Chemistry: An Atoms-Focused Approach (Second Edition)
Chemistry: An Atoms-Focused Approach (Second Edition)
2nd Edition
ISBN: 9780393614053
Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 2, Problem 2.93QA
Interpretation Introduction

To find:

To compare the number of atoms of uranium and diamond in a cube (having 1 cm on a side) made of each of these two elements.

Expert Solution & Answer
Check Mark

Answer to Problem 2.93QA

Solution:

The cube made up of diamond contains more atoms.

Explanation of Solution

Concept:

The volume of two cubes made up of Uranium and Carbon are same. Only these two elements have different values of densities.

So, the masses of uranium and diamond cubes will be different since

Mass = Volume x Density

So, here, firstly, the given masses of each cube have to be calculated separately by using the above formula.

Atomic masses of Uranium and Carbon are known.

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

Formula:

i) Number of moles of any substance = given mass of the substanceMolar mass of the substance

ii) The number of atoms of each element = Number of moles x Avogadro number

Given information:

Density of Uranium = 19.05 g/cm3

Density of Carbon (Diamond) = 3.514 g/cm3

Length of each side of cube = 1 cm

Atomic mass of Uranium (U) = 238.03 g/mol

Atomic mass of Diamond = 12.01 g/mol

Calculation:

We will calculate the volume of cube from the 1 cm edge length.

Volume of cube = 1 cm × 1 cm × 1 cm = 1 cm3

We will calculate the mass of uranium from volume and density.

1 cm3 × 19.05 gcm3=19.05 g Uranium

Now calculate the moles of uranium atom by using 19.05 g and atomic weight.

19.05 g U ×1 mol238.03 g U = 0.08003 moles U

We will calculate the number of atom by using the Avogadro’s number.

0.08003 mol U × 6.022 × 1023 atoms1 mol=4. 819 × 1022 atoms.

So 4. 819 × 1022 atoms present in uranium atom

Now we will calculate the mass of diamond from volume and density.

1 cm3 × 3.514 gcm3=3.514 g Diamond

Now calculate the moles of uranium atom by using 3.514 g and atomic weight.

3.514 g Diamond ×1 mol12.01 g = 0.2926 moles Diamond

We will calculate the number of atom by using the Avogadro’s number.

0.2926 mol Diamond × 6.022 × 1023 atoms1 mol=1.762 × 1023 atoms

So, the cube of diamond contains more number of atoms than uranium.

Conclusion:

We calculate the number of atom from the edge length of cube, atomic mass and Avogadro’s number.

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Chapter 2 Solutions

Chemistry: An Atoms-Focused Approach (Second Edition)

Ch. 2 - Prob. 2.11VPCh. 2 - Prob. 2.12VPCh. 2 - Prob. 2.13QACh. 2 - Prob. 2.14QACh. 2 - Prob. 2.15QACh. 2 - Prob. 2.16QACh. 2 - Prob. 2.17QACh. 2 - Prob. 2.18QACh. 2 - Prob. 2.19QACh. 2 - Prob. 2.20QACh. 2 - Prob. 2.21QACh. 2 - Prob. 2.22QACh. 2 - Prob. 2.23QACh. 2 - Prob. 2.24QACh. 2 - Prob. 2.25QACh. 2 - Prob. 2.26QACh. 2 - Prob. 2.27QACh. 2 - Prob. 2.28QACh. 2 - Prob. 2.29QACh. 2 - Prob. 2.30QACh. 2 - Prob. 2.31QACh. 2 - Prob. 2.32QACh. 2 - Prob. 2.33QACh. 2 - Prob. 2.34QACh. 2 - Prob. 2.35QACh. 2 - Prob. 2.36QACh. 2 - Prob. 2.37QACh. 2 - Prob. 2.38QACh. 2 - Prob. 2.39QACh. 2 - Prob. 2.40QACh. 2 - Prob. 2.41QACh. 2 - Prob. 2.42QACh. 2 - Prob. 2.43QACh. 2 - Prob. 2.44QACh. 2 - Prob. 2.45QACh. 2 - Prob. 2.46QACh. 2 - Prob. 2.47QACh. 2 - Prob. 2.48QACh. 2 - Prob. 2.49QACh. 2 - Prob. 2.50QACh. 2 - Prob. 2.51QACh. 2 - Prob. 2.52QACh. 2 - Prob. 2.53QACh. 2 - Prob. 2.54QACh. 2 - Prob. 2.55QACh. 2 - Prob. 2.56QACh. 2 - Prob. 2.57QACh. 2 - Prob. 2.58QACh. 2 - Prob. 2.59QACh. 2 - Prob. 2.60QACh. 2 - Prob. 2.61QACh. 2 - Prob. 2.62QACh. 2 - Prob. 2.63QACh. 2 - Prob. 2.64QACh. 2 - Prob. 2.65QACh. 2 - Prob. 2.66QACh. 2 - Prob. 2.67QACh. 2 - Prob. 2.68QACh. 2 - Prob. 2.69QACh. 2 - Prob. 2.70QACh. 2 - Prob. 2.71QACh. 2 - Prob. 2.72QACh. 2 - Prob. 2.73QACh. 2 - Prob. 2.74QACh. 2 - Prob. 2.75QACh. 2 - Prob. 2.76QACh. 2 - Prob. 2.77QACh. 2 - Prob. 2.78QACh. 2 - Prob. 2.79QACh. 2 - Prob. 2.80QACh. 2 - Prob. 2.81QACh. 2 - Prob. 2.82QACh. 2 - Prob. 2.83QACh. 2 - Prob. 2.84QACh. 2 - Prob. 2.85QACh. 2 - Prob. 2.86QACh. 2 - Prob. 2.87QACh. 2 - Prob. 2.88QACh. 2 - Prob. 2.89QACh. 2 - Prob. 2.90QACh. 2 - Prob. 2.91QACh. 2 - Prob. 2.92QACh. 2 - Prob. 2.93QACh. 2 - Prob. 2.94QACh. 2 - Prob. 2.95QACh. 2 - Prob. 2.96QACh. 2 - Prob. 2.97QACh. 2 - Prob. 2.98QACh. 2 - Prob. 2.99QACh. 2 - Prob. 2.100QACh. 2 - Prob. 2.101QACh. 2 - Prob. 2.102QACh. 2 - Prob. 2.103QACh. 2 - Prob. 2.104QACh. 2 - Prob. 2.105QACh. 2 - Prob. 2.106QACh. 2 - Prob. 2.107QACh. 2 - Prob. 2.108QACh. 2 - Prob. 2.109QACh. 2 - Prob. 2.110QACh. 2 - Prob. 2.111QACh. 2 - Prob. 2.112QACh. 2 - Prob. 2.113QACh. 2 - Prob. 2.114QACh. 2 - Prob. 2.115QACh. 2 - Prob. 2.116QA
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