Some Basic Concepts of Chemistry, on the exam screen.

Moles of urea = 5.4/60 = 0.09 mol. Urea, $\mathrm{NH}_{2}\mathrm{CONH}_{2}$, contains 4 hydrogen atoms per molecule. So number of H atoms = $0.09 \times 4 \times 6.022 \times 10^{23} \approx 2.168 \times 10^{23}$. The distractors arise from using the wrong number of H atoms per molecule or a wrong power of ten.

The reaction $\mathrm{CO}_{2} + \mathrm{C}(\text{s}) \rightarrow 2\mathrm{CO}$ converts x dm$^3$ of $\mathrm{CO}_{2}$ into 2x dm$^3$ of CO. Total final volume $=(1-x)+2x = 1+x = 1.4$, so x = 0.4 dm$^3$. Therefore $\mathrm{CO}_{2}$ remaining $=1-0.4=0.6$ dm$^3$ and CO formed $=2\times0.4=0.8$ dm$^3$, which is option (2).

Both numbers end in an exact 5 with nothing after it, which is the tie case, so the round-half-to-EVEN convention applies rather than always rounding up. In 17.0145 the digit to be retained is 4, already even, so it stays: 17.014. In 21.0235 the retained digit is 3, which is odd, so it is raised to the even value 4: 21.024. NCERT Class 11, Some Basic Concepts of Chemistry (significant figures).

n-Butane is $\mathrm{C}_{4}\mathrm{H}_{10}$ with molar mass 58 g/mol, so 116 g is exactly 2 mol. The balanced combustion $\mathrm{C}_{4}\mathrm{H}_{10} + 13/2 \mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 5\mathrm{H}_{2}\mathrm{O}$ yields 4 mol of $\mathrm{CO}_{2}$ per mole of butane, hence 8 mol here. Mass of $\mathrm{CO}_{2} = 8 \times 44 = 352\ \text{g}$. NCERT Class 11, Some Basic Concepts of Chemistry (stoichiometry).

To find the number of atoms, use $n = \frac{m}{M} \times N_A$, where m is mass, M is molar mass, and $N_A$ is Avogadro's number. For 1 g of each: - Ag: $\frac{1}{108} \times N_A$ - Mg: $\frac{1}{24} \times N_A$ - $\mathrm{O}_{2}$: $\frac{1}{32} \times 2 \times N_A = \frac{1}{16} \times N_A$ (since $\mathrm{O}_{2}$ has a molar mass of 32 g/mol and two atoms per molecule) - Li: $\frac{1}{7} \times N_A$ Li has the highest number of atoms. Chapter: Some Basic Concepts of Chemistry.