Chapter 7, Problem 7.135MP

(a)

Interpretation Introduction

Interpretation:

The moles of sulphur are present in the ${\text{MS}}_{\text{n}}$ sample should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two soluble salt solutions are combined together is known as precipitation reaction.

Mole:

Mole of the compound present in the solution is calculated by molarity concentration of the compound multiplied with litre volume of solution.

$\text{Mole}\text{=Molarity}\text{×Liter}$

Mass:

Mass of the compound is calculated by mole of the compound is multiplied with molar mass of the compound.

$\text{Mass}\text{=}\text{Molar}\text{mass}\text{×}\text{mole}$

Mass percent:

The mass percent of analyte in give sample is given by below equation,

$\text{Mass}\text{\%}\text{=}\frac{\text{Mass}\text{of}\text{analyte}}{\text{Mass}\text{of}\text{sample}}\text{×100}$

(b)

Interpretation Introduction

Interpretation:

The atomic weight of M should be calculated, when n is assumed to be $\left(\text{n = 1, 2, 3 }\text{. }\text{. }\text{.}\right).$

Concept Introduction:

Precipitation reaction:

If precipitate is formed, when two soluble salt solutions are combined together is known as precipitation reaction.

Mole:

Mole of the compound present in the solution is calculated by molarity concentration of the compound multiplied with litre volume of solution.

$\text{Mole}\text{=Molarity}\text{×Liter}$

Mass:

Mass of the compound is calculated by mole of the compound is multiplied with molar mass of the compound.

$\text{Mass}\text{=}\text{Molar}\text{mass}\text{×}\text{mole}$

Mass percent:

The mass percent of analyte in give sample is given by below equation,

$\text{Mass}\text{\%}\text{=}\frac{\text{Mass}\text{of}\text{analyte}}{\text{Mass}\text{of}\text{sample}}\text{×100}$

(c)

Interpretation Introduction

Interpretation:

The atomic weight of M should be calculated.

Concept introduction:

Precipitation reaction:

If precipitate is formed, when two soluble salt solutions are combined together is known as precipitation reaction.

Mole:

Mole of the compound present in the solution is calculated by molarity concentration of the compound multiplied with litre volume of solution.

$\text{Mole}\text{=Molarity}\text{×Liter}$

Mass:

Mass of the compound is calculated by mole of the compound is multiplied with molar mass of the compound.

$\text{Mass}\text{=}\text{Molar}\text{mass}\text{×}\text{mole}$

Mass percent:

The mass percent of analyte in give sample is given by below equation,

$\text{Mass}\text{\%}\text{=}\frac{\text{Mass}\text{of}\text{analyte}}{\text{Mass}\text{of}\text{sample}}\text{×100}$

(d-1)

Interpretation Introduction

Interpretation:

The balanced equation for the given reaction should be given.

Concept Introduction:

Balanced equation:

The same number of atoms present in both reactant and product side of the equation is known as balanced equation.

Hints to balancing chemical equations:

To find the charge (valance) of each species in the reaction then adjust the coefficients not a subscripts to make an equal counts of atoms present in both sides.

(d-2)

Interpretation Introduction

Interpretation:

The balanced equation for the given reaction should be given.

Concept introduction:

Balanced equation:

The same number of atoms present in both reactant and product side of the equation is known as balanced equation.

Hints to balancing chemical equations:

To find the charge (valance) of each species in the reaction then adjust the coefficients not a subscripts to make an equal counts of atoms present in both sides.

(d-3)

Interpretation Introduction

Interpretation:

The balanced equation for the given reaction should be given.

Concept Introduction:

Redox reaction:

The reduction and oxidation reactions are occur simultaneously with in a single reaction is known as redox reaction.

Balancing of redox reaction:

The unbalanced charge (electron) and atoms (ions) in the equation of redox reactions are commonly balanced by half reaction method.

In this method, the given redox reaction is separated by two half reactions, one is oxidation and another one is reduction.

The medium of the reaction is used to balance the Hydrogen and Oxygen ions in the half cell reactions.

Balancing of redox reaction in acid medium:

In the acid medium, the addition of ${\text{H}}^{+}$ ion to balance the Hydrogen and water is added to balance the Oxygen atom.

Charge is balanced by addition of electron in the sides with the greater positive charge.

Balancing of redox reaction in basic medium:

In the basic medium, the addition of ${\text{OH}}^{-}$ ion and electron to balance the charge and Hydrogen and Oxygen atoms are balanced by addition of water in the equation.

(d-4)

Interpretation Introduction

Interpretation:

The balanced equation for the given reaction should be given.

Concept Introduction:

Balanced equation:

The same number of atoms present in both reactant and product side of the equation is known as balanced equation.

Hints to balancing chemical equations:

To find the charge (valance) of each species in the reaction then adjust the coefficients not a subscripts to make an equal counts of atoms present in both sides.