Permanent hardness of water can be removed, by adding –

(1) Soda lime

(2) Sodium bicarbonate

(3) Washing soda

(4) Sodium chloride

The correct increasing order of the acidity of ${\mathrm{CO}}_2, {\mathrm{H}}_2\mathrm{O} \mathrm{and} {\mathrm{H}}_2{\mathrm{O}}_2$ is $-$

(1) ${\mathrm{CO}}_2<{\mathrm{H}}_2{\mathrm{O}}_2<{\mathrm{H}}_2\mathrm{O}$                         

(2) ${\mathrm{H}}_2\mathrm{O}<{\mathrm{H}}_2{\mathrm{O}}_2<{\mathrm{CO}}_2$

(3) ${\mathrm{H}}_2\mathrm{O}<{\mathrm{H}}_2{\mathrm{O}}_2>{\mathrm{CO}}_2$                         

(4) ${\mathrm{H}}_2{\mathrm{O}}_2>{\mathrm{CO}}_2>{\mathrm{H}}_2\mathrm{O}$

Metal which does not react with cold water but evolves H₂ with steam is –

(1) Na                           

(2) K

(3) Pt                             

(4) Fe

The oxidation states of the most electronegative element in the products of the reaction between ${\mathrm{BaO}}_2$ with dil. ${\mathrm{H}}_2{\mathrm{SO}}_4$ are–

1. 0, and –1                                   

2. –1, and –2

3. –2, and 0                                     

4. –2, and + 1

Hardness of water is due to presence of salts of –

(1) ${\mathrm{Na}}^{+} \mathrm{and} {\mathrm{K}}^{+}$                                  

(2) ${\mathrm{Ca}}^{2+} \mathrm{and} {\mathrm{Mg}}^{2+}$

(3) ${\mathrm{Ca}}^{2+} \mathrm{and} {\mathrm{K}}^{+}$                                

(4) ${\mathrm{Ca}}^{2+} \mathrm{and} {\mathrm{Na}}^{+}$

Ice melts at a temperature lower than its usual melting point when the pressure is increased. This is because-

1. Ice has less density than water

2. Pressure generates heat

3. The chemical bonds break under pressure

4. Ice is not a true solid

An ionic compound is dissolved simultaneously in heavy water and ordinary water, its solubility is –

1. Larger in heavy water

2. Smaller in heavy water

3. Solubility is same in both waters

4. Smaller in simple water

The oxide that gives hydrogen peroxide $\left({\mathrm{H}}_2{\mathrm{O}}_2\right)$ on the treatment with a dilute acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ is –

(A) ${\mathrm{MnO}}_2$                                  

(B) ${\mathrm{PbO}}_2$

(C) ${\mathrm{Na}}_2{\mathrm{O}}_2$                                 

(D) ${\mathrm{TiO}}_2$ 

The oxidising property of ${\mathrm{H}}_2{\mathrm{O}}_2$ is best explained by assuming that two oxygen atoms in its molecule are.

(1) Bonded differently

(2) Bonded similarly

(3) Bonded covalently

(4) Bonded by hydrogen bonds

${\mathrm{H}}_2\mathrm{O}$ has higher boiling point than ${\mathrm{H}}_2\mathrm{S}$ because.

(1) ${\mathrm{H}}_2\mathrm{O}$ is a smaller molecule and hence more closely packed

(2) The bond angle in ${\mathrm{H}}_2\mathrm{O}$ is more than in ${\mathrm{H}}_2\mathrm{S}$ and hence the former molecules are more tightly packed

(3) Of intermolecular hydrogen bonding in liquid ${\mathrm{H}}_2\mathrm{O}$

(4) The latent heat of vaporisation is higher for ${\mathrm{H}}_2\mathrm{O}$ than for ${\mathrm{H}}_2\mathrm{S}$