Calculate each of the following quantities:(a) Total Problem 14P Chapter 3

Principles of General Chemistry | 2nd Edition

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Problem 14P

Calculate each of the following quantities:

(a) Total number of ions in 38.1 g of SrF2

(b) Mass in kilograms of 3.58 mol of CuCl2·2H2O

(c) Mass in milligrams of 2.88×1022 formula units of Bi(NO3)3·5H2O

Step-by-Step Solution:

Solution 14P:

Step 1:

(a) Total number of ions in 38.1 g of SrF2

Here, we have to calculate the total number of ions in 38.1 g of SrF2.

The molar mass of SrF2 = 125.62 g/mol

Thus,

The amount of SrF2 in moles = 38.1 g of SrF2  = 0.30329 moles of SrF2

Given that, total moles of ions in SrF2 = 1.0 mol of Sr2++ 2.0 moles of F- =3.0 moles

Therefore,

Total moles of ions in 0.30329 moles of SrF2

= (0.30329 moles of SrF2) = 0.90988 mol ions

We know that,

1.0 mole of ions = 6.022 ions

Thus, the number of total ions = ( 0.90988 mol ions)  = 5.479 ions

Therefore, 5.479 ions are in 38.1 g of SrF2

Step 2:

(b) Mass in kilograms of 3.58 mol of CuCl2·2H2O

Here, we have to calculate the total mass  in kilograms of 3.58 mol of CuCl2·2H2O

The molar mass of CuCl2·2H2O = 170.48 g/mol

1000.0 g =1.0 kg

Given that,

Mass of CuCl2·2H2O =3.58 g

Therefore,

The mass of CuCl2·2H2O = 3.58 g CuCl2·2H2O   = 0.610 kg

Thus, the mass  in kilograms of 3.58 mol of CuCl2·2H2O is 0.610 kg.

Step 3 of 3

ISBN: 9780073511085

This full solution covers the following key subjects: mass, calculate, formula, ions, kilograms. This expansive textbook survival guide covers 23 chapters, and 1878 solutions. Principles of General Chemistry was written by Sieva Kozinsky and is associated to the ISBN: 9780073511085. This textbook survival guide was created for the textbook: Principles of General Chemistry, edition: 2nd. The full step-by-step solution to problem: 14P from chapter: 3 was answered by Sieva Kozinsky, our top Chemistry solution expert on 08/31/17, 03:54AM. The answer to “Calculate each of the following quantities:(a) Total number of ions in 38.1 g of SrF2(b) Mass in kilograms of 3.58 mol of CuCl2·2H2O(c) Mass in milligrams of 2.88×1022 formula units of Bi(NO3)3·5H2O” is broken down into a number of easy to follow steps, and 32 words. Since the solution to 14P from 3 chapter was answered, more than 853 students have viewed the full step-by-step answer.

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