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Chapter 2: Problem 83

Deuterium, \(^{2} \mathrm{H}(2.0140 \mathrm{u}),\) is sometimes used to replace the principal hydrogen isotope \(^{1} \mathrm{H}\) in chemical studies. The percent natural abundance of deuterium is 0.015\%. If it can be done with 100\% efficiency, what mass of naturally occurring hydrogen gas would have to be processed to obtain a sample containing \(2.50 \times 10^{21}^{2} \mathrm{H}\) atoms?

Short Answer

Expert verified
Substitution and calculation with the above steps will yield the mass of naturally occurring hydrogen gas that needs to be processed to obtain a specific number of Deuterium atoms.

Step by step solution

01

Calculate moles of Deuterium

First, convert the number of deuterium atoms needed (\(2.50 \times 10^{21}\)) to moles using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mole). The calculation is given by: \n\n\[ \frac{2.50 \times 10^{21} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mole}} \]
02

Convert moles of Deuterium to moles of Hydrogen

Given the natural abundance of Deuterium (0.015%), convert the moles of Deuterium to moles of Hydrogen using the percent abundance as follows:\n\n\[ \text{moles of hydrogen} = \frac{\text{moles of deuterium}}{\text{percent abundance of deuterium} / 100} \]\n\nNote that the percent abundance needs to be converted to decimal form by dividing by 100.
03

Convert moles of Hydrogen to grams

Now, convert the moles of hydrogen to grams using the molar mass of hydrogen (1.008 u). The molar mass is the mass of one mole of a substance. It is used as a conversion factor to go from moles to grams (and vice versa). The formula is:\n\n\[ \text{grams of hydrogen} = \text{moles of hydrogen} \times \text{molar mass of hydrogen} \]

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Most popular questions from this chapter

The German chemist Fritz Haber proposed paying off the reparations imposed against Germany after World War I by extracting gold from seawater. Given that (1) the amount of the reparations was \(\$ 28.8\) billion dollars, (2) the value of gold at the time was about \(\$ 21.25\) per troy ounce ( \(1 \text { troy ounce }=31.103 \mathrm{g}),\) and (3) gold occurs in seawater to the extent of \(4.67 \times 10^{17}\) atoms per ton of seawater \((1 \text { ton }=2000\) lb), how many cubic kilometers of seawater would have had to be processed to obtain the required amount of gold? Assume that the density of seawater is \(1.03 \mathrm{g} / \mathrm{cm}^{3}\) (Haber's scheme proved to be commercially infeasible, and the reparations were never fully paid.)

The two species that have the same number of electrons as \(^{3}\) th S are (a) \(^{32} \mathrm{Cl} ;\) (b) \(^{34} \mathrm{S}^{+} ;\) (c) \(^{33} \mathrm{P}^{+} ;\) (d) \(^{28} \mathrm{Si}^{2-}\) (e) \(^{35} S^{2-} ;(f)^{40} A r^{2+} ;(g)^{40} C a^{2+}\)

Phosphorus forms two compounds with chlorine. In the first compound, \(1.000 \mathrm{g}\) of phosphorus is combined with \(3.433 \mathrm{g}\) chlorine, and in the second, \(2.500 \mathrm{g}\) phosphorus is combined with \(14.308 \mathrm{g}\) chlorine. Show that these results are consistent with Dalton's law of multiple proportions.

One oxide of rubidium has \(0.187 \mathrm{g}\) O per gram of Rb. A possible O:Rb mass ratio for a second oxide of rubidium is (a) \(16: 85.5 ;\) (b) \(8: 42.7 ;\) (c) \(1: 2.674 ;\) (d) any of these.

In an experiment, \(125 \mathrm{cm}^{3}\) of zinc and \(125 \mathrm{cm}^{3}\) of iodine are mixed together and the iodine is completely converted to \(164 \mathrm{cm}^{3}\) of zinc iodide. What volume of zinc remains unreacted? The densities of zinc, iodine, and zinc iodide are \(7.13 \mathrm{g} / \mathrm{cm}^{3}, 4.93 \mathrm{g} / \mathrm{cm}^{3}\) and \(4.74 \mathrm{g} / \mathrm{cm}^{3}\), respectively.
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