Correct Answer: B. Sodium trioxonitrate (V)

Explanation

(a)(i) Postulates of the kinetic theory of gases: Gas molecules move randomly in straight lines. They collide with one another and with the walls of the container. The temperature of the gas is a measure of the average kinetic energy of the molecule. The forces of attraction between gas molecules are negligible. The volume occupied by gas molecules is negligible compared to the volume of the container. The collision of the molecules is perfectly elastic (any three).

(ii) Sketches to illustrate Boyle's Law

(iii) Absolute zero of temperature is the temperature at which the volume of a gas would be (theoretically) reduced to zero. OR It is the temperature at which all motion is assumed to have ceased. OR It is the lowest temperature theoretically possible OR It is temperature at which all gases liquefy or solidify OR the entropy of the system is at rest.

(b) If equal volumes of gas X and oxygen are at the same pressure but at 200 K and 800 K respectively:

(i) X contains the greater number of molecules;Reason: Since X is at a lower temperature, it must contain more molecules to be able to exert the same pressure Alternatively n = \(\frac{PV}{RT}\)

n\(_x\) = \(\frac{PV}{200R}\)

nO\(_2\) = \(\frac{PV}{800R}\)

Hence, n\(_x\) is greater

(ii) To calculate molar mass of X if one molecule of it has a mass of 2.19 x 10\(^{-22}\)g. Calculation: 1 mole of x contains 6.02 x 10\(^{23}\) molecule.

1 mole of x will have a mass of (2.19 x 10\(^{-22}\) x 6.02 x 10\(^{23}\))g = 131.84 g mo1\(^{-1}\) ; approximately 132g mol\(^{-1}\).

(iii) Comparison of rate of diffusion of X with that of oxygen: O\(_2\) will diffuse faster. The lighter the gas, the faster the rate of diffusion. The molar mass of O\(_2\) is 32g. Hence it is lighter than X and will diffuse faster.

(c)(i) Increasing the pressure or increasing the concentration of reactions of the left hand side.

-lowering the temperature of the system.

-Continuous removal of NH\(_3\) as it is produced. (any two)

(ii) Energy profile diagram for an endothermic reaction.

(iii) in the reaction represented by the following equation W_(s)\) —> Energy Yv + Z\(_(s)\). Reaction rate can be affected by the following factors apart from temperature.

- Surface area / particle size

- Presence of catalyst.

Correct Answer: C. 4

Correct Answer: A. C2H2CI2

Correct Answer: B. losing two electrons

Correct Answer: C. Ethyne

Correct Answer: A. 1s₂2s₂2p₅

Correct Answer: D. esterification

Correct Answer: A. 4.0g

Explanation

(a)(i) - Covalent compounds contain only molecules while electrovalent compounds do not contain molecules

- Covalent compounds usually dissolve in covalent organic Products compounds but not in water but electrovalent compounds are soluble in water

- Covalent compounds are usually volatile with low melting and boiling points but electrovalent compounds usually have high melting and boiling points.

(ii) \(_{12}\)K has an electronic structure of 1s\(^2\)2\(^2\)2p\(^6\)3s\(^2\); L has the structure of 1s\(^2\)2s\(^2\)2p\(^4\); belongs to group II and L to group VI

(iii) Alkali metal M cannot be extracted by reduction of the oxide with oxide coke because of its high stability and reactivity.

(b)(i) The half-life of a radioactive element is the time it takes for one half of any quantity of a radioactive molecule to decay. It is therefore also equal to the time for the amount of the radioactive nuclide present to fall from one half to one quarter, from one quarter to one eight, etc.

(ii) The radioactive isotope normally used in dating archaelogical specimens is carbon - 14.

(iii) \(^{28}_{13}Al\) -> \(^{28}_{14} Si\) + \(^0_{-1}Q\)

(c)(i) - electropositive

- form electrovalent chlorides

- Replace the hydrogen in acid if more electropositive than hydrogen

-Form metallic hydrides/ionic hydrides.

(ii) Duralumin is preferred to steel in aircraft manufacture because duralumin is lighter and more resistant to corrosion.

(iii) The ability of metals to be hammered into thin sheet is described as malleability.

(iv) To calculate the number of mole of electrons involved in the oxidation of 2.8g of iron filing to Fe\(^{2+}\) + Fe —> Fe\(^{2+}\) + 2e\(^{-}\) ; 56g of Fe loses 2 mole of electrons.

2.8 of Fe would lose \(\frac{2.8 \times 2}{56}\) = 0.10 mole