Which of the following compounds will give a silver mirror with Tollens' reagent?

Tollens' reagent (ammoniacal silver nitrate) is used to test for the presence of an aldehyde functional group ($\text{-CHO}$). Aldehydes are oxidized to carboxylic acids, and $\text{Ag}^+$ ions are reduced to metallic silver, forming a silver mirror. Ketones, alcohols, and carboxylic acids do not give a positive Tollens' test.

• A) $\text{CH}_3\text{CH(OH)CHO}$: This compound contains a $\text{-CHO}$ group, which is an aldehyde.
• B) $\text{CH}_3\text{CH}_2\text{OH}$: This is a primary alcohol.
• C) $\text{CH}_3\text{COCH}_3$: This is a ketone.
• D) $\text{CH}_3\text{CH(OH)CH}_3$: This is a secondary alcohol.

Only option A, $\text{CH}_3\text{CH(OH)CHO}$, is an aldehyde and will therefore give a silver mirror.

The final answer is $\boxed{\text{A}}$.

Sulphuric acid ($\text{H}_2\text{SO}_4$) is a strong diprotic acid, meaning it dissociates to release two protons per molecule. For strong acids, we can assume complete dissociation.

Given concentration of $\text{H}_2\text{SO}_4 = 0.05 \text{ mol dm}^{-3}$.
Therefore, $[\text{H}^+] = 2 \times 0.05 \text{ mol dm}^{-3} = 0.10 \text{ mol dm}^{-3}$.

The pH is calculated using the formula:

Substitute the calculated $[\text{H}^+]$:

The calculated pH of 1 matches option A.

The final answer is $\boxed{\text{A}}$.

Given half-life ($t_{1/2}$) = 8 days.
Given total time ($t$) = 24 days.
Number of half-lives ($n$) = $\frac{\text{Total time}}{\text{Half-life}}$

The fraction remaining after $n$ half-lives is given by the formula:

Substitute $n=3$:

The calculated fraction of 1/8 matches option B.

The final answer is $\boxed{\text{B}}$.

At s.t.p (standard temperature and pressure), 1 mole of any gas occupies $22.4 \text{ dm}^3$.
Given volume of gas = $0.112 \text{ dm}^3$.

Given mass of sample = $0.30 \text{ g}$.

Relative atomic masses (R.A.M): $\text{H}=1$, $\text{C}=12$, $\text{O}=16$.

• A) $\text{CH}_3\text{OH}$: $(1 \times 12) + (4 \times 1) + (1 \times 16) = 12 + 4 + 16 = 32 \text{ g mol}^{-1}$
• B) $\text{CH}_3\text{COCH}_3$ (Propanone, $\text{C}_3\text{H}_6\text{O}$): $(3 \times 12) + (6 \times 1) + (1 \times 16) = 36 + 6 + 16 = 58 \text{ g mol}^{-1}$
• C) $\text{CH}_3\text{CHO}$ (Ethanal, $\text{C}_2\text{H}_4\text{O}$): $(2 \times 12) + (4 \times 1) + (1 \times 16) = 24 + 4 + 16 = 44 \text{ g mol}^{-1}$
• D) $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ (Propan-1-ol, $\text{C}_3\text{H}_8\text{O}$): $(3 \times 12) + (8 \times 1) + (1 \times 16) = 36 + 8 + 16 = 60 \text{ g mol}^{-1}$

The molar mass of $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ (or any isomer with formula $\text{C}_3\text{H}_8\text{O}$) is $60 \text{ g mol}^{-1}$, which matches the calculated molar mass.

The final answer is $\boxed{\text{D}}$.

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.

• A) $\