CHEMISTRY_SOLUTION_9IC

GENERAL SOLUTIONSfull-width 

ANSWER 1

ANSWER 2

QUESTION 16 - SOLUTION

A) Tricalcium Phosphate

b) Potassium Carbonate

c) Potassium Manganate

d) Manganese Borate

e) Magnesium Hydrogen Carbonate

f) Sodium Pentacyanonitrosylferrate

g) Barium Chlorate

h) Silver Sulfite

I) (i) - Lead (ii) - Acetate

j) Sodium Metasliciate

QUESTION 17 SOLUTIONS 

1- BaSo4
2- Bi(No3)3 5H2O
3- CaBr2
4- Cr2(SO4)3
5- FeS
6- Ca2SiO4
7- C6H6FeK4N6O3
8- SnO2
9- H4Na2O4Zn
10- Mg3(PO4)2
11- NiSO4(H2O)6
12- K2MnO4

QUESTION 18 SOLUTIONS 

A)  Naclo = Sodium hypochlorite
.
B)        Naclo2 = Sodium chlorite

C)       Naclo3 = Sodium chlorate

D)  Naclo4 = Sodium perchlorate

QUESTION 20 SOLUTIONS

A) - NaOH + H2SO4 → Na2SO4 +H2O2 by balancing the equation you get, 

2NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O

B) - H2SO4 + 2KHCO3 → K2SO4 +2H2O +2CO2 by balancing the equation you get, 

2KHCO3 + H2SO4 → K2SO4 + 2CO2 + 2H2O

C) - Fe + H2SO4  → FeSO4 +H2 by balancing the equation you get, 

Fe + H2SO4 → FeSO4 +H2
D) - Al + NaOH + H2O --> NaAlO2 + H2 To balance any equation, we have to equalize no of each entity on both sides.
2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

E) - Silver nitrate → silver + nitrogen dioxide + oxygen 

Balanced Equations is 2AgNO3 →2 Ag + 2NO2 + O2

H) - Barium chloride + sulphuric acid → Barium sulphate + hydrochloric acid 

To balance equation is BACL2+H2SO4 → BASO4+2HCL

QUESTION 21 SOLUTIONS

a) HgO = Solution

The first step is finding the total mass 

Calculate the number of each element in the formula 

Here we can see that ,there is 1 Hg and 1 O 

So adding their mass together will give you the total mass 

Hg+O 

200+16= 216 

> Second step 

mass of the required element/total mass 

mass of oxygen / total mass 

16/216 

> Third step, find the percentage 

16/216 multiply by 100%

= 7.40 Answer

b) k₂Cr₂O₇ = Solution

Mass = 2(39) + 2(52) + 7(16)

= 78 + 104 + 112

= 294

% of O₂ = 112/294 multiply by 100

= 11200/249

= 38.09 Answer

c) Al₂(SO₄)₃ = Solution

To calculate the percentage of oxygen in aluminum sulfate, Al₂(SO₄)₃  , we need to determine the molar mass of this salt.

The atomic masses of aluminum (Al), sulfur (S) and oxygen (O) are 27, 32 and 16, respectively.

Thus, the molar mass of aluminum sulfate = 2 x 27 + 3 x 32 + 4 x 3 x 16

= 342 gm/mole

The mass contributed by oxygen = 4 x 3 x 16 = 192 gm

% contribution of oxygen = mass of oxygen x 100/ mass of aluminum sulfate

= 192/342 x 100 = 56.14% Answer

Thus, oxygen contributes about 56% of mass to aluminum sulfate. Similarly, sulfur contributes 28.1% (96/342 x 100) and aluminum contributes 15.8% to the total mass of aluminum sulfate.  

This salt is generally sold as alum and is used as a coagulating agent in water treatment process.

Question 24 Solution

1.Compound contains 17.7% hydrogen and 82.3% Nitrogen
i.e if 100g of the compound is present, 17.7g of Hydrogen and 82.3g of Nitrogen is present

2. Converting the weight into number of moles of each element: We divide the mass of each element by the molar mass
Moles of hydrogen = 17.71=17.7$\frac{17.7}{1}=17.7$
Moles of Nitrogen =82.314=5.88$\frac{82.3}{14}=5.88$

3. We divide the mole value obtained by the smallest number ;
i.e   N: H5.885.88:17.75.88  1: 3$$\begin{gathered} N: H \\ \frac{5.88}{5.88}:\frac{17.7}{5.88} \\ 1: 3 \end{gathered}$$

Hence the formula of the compound is NH3

[Please note that you need to divide mole value by the lowest /smallest number, to get the correct answer. in response to your query , I would like to mention that In order to know the formula, in case of doubts you need to arrange the atoms in the form of common existing order, like H3N is not a common compound , but NH3 is. Though both of these have the same number of atoms]

Question 29 Solution

Percent composition of N is 35%.

Percent composition of H is 5%.

Percent composition of O is 60%.

Explanation:

= 2×14g+4×1g+3×16g=80g

The mass of 1 mol of NH4NO3 is 80g.

= 2×14g=28g

The mass of N is 28g.

Percent composition of N is 28g80g×100%=35%.