This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Welcome back Hope — missed you this week. Step 1: Identify the functional group that gives a silver mirror with Tollens' reagent. Tollens' reagent (ammoniacal silver nitrate) is used to test for the presence of an aldehyde functional group (-CHO). Aldehydes are oxidized to carboxylic acids, and Ag^+ ions are reduced to metallic silver, forming a silver mirror. Ketones, alcohols, and carboxylic acids do not give a positive Tollens' test. Step 2: Examine the given options to find the compound with an aldehyde group. A) CH_3CH(OH)CHO: This compound contains a -CHO group, which is an aldehyde. B) CH_3CH_2OH: This is a primary alcohol. C) CH_3COCH_3: This is a ketone. D) CH_3CH(OH)CH_3: This is a secondary alcohol. Step 3: Conclude which compound will give a silver mirror. Only option A, CH_3CH(OH)CHO, is an aldehyde and will therefore give a silver mirror. The final answer is A. --- Step 1: Determine the concentration of hydrogen ions ([H^+]). Sulphuric acid (H_2SO_4) is a strong diprotic acid, meaning it dissociates to release two protons per molecule. For strong acids, we can assume complete dissociation. H_2SO_4 2H^+ + SO_4^2- Given concentration of H_2SO_4 = 0.05 mol dm^-3. Therefore, [H^+] = 2 × 0.05 mol dm^-3 = 0.10 mol dm^-3. Step 2: Calculate the pH using the hydrogen ion concentration. The pH is calculated using the formula: pH = -_10[H^+] Substitute the calculated [H^+]: pH = -_10(0.10) pH = 1 Step 3: Compare the calculated pH with the given options. The calculated pH of 1 matches option A. The final answer is A. --- Step 1: Calculate the number of half-lives that occur in 24 days. Given half-life (t_1/2) = 8 days. Given total time (t) = 24 days. Number of half-lives (n) = Total timeHalf-life n = 24 days8 days = 3 Step 2: Calculate the fraction of the original isotope remaining after 3 half-lives. The fraction remaining after n half-lives is given by the formula: Fraction remaining = ((1)/(2))^n Substitute n=3: Fraction remaining = ((1)/(2))^3 = (1)/(2 × 2 × 2) = (1)/(8) Step 3: Compare the calculated fraction with the given options. The calculated fraction of 1/8 matches option B. The final answer is B. --- Step 1: Calculate the number of moles of the volatile organic liquid. At s.t.p (standard temperature and pressure), 1 mole of any gas occupies 22.4 dm^3. Given volume of gas = 0.112 dm^3. Moles = VolumeMolar volume at s.t.p Moles = 0.112 dm^322.4 dm^3 mol^-1 = 0.005 mol Step 2: Calculate the molar mass of the liquid. Given mass of sample = 0.30 g. Molar mass = MassMoles Molar mass = 0.30 g0.005 mol = 60 g mol^-1 Step 3: Determine the molecular formula by calculating the molar mass for each option. Relative atomic masses (R.A.M): H=1, C=12, O=16. A) CH_3OH: (1 × 12) + (4 × 1) + (1 × 16) = 12 + 4 + 16 = 32 g mol^-1 B) CH_3COCH_3 (Propanone, C_3H_6O): (3 × 12) + (6 × 1) + (1 × 16) = 36 + 6 + 16 = 58 g mol^-1 C) CH_3CHO (Ethanal, C_2H_4O): (2 × 12) + (4 × 1) + (1 × 16) = 24 + 4 + 16 = 44 g mol^-1 D) CH_3CH_2CH_2OH (Propan-1-ol, C_3H_8O): (3 × 12) + (8 × 1) + (1 × 16) = 36 + 8 + 16 = 60 g mol^-1 Step 4: Identify the option with the matching molar mass. The molar mass of CH_3CH_2CH_2OH (or any isomer with formula C_3H_8O) is 60 g mol^-1, which matches the calculated molar mass. The final answer is D. --- Step 1: Understand the conditions for cis-trans (geometric) isomerism. Cis-trans isomerism occurs in molecules with restricted rotation around a bond (typically a carbon-carbon double bond or a ring structure) and where each carbon atom involved in the restricted rotation is bonded to two different groups. Step 2: Analyze each option for the presence of a carbon-carbon double bond and different groups on each carbon. A) $\