Answer

Mass of the compound $=0.24g$

Mass of boron $=0.096g$

Mass of oxygen $=0.144g$

$\text{Percentage of boron }=\frac{\text{Mass of boron}}{\text{Mass of compound}}\times 100=\frac{0.096g}{0.240g}\times 100=40\mathrm{\%}$

$\text{Percentage of oxygen }=\frac{\text{Mass of oxygen}}{\text{Mass of compound}}\times 100=\frac{0.144g}{0.240g}\times 100=60\mathrm{\%}$

Alternative method

Percentage of oxygen $=100-$ percentage of boron

$=100-40=60\mathrm{\%}$

Answer

First we find the proportion of mass of carbon and oxygen in carbon dioxide.

In $CO2,C:O=12:32$ or $3:8$

In other words, we can say that

$12.00g$ carbon reacts with oxygen $=32.00g$

$3.00g$ carbon will react with oxygen $=32\times 3=8g$

Therefore, $3.00g$ of carbon will always react with $8.00g$ of oxygen to form CO2 $(11g)$, even if large amount $(50.00g)$ of oxygen is present. This answer will be governed by ’the law of constant proportions'.

Answer

The group of atoms which carry a fixed charge (either positive or negative) on them and behave as ions are called polyatomic ions.

Example

(i) Carbonate ion

(ii) Sulphate ion

(iii) Ammonium ion

(iv) Phosphate ion

Answer

(i) Formula $=MgC{l}_2$ (Magnesium chloride)

(ii) Formula $=CaO$ (Calcium oxide)

(iii) Formula $=Cu(N{O}_3{)}_2$ (Copper nitrate)

(iv) Formula $=AIC{l}_3$ (Aluminium Chloride)

(v) Formula $=CaC{O}_3$ (Calcium carbonate)

Answer

(a) Quick lime- Calcium oxide - $CaO$

Elements-Calcium, oxygen.

(b) Hydrogen bromide- $HBr$

Elements- Hydrogen, bromine.

(c) Baking powder-Sodium hydrogen carbonate - $NaHC{O}_3$

Elements- Sodium, hydrogen, carbon, oxygen.

(d) Potassium sulphate- $K_{2}S{O}_4$

Elements- Potassium, sulphur, oxygen.

Answer

(a) Molar mass of $C_2{H}_2$

$=(2\times$ Atomic mass of $C)+(2\times$ Atomic mass of $H)$

$=(2\times 12)+(2\times 1)$

$=26u$

(b) Molar mass of $S_8$

$=(8\times$ Atomic mass of $S)$

$=8\times 32=256u$

(c) Molar mass of $P_4$

$=4\times$ Atomic mass of $P$

$=4\times 31=124u$

(d) Molar mass of $HCl$

$=$ Atomic mass of hydrogen + Atomic mass of $Cl$

$=1+35.5=36.5u$

(e) Molar mass of $HN{O}_3$

$=$ Atomic mass of $H+$ Atomic mass of $N+(3\times \text{ Atomic mass of }0)$

$=1+14+(3\times 16)=15+48=63u$