Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 8

Determine the number of moles of \(\mathrm{Br}_{2}\) in a sample consisting of (a) \(8.08 \times 10^{22} \mathrm{Br}_{2}\) molecules; (b) \(2.17 \times 10^{24}\) Br atoms; (c) 11.3 kg bromine; (d) \(2.65 \mathrm{L}\) liquid bromine \((d=3.10 \mathrm{g} / \mathrm{mL})\)

Short Answer

Expert verified
The answer will be the calculated value for each part (a), (b), (c), and (d) obtained after performing the calculations outlined in the steps above.

Step by step solution

01

- Calculate Moles from Molecules

Given that there are \(8.08 \times 10^{22}\) molecules of bromine \(\mathrm{Br}_{2}\), the number of moles can be calculated by dividing by Avogadro's number \(6.022 \times 10^{23}\) mol\(^{-1}\). Thus, Moles = \(\frac{8.08 \times 10^{22}}{6.022 \times 10^{23}}\)
02

- Calculate Moles from Atoms

Given that there are \(2.17 \times 10^{24}\) atoms of bromine, we have to keep in mind that one molecule of \(\mathrm{Br}_{2}\) has two atoms. So first calculate the number of molecules and then calculate the moles. Thus, Molecules = \(\frac{2.17 \times 10^{24}}{2}\), and Moles = \(\frac{Molecules}{6.022 \times 10^{23}}\)
03

- Calculate Moles from Mass

For 11.3 kg of bromine, first convert the mass from kg to g, then use the molar mass of \(\mathrm{Br}_{2} = 159.808 g/mol\). Thus, Mass = \(11.3 \times 10^{3}\) g, and Moles = \(\frac{Mass}{159.808}\)
04

- Calculate Moles from Volume

Given a volume 2.65 L of liquid bromine and the density \(3.10 g/cm^{3}\), convert the volume from L to mL, find the mass using the density, then use the molar mass of \(\mathrm{Br}_{2} = 159.808 g/mol\) to find the moles. Thus, Volume = \(2.65 \times 10^{3}\) mL, Mass = Volume \(\times\) Density, and Moles = \(\frac{Mass}{159.808}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the empirical formula of (a) the rodenticide (rat killer) warfarin, which consists of \(74.01 \% \mathrm{C}\) \(5.23 \% \mathrm{H},\) and \(20.76 \%\) O, by mass; (b) the antibacterial agent sulfamethizole, which consists of \(39.98 \%\) C. \(3.73 \% \mathrm{H}, 20.73 \% \mathrm{N}, 11.84 \% \mathrm{O},\) and \(23.72 \% \mathrm{S},\) by mass.

Write a formula for (a) the chloride of titanium having Ti in the O.S. \(+4 ;\) (b) the sulfate of iron having Fe in the O.S. \(+3 ;(c)\) an oxide of chlorine with Cl in the O.S. \(+7 ;\) (d) an oxoanion of sulfur in which the apparent O.S. of \(S\) is +7 and the ionic charge is \(2-\).

The element \(X\) forms the chloride \(X C l_{4}\) containing \(75.0 \% \mathrm{Cl},\) by mass. What is element \(\mathrm{X} ?\)

Which answer is correct? One mole of liquid bromine, \(\mathrm{Br}_{2},\) (a) has a mass of \(79.9 \mathrm{g} ;\) (b) contains \(6.022 \times 10^{23}\) Br atoms; (c) contains the same number of atoms as in \(12.01 \mathrm{g} \mathrm{H}_{2} \mathrm{O} ;\) (d) has twice the mass of 0.500 mole of gaseous \(\mathrm{Cl}_{2}\)

Para-cresol ( \(p\) -cresol) is used as a disinfectant and in the manufacture of herbicides. A 0.4039 g sample of this carbon-hydrogen-oxygen compound yields \(1.1518 \mathrm{g} \mathrm{CO}_{2}\) and \(0.2694 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) in combustion analysis. Its molecular mass is 108.1 u. For \(p\) -cresol, determine its (a) mass percent composition; (b) empirical formula; (c) molecular formula.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free