a) first box is it n=MxV or n=MxV/1000 or n=M+V or n=M+V/1000 or n=V+M or n=V+M/1000second box is it 0.001 or 0.268 or 3.732 or 0.837 or 3731.524 or 0.000third box is it Unit = g/mol or L or mL or atoms or mol/L or g or molecules b) first box is it 1.674x0.474 or 1.674x(0.674-0.474) or 1.674x(0.674+0.474)or 1.674x0.674 or 1.674-(0.674+0.474)second box is it = 0.793 or 1.459 or 1.128 or 0.335 or 8.371 or 1.921 the third box is it g/mol or L or mL or atoms or mol/L or g or molecules Consider the following reaction between bariumchloride and sodium sulfate:BaCl (aq) + NazSo, (aq)Baso, (s) + 2NACI (aq)Suppose that 0.474 L of 1. 767 mol/L of bariumchlotide is added to excess amount of sodium sulfate(0.674 L of 2. 067 mol/L).a To calculate the number of moles of bariumchloride which were present initially before thereaction, we should use the following formula:The number of moles of barium chloride is::Unit =b) Suppose that 1.674 mol ot sodium chloride wasproduced from the reaction above, calculate theconcentration of chtoride ions in the final solution.The method of calculation is:* AnswerUnit =

The given reaction is : BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) The volume of BaCl2 =…

Answered: Consider the following reaction between…

Introductory Chemistry: An Active Learning Approach

6th Edition

ISBN:9781305079250

Author:Mark S. Cracolice, Ed Peters

Publisher:Mark S. Cracolice, Ed Peters

Chapter16: Solutions

Section: Chapter Questions

Problem 37E

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a) first box is it n=MxV or n=MxV/1000 or n=M+V or n=M+V/1000 or

n=V+M or n=V+M/1000

second box is it 0.001 or 0.268 or 3.732 or 0.837 or 3731.524 or 0.000

third box is it Unit = g/mol or L or mL or atoms or mol/L or g or molecules

b) first box is it 1.674x0.474 or 1.674x(0.674-0.474) or 1.674x(0.674+0.474)
or 1.674x0.674 or 1.674-(0.674+0.474)

second box is it = 0.793 or 1.459 or 1.128 or 0.335 or 8.371 or 1.921

the third box is it g/mol or L or mL or atoms or mol/L or g or molecules

Consider the following reaction between barium
chloride and sodium sulfate:
BaCl (aq) + NazSo, (aq)
Baso, (s) + 2NACI (aq)
Suppose that 0.474 L of 1. 767 mol/L of barium
chlotide is added to excess amount of sodium sulfate
(0.674 L of 2. 067 mol/L).
a To calculate the number of moles of barium
chloride which were present initially before the
reaction, we should use the following formula:
The number of moles of barium chloride is:
:Unit =
b) Suppose that 1.674 mol ot sodium chloride was
produced from the reaction above, calculate the
concentration of chtoride ions in the final solution.
The method of calculation is:
* Answer
Unit =

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