Solved: Calculate the number of cations and anions in each

Chemistry | 11th Edition | ISBN: 9780073402680 | Authors: Raymond Chang

ISBN: 9780073402680 118

Solution for problem 121P Chapter 3

Chemistry | 11th Edition

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Chemistry | 11th Edition

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Problem 121P

Calculate the number of cations and anions in each of the following compounds: (a) 0.764 g of CsI, (b) \(72.8\ \text{g}\ K_2Cr_2O_7\), (c) \(6.54\ \text{g}\ \text{of}\ Hg_2(NO_3)_2\).

Step-by-Step Solution:

Step 1 of 3

Here, we are asked to calculate the number of cations and anions in the given compounds

(a) Given: 0.764 g of CsI

First, let’s find the number of moles of CsI:

Molar mass of CsI (Caesium iodide) = 259.81 g/mol

We know 1 mole of any unit contains Avogadro’s number of particles. Here, 1 mole of the CsI element contains Avogadro’s number of ions, .

Therefore, number of ions will be:

= ions

Cs (Caesium) is a metal and hence these ions are the cations. Hence, there are ions.

The compound, CsI contains 1 atom of Cs and 1 atom of I (Iodine). A compound always contains an equal number of cations and anions in the neutral state. Hence, CsI has anions, i.e ions.

Step 2 of 3

Chapter 3, Problem 121P is Solved

Step 3 of 3

Textbook: Chemistry

Edition: 11

Author: Raymond Chang

ISBN: 9780073402680

This full solution covers the following key subjects: anions, calculate, cations, compounds, CSI. This expansive textbook survival guide covers 25 chapters, and 3591 solutions. The answer to “?Calculate the number of cations and anions in each of the following compounds: (a) 0.764 g of CsI, (b) \(72.8\ \text{g}\ K_2Cr_2O_7\), (c) \(6.54\ \text{g}\ \text{of}\ Hg_2(NO_3)_2\).” is broken down into a number of easy to follow steps, and 27 words. This textbook survival guide was created for the textbook: Chemistry, edition: 11. Chemistry was written by and is associated to the ISBN: 9780073402680. The full step-by-step solution to problem: 121P from chapter: 3 was answered by , our top Chemistry solution expert on 11/08/17, 03:59AM. Since the solution to 121P from 3 chapter was answered, more than 10129 students have viewed the full step-by-step answer.