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When chlorine reacts with water in the presence of sunlight, it produces oxygen gas and hydrochloric acid as the primary products; essentially, the sunlight breaks down the chlorine in water, liberating oxygen gas. 2Cl\(_2\) + 2H\(_2\)O → 4HCl + O\(_2\)

In an exothermic reaction, an increase in temperature will favour the reactants (shift the reaction towards the left), meaning the reaction will produce less product as heat is considered a product in an exothermic reaction; according to Le Chatelier's principle, the system will try to counteract the added heat by favouring the reactant side. Thus, the reverse reaction will be favoured.

A Daniell cell undergoes a redox reaction where zinc is oxidized at the anode, releasing electrons that travel through a circuit to the cathode where copper ions are reduced to solid copper, generating an electric current. Oxidation half-reaction: Zinc metal (Zn) is oxidized to zinc ions (Zn\(^{2+}\)) at the anode, releasing electrons. Zn(s) → Zn\(^{2+}\) + 2e\(^-\) Reduction half-reaction: Copper ions Cu\(^{2+}\) are reduced to solid copper metal (Cu) at the cathode, accepting electrons. Cu\(^{2+}\) + 2e\(^-\) → Cu Overall reaction: Zn (s) + Cu\(^{2+}\) (aq) → Zn\(^{2+}\) (aq) + Cu (s).

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option B.

Tartaric acid is used industrially as an acidulant in food products like carbonated drinks, jellies, and baking powder, as well as in metal cleaning and polishing processes due to its chelating properties; it also finds applications in textile processing and certain photographic development procedures. The major industrial use of tartaric acid is as an acidulant in the food and beverage industry, primarily used to add a tart flavour to products like carbonated drinks, fruit jellies, candies, and wine, often in combination with other acids to achieve the desired taste profile; it is also used as a leavening agent in baking due to its reaction with baking soda when heated.

No expanded working has been attached for this question yet. The verified answer is Option D.

Number of moles = (conc. in mol/dm3*volume in cm3)/1000 (0.75 * 20)/100 = 0.015 mole

H2SO4 (O.N of hydrogen * 2 ) + O.N of sulphur + O.N of oxygen * 4 = 0 ∴ 1 * 2 + 6 + 4x = 0 2 + 6 + 4x = 0 8 + 4x = 0 4x 0 - 8 = -8 4x = -8 x = -8/4 = -2

MnO\(_2\)(s) + xHCl(aq) → MnCl\(_2\)(aq) + yH\(_2\)O(l) + zCl\(_2\)(g) First, balance the chemical equation, then equate each of the letters with the coefficients. On balancing, we have : MnO\(_2\)(s) + 4HCl(aq) → MnCl\(_2\)(aq) + 2H\(_2\)O(l) + Cl\(_2\)(g) Therefore, x = 4,y = 2 and z = 1

2 * 35.5g of chlorine has a volume of 22.4 dm3 at s.t.p ∴ xg of chlorine will have a volume of 1.12 dm3 x = (1.12 * 35.5 * 2)/22.4 = 3.55g

In 1909, Robert Millikan and Harvey Fletcher conducted the oil drop experiment to determine the charge of an electron.

CH\(_4\)(g) + 2O\(_2\)(g) → 2H\(_2\)O(g) + CO\(_2\)(g) First, ensure the chemical equation is balanced, stoichiometrically. i.e CH\(_4\) reacts with 2O\(_2\) in ratio 1 : 2 If 1 mole of CH\(_4\) reacts with 2 moles of O\(_2\) Then 45 cm\(^3\) CH\(_4\) react with X cm\(^3\) O\(_2\) i.e 1cm\(^3\) = 2cm\(^3\) 45cm\(^3\)= X cm\(^3\) X = 45 x 2 = 90 cm\(^3\)

N.B : Atomic number = no. of protons = no. of electrons = 15 Orbital configuration of the neutral atom of the element = 2, 8, 5 which represents the shells starting from the first which is K, L, M respectively. i.e K = 2, L = 8 , M = 5. Therefore, the L - shell has 8 electrons.