(a) Considering the following table.

Element	Atomic Number	Mass Number
J	9	19
Q	13	27
R	16	32
X	19	39
Y	24	52

(i) I. J is Halogen.
II. Y is most likely to be attracted by magnet.
III. X belongs to group I.
IV. R wduld readily form an ion with a double negative charge.
(ii) Type of bond that can exist between J and X when they combine: – Ionic/Electrovalent bond.
(iii) Neutrons in Q – 14
(iv) Compound formed when R combines with X: – X\(_2\)R.
(v) Element that exists as diatomic molecule: – Element J.
(vi) D-block element of the periodic table: Element Y.

(b)(i) Explain the term Atomic Orbital: This is a region/space within a principal energy level where there is high probability of finding an electron or an area where the electron lives.
(ii) I. Postulates of Dalton's atomic theory: –Matters are made up of small indivisible particles called atoms. –Atom can neither be destroyed nor created. –Atoms of the same element are identical but different from atoms of other elements. –Atoms combine (to form compounds) in simple whole number ratio.
II. Limitations of Dalton's Atomic Theory: –Atoms are divisible (protons neutrons and electrons). –Atoms can be created and destroyed (in nuclear reaction). –Existence of polymers (large) organic molecules. –Existence of isotopes.
(iii) Structure of sodium chloride in solid state: Soldium chloride (Nacl) as an ionic solid compound exists in crystal lattic/face centred cube which is made up of positive and negative ions.

(c) A sample of carbon is burnt at a rate of 0.50 g per second for 30 minutes to generate heat:
(i) Balance equation for the reaction: C\(_{(s)}\) + O\(_{2(g)}\) -> CO\(_{2(g)}\)
(ii) Determine the: I. Volume of carbon (IV) oxide produced at s.t.p. Mass of carbon burnt = 0.5 x 30 x 60 = 900 g from the reaction equation. 12 g of carbon at s.t.p produces 22.4 dm\(^3\) of CO\(_2\)
900 g of carbon = \(\frac{22.4}{12}\) x 900 = 1680 dm\(^3\)
II. Moles of oxygen used up in the process at s.t.p. from the reaction equation:
I mole of O\(^2\) used = 1 mol of CO\(_2\) produced.
But 22.4 of O\(_2\) = 1 mole of O\(_2\).
:. 1680 dm\(^3\) of O\(_2\) = \(\frac{1}{22.4}\) x 1680 = 75.0 mol
OR
Mol of O\(_2\) = \(\frac{900}{12}\) = 75.0 mol.