NSMQ 2017 past questions with answers

Answer

Hydrogen bonding and van der Waals forces.

1. 1.Glucose has molecular formula $C_6{H}_{12}{O}_6$ and each molecule contains several hydroxyl (–OH) groups.
2. 2.The –OH groups make each molecule both a hydrogen-bond donor and acceptor, so strong intermolecular hydrogen bonding occurs in the solid.
3. 3.Besides hydrogen bonds, neutral molecular solids show van der Waals interactions (London dispersion and dipole interactions) between molecules.
4. 4.Therefore the main forces present in solid glucose are hydrogen bonding and van der Waals forces.

Answer

$6.5\times 10^{14}\text{Hz}$

1. 1.Recall the photoelectric equation: $K_{\text{max}}=hf-\varphi$, where $\varphi =h{f}_0$ is the work‐function energy and $f_0$ is the threshold frequency.
2. 2.Substitute $\varphi =h{f}_0$ into the equation to get $K_{\text{max}}=h(f-f_0)$, so $f=f_0+\frac{K_{\text{max}}}{h}$.
3. 3.Use the given threshold frequency $f_0=5.5\times 10^{14}\mathrm{Hz}$ (from the preamble) and $K_{\text{max}}=6.4\times 10^{-20}\mathrm{J}$, and Planck’s constant $h=6.63\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$.
4. 4.Compute the extra frequency $\Delta f=\frac{6.4\times 10^{-20}}{6.63\times 10^{-34}}\approx 9.65\times 10^{13}\mathrm{Hz}$, then add to $f_0$: $f=5.5\times 10^{14}+9.65\times 10^{13}\approx 6.5\times 10^{14}\mathrm{Hz}$.

Answer

$\tau =\frac{\Delta L}{t}=2\text{N m}$

1. 1.Calculate the change in angular momentum: $\Delta L=L_f-L_i=12\mathrm{J}\mathrm{s}-2\mathrm{J}\mathrm{s}=10\mathrm{J}\mathrm{s}$.
2. 2.Recall the definition of torque as the rate of change of angular momentum: $\tau =\frac{\Delta L}{\Delta t}$.
3. 3.Substitute $\Delta L=10\mathrm{J}\mathrm{s}$ and $\Delta t=5\mathrm{s}$ into the formula: $\tau =\frac{10\mathrm{J}\mathrm{s}}{5\mathrm{s}}=2\mathrm{N}\mathrm{m}$.

Answer

$200\mu \text{s}$

1. 1.Write the formula for the time constant: $\tau =RC$.
2. 2.Convert capacitance from microfarads to farads: $C=20\mu \mathrm{F}=20\times 10^{-6}\mathrm{F}$.
3. 3.Substitute $R=10\Omega$ and $C=20\times 10^{-6}\mathrm{F}$ into $\tau =RC$ and calculate: $\tau =10\times 20\times 10^{-6}=200\times 10^{-6}\mathrm{s}=200\mu \mathrm{s}$.

Answer

A buffer solution.

1. 1.Identify the species present: NH₃ (weak base), NH₄⁺ (conjugate acid), and Cl⁻ (spectator ion)
2. 2.Observe that [NH₃] = [NH₄⁺] = 0.036 M
3. 3.Conclude that equal concentrations of a weak base and its conjugate acid form a buffer solution

Answer

$-14\text{J}$

1. 1.Compute initial kinetic energy $K_i$ using $p_i=12$N·s: $K_i=\frac{12^2}{2\times 5}=\frac{144}{10}=14.4$J.
2. 2.Compute final kinetic energy $K_f$ using $p_f=2$N·s: $K_f=\frac{2^2}{2\times 5}=\frac{4}{10}=0.4$J.
3. 3.Compute change in kinetic energy: $\Delta K=K_f-K_i=0.4\text{J}-14.4\text{J}=-14$J.

Answer

False

Congruent diagonals occur in rectangles, but not necessarily with two axes of symmetry.

Answer

True

Nitrogen’s higher electronegativity relative to carbon causes unequal sharing of electrons in the C=N bond, creating a dipole.

Answer

Electromagnetic induction

Answer

Proof