(a) Mention one oxide in each case, which (i) used in bleaching (ii) is a redish-brown gas (iii) reacts with NaOH and also with HCI; (iv) dissolves in water to give a solution with pH greater than 7; (v) oxidizes hot, concentrated HCI to chlorine
(b)(i) State three methods that can be used to removo hardness in a sample of water that contains calcium hydrogentrioxocarbonate (IV). (ii) Explain with the aid of appropriate equation, why it is not advisable to build a house with limestone in an environment polluted by sulphur (IV) oxide.
(c)(i) List two compounds of potassium which yield oxygen when heated strongly (ii) Calculate the amount (in moles) of gas which occupies 250 cm\(^3\) at s.t.p. [1 mole of gas occupies 22.4 dm\(^3\) at s.t.p] (iii) If 250 cm\(^3\) of a gas at s.t.p. is heated to 27°C at constant pressure, calculate its new volume. (iv) Explain in terms of the collision theory what happens as a gas is heated at constant pressure.
Explanation
(a) An oxide which fits the following descriptions: (i) Used in bleaching - SO\(_2\) (ii) Is a reddish-brown gas NO\(_2\) (iii) Reacts with NaOH and also with HCI - ZnO/Al\(_2\)O\(_3\)/SiO\(_2\)/BeO/PbO/SnO\(_2\) (iv) Dissolves readily in water to give a solution with pH greater than 7 - Na\(_2\)O/K\(_2\)O or CaO and MgO. (v) Oxidizes hot concentrated HCI to chlorine -MnO\(_2\) or PbO\(_4\) or Pb\(_3\)O\(_2\)
(b)(i) Removal of hardness in a sample of water that contains Ca(CO\(_3\))\(_2\). - By boiling/distillation - Adding washing soda (N\(_2\)CO\(_3\).10H\(_2\)O) -Passing through ion-exchange resin/permutit process -Adding a calculated amount of slacked lime -Adding sodium trioxophosphate (V) NaPO\(_3\). (ii) It is not advisable to build a house with limestone in an environment polluted by SO\(_2\) for the following reason. SO\(_2\) forms acid rain which reacts.with the limestone and wears away the building. CaCO\(_3\) + SO\(_2\) + H\(_2\)O -> CaSO\(_3\) + H\(_2\)O + CO\(_2\)
(c)(i) KNO\(_3\)/KCIO\(_3\)/K\(_2\)O\(_2\) yield oxygen when heated strongly. (ii) To calculate the cmount oT gas which occupies 250cm\(^3\) at s.t.p. 1 mole occupies 22400cm\(^3\) at s.t.p.Hence \(\frac{250}{22400}\) mole occupies 0.00112 moles (iii) \(\frac{V_1}{T_1} = \frac{V_2}{T_2}; \frac{250}{273} = \frac{V_2}{300}; v_2 = \frac{250 \times 300}{273} = 275cm^3\) (iv) When a gas is heated at constant pressure, the molecules acquire more kinetic energy and move faster. They collide with one another and with the walls of the container more frequently. To maintain the Same number of collisions on the walls of the container (i.e keep pressure constant) the volume of the gas increases.
