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Chapter 3: Problem 21

Which of the following has more atoms: \(1.10 \mathrm{~g}\) of hydrogen atoms or \(14.7 \mathrm{~g}\) of chromium atoms?

Short Answer

Expert verified
1.10 grams of hydrogen has more atoms than 14.7 grams of chromium

Step by step solution

01

Calculate the number of moles for hydrogen and chromium

The number of moles can be calculated using the formula \[ n = \frac{m}{M} \] where, \( n \) is the number of moles, \( m \) is the mass of the sample and \( M \) is the atomic mass of the substance. In this case, the mass of hydrogen is 1.10 g and the atomic mass is approximately 1. Therefore the number of moles for hydrogen is approximately \[ n_{H} = \frac{1.10}{1} = 1.10 moles \] The mass of chromium is 14.7 g and the atomic mass is approximately 52. Therefore the number of moles for chromium is approximately \[ n_{Cr} = \frac{14.7}{52} = 0.283 moles \]
02

Calculate the number of atoms

The number of atoms can be calculated by multiplying the number of moles with Avogadro's number \((6.02 \times 10^{23})\). The number of hydrogen atoms will be \(1.10 \times 6.02 \times 10^{23} = 6.62 \times 10^{23}\). The number of chromium atoms will be \(0.283 \times 6.02 \times 10^{23} = 1.70 \times 10^{23}\)
03

Compare the number of atoms

The number of atoms in hydrogen is larger than the number of atoms in chromium, therefore the 1.10 g of hydrogen has more atoms than the 14.7 g of chromium

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Most popular questions from this chapter

The molar mass of caffeine is \(194.19 \mathrm{~g}\). Is the molecular formula of caffeine \(\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{O}\) or \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} ?\)

A major industrial use of hydrochloric acid is in metal pickling. This process involves the removal of metal oxide layers from metal surfaces to prepare them for coating. (a) Write an equation between iron(III) oxide, which represents the rust layer over iron, and \(\mathrm{HCl}\) to form iron(III) chloride and water. (b) If 1.22 moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(289.2 \mathrm{~g}\) of HCl react, how many grams of \(\mathrm{FeCl}_{3}\) will be produced?

Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is the gas that is mainly responsible for global warming (the greenhouse effect). The burning of fossil fuels is a major cause of the increased concentration of \(\mathrm{CO}_{2}\) in the atmosphere. Carbon dioxide is also the end product of metabolism (see Example 3.13 ). Using glucose as an example of food, calculate the annual human production of \(\mathrm{CO}_{2}\) in grams, assuming that each person consumes \(5.0 \times 10^{2} \mathrm{~g}\) of glucose per day. The world's population is 7.2 billion, and there are 365 days in a year.

Lysine, an essential amino acid in the human body, contains \(\mathrm{C}, \mathrm{H}, \mathrm{O},\) and \(\mathrm{N} .\) In one experiment, the complete combustion of \(2.175 \mathrm{~g}\) of lysine gave \(3.94 \mathrm{~g}\) \(\mathrm{CO}_{2}\) and \(1.89 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In a separate experiment, \(1.873 \mathrm{~g}\) of lysine gave \(0.436 \mathrm{~g} \mathrm{NH}_{3}\). (a) Calculate the empirical formula of lysine. (b) The approximate molar mass of lysine is \(150 \mathrm{~g}\). What is the molecular formula of the compound?

A sample of a compound of \(\mathrm{Cl}\) and \(\mathrm{O}\) reacts with an excess of \(\mathrm{H}_{2}\) to give \(0.233 \mathrm{~g}\) of \(\mathrm{HCl}\) and \(0.403 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Determine the empirical formula of the \(\mathrm{com}-\) pound.
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