Question Library

The given reaction is endothermic, hence, a decrease in temperature will favor the exothermic side which is the backward reaction.

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

Ammonia \((NH_3\)) has a trigonal pyramidal structure with a bond angle of approximately 107 degrees. Methane (\(CH_4\)) has a tetrahedral structure with a bond angle of 109.5 degrees. Carbon dioxide is a linear molecule, with a 180-degree bond angle. Water \((H_2O\)) has a bent or V-shaped structure with a bond angle of approximately 104.5 degrees.

The azimuthal quantum number determines an atom's orbital angular momentum and the shape of the orbital. The principal quantum number determines its number of shells. Magnetic quantum number determines how many orbitals there are as well as their orientation within a subshell.

Ammonium nitrite = NH\(_4\)NO\(_2\) NH\(_4 ^+\): Let the oxidation number of Nitrogen = x x + 4 = 1 \(\implies\) x = 1 - 4 x = -3 NO\(_2 ^-\): x - 4 = -1 x = -1 + 4 \(\implies\) x = +3. The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3.

Hydrogen bonding is a type of Van der Waals force, specifically a strong dipole-dipole interaction. It's a distinct type of intermolecular attraction, not a covalent bond, though it's stronger than most other Van der Waals forces like London dispersion forces.

Soda lime is a mixture of caustic soda (NaOH) and lime water (Ca(OH)\(_2\)).

Process: Add water to the mixture and then filter. The resultant is a salt solution as your filtrate and a mixture of iodine and sand as your residue. Evaporate the filtrate to dryness to get your sodium chloride (salt). Heat the residue to sublime the iodine from the sand. Addition of water \(\to\) Filtration \(\to\) Evaporation \(\to\) Sublimation.

Equation: Ca(HCO\(_3\))\(_2\) \(\to\) CaCO\(_3\) (precipitate) + H\(_2\)O + CO\(_2\) (in the presence of heat) V = 200 cm\(^3\) = 0.2 dm\(^3\) C = 0.5 mol/dm\(^3\) = 0.5M N = CV \(\implies\) N = 0.5 \(\times\) 0.2 = 0.1 mole From the equation 1 mole of Ca(HCO\(_3\))\(_2\) gives 100g of CaCO\(_3\) (1 mole CaCO\(_3\) = 40 + 12 + (3 x 16) = 100g\) 0.1 mole of Ca(HCO\(_3\))\(_2\) gives x g of CaCO\(_3\). \(\frac{1}{0.10} = \frac{100}{x}\) \(x = 100 \times 0.10 = 10 g\)

A reaction is spontaneous when the Gibbs free energy change (\DeltaG) is negative. This is determined by the equation \DeltaG = \DeltaH - T\DeltaS, where \DeltaH is the enthalpy change, T is the temperature, and \DeltaS is the entropy change. Therefore, a reaction is spontaneous if \DeltaH is negative and T\DeltaS is negative, or if \DeltaH is positive and T\DeltaS is positive and greater than \DeltaH. In other words, for a reaction to be spontaneous, the decrease in enthalpy (\DeltaH) must be greater than the increase in entropy (T\DeltaS). In summary: For a reaction to be spontaneous, either \DeltaH must be negative (exothermic) and T\DeltaS must be smaller than \DeltaH, or T\DeltaS must be positive and larger than \DeltaH. The key is to ensure that the decrease in Gibbs free energy (\DeltaG < 0) is negative. Applying values to those parameters, options A, B and C appear correct because they'll all make \DeltaG negative but the one that will mostly enhance spontaneity is option A because \DeltaG = -ve, \DeltaH = - ve.

It should be noted that for X to liberate CO\(_2\), X must be a carbonate or an oxalate. Since X decolorizes KMnO\(_4\), X must be an oxalate. Therefore, X is H\(_2\)C\(_2\)O\(_4\).

\(^{32} _{15}\)P \(\to\) 2\(^{0} _{-1}\)e + \(^{a} _{b}\)X 32 = 2(0) + a a = 32 15 = 2(-1) + b b = 15 + 2 b = 17 \(\therefore\) \(^{a} _{b}\)X = \(^{32} _{17}\)Cl

Using Graham's Law of Diffusion, \(\frac{R_1}{R_2} = \sqrt{\frac{m_2}{m_1}}\) \(\frac{R_H}{R_O} = \sqrt{\frac{m_O}{m_H}}\) \(\frac{R_H}{R_O} = \sqrt{\frac{32}{2}}\) \(\frac{R_H}{R_O} = \sqrt{16} = 4\) \(\therefore R_H = 4 \times R_O\) Therefore, hydrogen diffuses 4 times faster than oxygen.

X\(^{3+}\) + 3e\(^-\) \(\to\) X 3F = 27g xF = 10g \(\frac{x}{3} = \frac{10}{27} \implies x = \frac{10}{9} F\) \(\frac{10}{9}\)F \(\equiv\) N20.00 1F is equivalent to x \(\frac{1}{\frac{10}{9}} = \frac{x}{20}\) \(\frac{9}{10} = \frac{x}{20} \implies x = N18.00\) 1F is equivalent to N18.00. Y\(^{2+}\) + 2e\(^-\) \(\to\) Y 2F = 24g xF = 6g x = \(\frac{6 \times 2}{24} = \frac{1}{2} F\) 1F = N18.00 \(\frac{1}{2}\)F = \(\frac{1}{2} \times N18.00\) = N9.00

For X: X\(^{2+}\) + 2e\(^-\) \(\to\) X 2F = 63g xF = 6g x = \(\frac{6 \times 2}{63} = \frac{4}{21} F\) \(\frac{4}{21}\)F = N12.00 1F = \(\frac{12}{\frac{4}{21}}\) = N63.00 1F is equivalent to N63.00. For Y: Y\(^{3+}\) + 3e\(^-\) \(\to\) Y 3F = 27g xF = 9g x = \(\frac{3 \times 9}{27}\) = 1F 1F = N63.00