Science Answer the questions.[Che-5M]

1. Solid P: Calcium Carbonate (CaCO₃)

Gas Q: Carbon dioxide (CO₂)

Solid R: Calcium oxide (CaO)

2. Total mass of Q and R = 22gm + 28gm = 50gm
3. Total mass of Q and R (50gm) is equal to mass of reactant (50gm).
4. The law of conservation of mass is followed, i.e. total mass of product is equal to mass of reactant.
5. Law of conservation of mass is illustrated by the example.

$\ \ 2\text{Cu} \ \ \ \ \ \ +\ \ \ \ \ \ \ \ \ \ \ \text{O}_2\ \ \ \ \ \ \ \ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ }\ \ \ \ 2\text{CuO}\\ ^{(3.92\text{gm})} \ \ \ \ \ \ \ \ ^{(4.90-3.92\ =\ 0.98\text{gm})}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{(4.90\text{gm})}$

$\frac{3.92}{3.92}=1,\ \frac{0.98}{3.92}=0.25,\ \frac{4.90}{3.92}=1.25$

So, 1 equivalent of Cu reacts with 0.25 equivalent of O₂ to from 1.25 equivalent of copper oxide.

Reaction 2:

$2\text{CuO}\ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ }\ \ \ \ 2\text{Cu}\ \ \ +\ \ \ \ \ \ \ \ \ \ \text{O}_2\\ ^{(4.55\text{gm})}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{(3.64\text{gm})}\ \ \ \ \ \ ^{(4.55-3.64\ =\ 0.91\text{gm})}$

$\frac{4.55}{3.64}=1.25,\ \frac{3.64}{3.64}=1,\ \frac{0.91}{3.64}=0.25$

Here again, one can see that 1.25 equivalent of CuO decomposed to form 1 equivalent of Cu and 0.25 equivalent of oxygen.

Hence, law of constant proportion is verified.

Number of moles in 1g of sulphur dioxide:

Number of moles in 1g of oxygen:

$\frac{1}{32}$ mole of oxygen will contain $=\frac{\text{x}\ \times\ 64}{32}=2\text{x}$

1. $\ \text{Element}:\ \ \ \ \ \text{Ca}\ \ \ \ \ \ \ \ \ \ \text{F}\\ {\text{Valencies}}:\ \ +2\ \ \ \ \ \ -2$

Thus, the resulting compound is CaF₂

2. $\ \text{Element}:\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \text{S}\\ {\text{Valencies}}:\ \ +1\ \ \ \ \ -2$

Thus, the resulting compund is H₂S

3. $\ \text{Element}:\ \ \ \ \ \ \ \text{N}\ \ \ \ \ \ \ \ \ \ \text{H}\\ {\text{Valencies}}:\ \ -3\ \ \ \ \ \ +1$

Thus, the resulting compound is NH₃

4. $\ \text{Element}:\ \ \ \ \ \ \text{C}\ \ \ \ \ \ \ \ \ \ \ \text{Cl}\\ {\text{Valencies}}:\ \ +4\ \ \ \ \ \ -1$

Thus, the resulting compound is CCl₄

5. $\ \text{Element}:\ \ \ \ \ \text{Na}\ \ \ \ \ \ \ \ \ \ \text{O}\\ {\text{Valencies}}:\ \ +1\ \ \ \ \ \ -2$

Thus, the resulting compound is Na₂O

6. $\ \text{Element}:\ \ \ \ \ \ \ \text{C}\ \ \ \ \ \ \ \ \ \ \text{O}\\ {\text{Valencies}}:\ \ +4\ \ \ \ \ \ -2$

Thus, the resulting compound is CO₂

$\text{Element/ Ion}\ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \text{X}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ 2$

To stabilize the compound with hydrogen, two atoms of X are required. Therefore, the valency of X in this compound will be two.

CX₂:

 $\text{Element/ Ion}\ \ \ \ \ \ \ \text{C}\ \ \ \ \ \ \text{X}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\ \ \ \ \ \ 2$

To stabilize the compound with hydrogen, two atoms of X are required. Thus, the valency of X in this compound is two.

XO₂:

$\text{Element/ Ion}\ \ \ \ \ \ \ \text{X}\ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 4\ \ \ \ \ \ -2$

To stabilize this compound with oxygen, four atoms of X are required. Thus the valency of X in this compound if four.

XO₃:

$\text{Element/ Ion}\ \ \ \ \ \ \ \text{X}\ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \text{Valency}\ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\ \ \ \ \ \ -2$

To stabilize this compound with oxygen, six atoms of X are required. Thus, the valency of X in this compound is six.

1. Liquid X: Water.

Gas Y: Oxygen.

Gas Z: Hydrogen.

2. $\frac{\text{mass of Z}}{\text{mass of Y}}=\frac{22\text{gm}}{178\text{gm}}=1\ :\ 8$

3. Law of constant proportion is illustrated by this example.
4. Two sources of liquid X– Sea, Well.
5. Gas Y(oxygen) is necessary for breathing.