Chapter 4, Problem 152CP

a)

Interpretation Introduction

Interpretation: The total stream’s down flow rate has to be calculated.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

b)

Interpretation Introduction

Interpretation: the concentration of $\text{HCl}$ has to be determined.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

c)

Interpretation Introduction

Interpretation: The mass of $\text{CaO}$ consumed by the plant has to be determined.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$

d)

Interpretation Introduction

Interpretation: the concentration of ${\text{Ca}}^{\text{2+}}$ in ppm of second plant downstream has to be determined.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

$\text{Molarity(in}\text{M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{grams)}}{\text{Volume}\text{of}\text{solution(in}\text{litres)}}$