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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.

When the electrons are drawn out in the orbitals, following the Hund's rule, the diagram above is what will be obtained.

When a gas is compressed to half its volume, the average intermolecular distance decreases because the same number of gas molecules are now confined to a smaller space, forcing them closer together, resulting in a smaller average distance between each molecule. Conversely, when the volume of a gas is reduced, the frequency of collisions between molecules increases; this is because the molecules have less space to move around, leading to more frequent interactions with each other and the container walls.

If the metal X forms two bromides with the formulae XBr\(_2\) and XBr\(_3\), it simply means it has variable oxidation number or states which is one of the characteristics of transition metals. This implies X is a metal. However, whenever a metal combines with a non-metal, there is usually a transfer of electrons from the metals to the non-metals, and the type of bond formed is the ionic or electrovalent bond.

Petrol and water are immiscible liquids, so the best method of separation of petrol and water is by a separating funnel.

Aromatic compounds are chemical compounds that consist of conjugated planar ring systems accompanied by delocalized pi-electron clouds in place of individual alternating double and single bonds. In a simpler term, aromatic compounds are compounds of Benzene and its derivatives.e.g Benzene, Phenol, Toluene, xylene, anthracene, Naphthalene etc all cyclically conjugated. The compound in option A is called phenol - a benzene ring with an hydroxyl group (OH) attached to C-1.