.0 \mathrm{~g}\) of a triatomic gaseous element was found to occupy a volume of \(448 \mathrm{ml}\) at \(76 \mathrm{~cm}\) of \(\mathrm{Hg}\) and \(273 \mathrm{~K}\). The mass of its each atom is (a) \(100 \mathrm{amu}\) (b) \(5.53 \times 10^{-23} \mathrm{~g}\) (c) \(33.3 \mathrm{~g}\) (d) \(5.53\) amu

A gaseous alkane is exploded with oxygen. The volume of \(\mathrm{O}_{2}\) for complete combustion to the volume of \(\mathrm{CO}_{2}\) formed is in \(7: 4\) ratio. The molecular formula of alkane is (a) \(\mathrm{CH}_{4}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{8}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{10}\)

A compound has carbon, hydrogen, and oxygen in \(3: 3: 1\) atomic ratio. If the number of moles in \(1 \mathrm{~g}\) of the compound is \(6.06 \times 10^{-3}\), the molecular formula of the compound will be (a) \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{9} \mathrm{H}_{9} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{O}_{4}\)

What is the percentage of 'free \(\mathrm{SO}_{3}\) 'in a sample of oleum labelled as '104.5\%"? (a) \(20 \%\) (b) \(40 \%\) (c) \(60 \%\) (d) \(80 \%\)

The concentration of same aqueous solution of glucose is determined by two students-Sawan and Gautam. Sawan reported the concentration as \(20 \%\) (w/w) and Gautam reported the concentration as \(25 \%(\mathrm{w} / \mathrm{v}) .\) If both the concentrations are correct, then the density of solution is (a) \(0.8 \mathrm{~g} / \mathrm{ml}\) (b) \(1.0 \mathrm{~g} / \mathrm{ml}\) (c) \(1.25 \mathrm{~g} / \mathrm{ml}\) (d) \(1.33 \mathrm{~g} / \mathrm{m} 1\)
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