Here are the solutions to the questions: 1) B) Ba > Mg Step 1: Atomic size increases down a group in the periodic table. Step 2: Barium (Ba) is below Magnesium (Mg) in Group 2. Step 3: Therefore, Ba has a larger atomic radius than Mg. 2) D) Mg$^{2+}$ Step 1: Heat of hydration is directly proportional to ionic charge and inversely proportional to ionic size. Step 2: Among the given ions, Mg$^{2+}$ has the highest charge (+2) and is relatively small compared to Ba$^{2+}$, Na$^+$, and Cs$^+$. Step 3: Smaller size and higher charge lead to stronger electrostatic attraction with water molecules, resulting in a higher heat of hydration. 3) B) B Step 1: Tincal is a natural mineral. Step 2: Its chemical formula is Na$_2$B$_4$O$_7 \cdot 10$H$_2$O, which is hydrated sodium borate. Step 3: This mineral is a primary source of Boron. 4) B) N$_2$O Step 1: Laughing gas is the common name for dinitrogen monoxide. Step 2: The chemical formula for dinitrogen monoxide is N$_2$O. 5) B) Cl$_2$O$_7$ Step 1: An acid anhydride is formed by removing water from an acid molecule. Step 2: For perchloric acid (HClO$_4$), two molecules combine to lose one water molecule: $$2\text{HClO}_4 - \text{H}_2\text{O} \rightarrow \text{H}_2\text{Cl}_2\text{O}_8 - \text{H}_2\text{O} = \text{Cl}_2\text{O}_7$$ Step 3: Therefore, the anhydride of HClO$_4$ is Cl$_2$O$_7$. 6) D) Co Step 1: A typical transition metal forms at least one ion with a partially filled d subshell. Step 2: Scandium (Sc) and Yttrium (Y) form Sc$^{3+}$ and Y$^{3+}$ ions, which have empty d-orbitals. Step 3: Radium (Ra) is an alkaline earth metal. Step 4: Cobalt (Co) forms ions like Co$^{2+}$ ([Ar]3d$^7$) and Co$^{3+}$ ([Ar]3d$^6$), both of which have partially filled d-orbitals. 7) C) sp$^2$ Step 1: The hybridization of atomic orbitals determines the molecular geometry. Step 2: sp hybridization results in a linear shape. Step 3: sp$^2$ hybridization results in a planar triangular (trigonal planar) shape. Step 4: sp$^3$ hybridization results in a tetrahedral shape. Step 5: dsp$^2$ hybridization results in a square planar shape. 8) A) CHCl$_3$ Step 1: Chloroform is a common organic solvent. Step 2: Its chemical name is trichloromethane. Step 3: The formula for trichloromethane is CHCl$_3$. 9) B) NO$_2^+$ Step 1: Nitration of benzene is an electrophilic aromatic substitution reaction. Step 2: The nitrating mixture (concentrated nitric acid and sulfuric acid) generates the electrophile. Step 3: The active nitrating agent is the nitronium ion, NO$_2^+$. 10) D) E$_1$ and SN$_1$ Step 1: E$_1$ (unimolecular elimination) and SN$_1$ (unimolecular nucleophilic substitution) mechanisms are both two-step processes. Step 2: The first step in both E$_1$ and SN$_1$ is the dissociation of the leaving group to form a carbocation intermediate. 11) C) Urease Step 1: Fermentation of starch involves enzymes that break down starch into sugars and then ferment those sugars. Step 2: Diastase (amylase) breaks down starch. Zymase ferments sugars. Invertase hydrolyzes sucrose. Step 3: Urease is an enzyme that catalyzes the hydrolysis of urea into ammonia and carbon dioxide. It is not involved in starch fermentation. 12) C) 40% Step 1: Formalin is an aqueous solution of formaldehyde. Step 2: It typically contains 37-40% formaldehyde by mass, often with a small amount of methanol as a stabilizer. Step 3: The most common concentration is approximately 40%. 13) D) LiAlH$_4$ Step 1: Reducing a carboxylic acid to an alcohol requires a strong reducing agent. Step 2: H$_2$/Ni and H$_2$/Pt are catalytic hydrogenation reagents, generally not strong enough for carboxylic acids under typical conditions. Step 3: NaBH$_4$ (sodium borohydride) is a mild reducing agent, effective for aldehydes and ketones but not carboxylic acids. Step 4: LiAlH$_4$ (lithium aluminum hydride) is a powerful reducing agent capable of reducing carboxylic acids to primary alcohols. 14) D) Sulphur Step