NSMQ 2019 past questions with answers

Answer

A proton converts to a neutron; atomic number decreases by 1; mass unchanged.

1. 1.Recall that in electron capture an inner electron is absorbed by the nucleus and interacts with a proton.
2. 2.Recognize that a proton + electron → neutron + neutrino, so the nucleus loses one proton and gains one neutron.
3. 3.Conclude that the atomic number decreases by 1 while the mass number remains unchanged.

Answer

$8\sqrt{\frac{10}{11}}$ minutes

1. 1.Calculate molar masses: M_CO2 = 12 + 2×16 = 44 g/mol, M_Ar = 40 g/mol.
2. 2.Use Graham’s law relating effusion times: t₂/t₁ = √(M₂/M₁), where t₁ = time for CO2 and t₂ = time for Ar.
3. 3.Substitute values: t_Ar = 8.00 min × √(40/44) = 8.00 × √(10/11) min.

Answer

$A=-4$, other root $-1$

1. 1.Substitute the known root x=5 into x²+Ax−5=0: 5²+5A−5=0 ⇒ 25+5A−5=0 ⇒ 5A+20=0 ⇒ A=−4
2. 2.Use the sum of roots formula α+β=−b/a for x²+Ax−5=0, where b=A and a=1: sum=−A=−(−4)=4
3. 3.Let the other root be r; then 5+r=4 ⇒ r=−1

Answer

$x=-\frac{5}{4},1$

1. 1.Identify coefficients: a = 4, b = 1, c = -5.
2. 2.Compute the discriminant: Δ = b² - 4ac = 1² - 4(4)(-5) = 1 + 80 = 81.
3. 3.Apply the quadratic formula: x = [-1 ± √81]/(2·4) = [-1 ± 9]/8, giving x = (8/8) = 1 or x = (-10/8) = -5/4.

Answer

$11.7\text{g}$

1. 1.Write the half-reaction: ${\mathrm{K}}^{+}+e^{-}\to \mathrm{K}$, so z = 1 electron per K atom.
2. 2.Use Faraday's law: moles of K deposited $n_{\text{K}}=\frac{Q}{zF}=\frac{0.300\text{F}}{1\times 1\text{F/mol}}=0.300\text{mol}$.
3. 3.Convert moles to mass: $m=n_{\text{K}}\times M_{\text{K}}=0.300\text{mol}\times 39.0\text{g/mol}=11.7\text{g}$.

Answer

${\left(\frac{10}{9}\right)}^4\approx 1.52$

1. 1.For laminar flow in a cylindrical vessel with the same length and pressure difference, Poiseuille's law gives $Q\propto r^4$.
2. 2.Let the first artery have radius $r_1=r$ and the second $r_2=\frac{9}{10}r$; then $\frac{Q_1}{Q_2}={\left(\frac{r_1}{r_2}\right)}^4$.
3. 3.Compute $\frac{Q_1}{Q_2}={\left(\frac{r}{(9/10)r}\right)}^4={\left(\frac{10}{9}\right)}^4$.
4. 4.Evaluate numerically ${\left(\frac{10}{9}\right)}^4\approx \mathrm{1.524...}\approx 1.52$.

Answer

False

While most Period 3 oxides are basic or acidic, Al2O3 is amphoteric, so the statement that they are only basic or acidic is false.

Answer

False

Mapping (x,y)→(x,−y) reflects across the x-axis while y=f(−x) reflects across the y-axis, so they do not coincide unless f is odd, which is not generally true.

Answer

Axiom

Answer

SnCl4