In the van der Waals equation, the term $\dfrac{an^2}{V^2}$ must have units of pressure.

So: $a = \dfrac{P \cdot V^2}{n^2}$

Units of $a = \dfrac{\text{atm} \cdot (\text{dm}^3)^2}{(\text{mol})^2} = \mathbf{\text{atm dm}^6\ \text{mol}^{-2}}$

The constant $b$ has units of dm$^3$ mol$^{-1}$ (it represents excluded molar volume). 'a' corrects for intermolecular attractions.

When alkali metals dissolve in liquid ammonia: $\text{Na} \to \text{Na}^+ + e^-$

The released electrons are trapped by cavities in the solvent and become ammoniated electrons ($e^-_\text{am}$). These solvated electrons absorb light in the visible range and are responsible for the characteristic deep blue colour of the solution.

At higher concentrations, the colour changes to bronze — indicating electron pairing.

Sodamide (NaNH$_2$) forms on prolonged standing or with catalysts, not immediately.

The environmental-organizational relationship goes only one way (few organizations can impact the environment), while organizational-individual challenges have a two-way relationship (each affects the other reciprocally).

Let $f'(1)=a$, $f''(2)=b$, $f'''(3)=c$. Then $f(x)=x^3+ax^2+bx+c$.

$f'(x)=3x^2+2ax+b \Rightarrow f'(1)=3+2a+b=a \Rightarrow a+b=-3$ ...(i)

$f''(x)=6x+2a \Rightarrow f''(2)=12+2a=b$ ...(ii)

$f'''(x)=6 \Rightarrow f'''(3)=6=c$ ...(iii)

From (ii): $b=12+2a$. Sub in (i): $a+12+2a=-3\Rightarrow3a=-15\Rightarrow a=-5$, $b=2$, $c=6$.

$f(2)=8+4(-5)+2(2)+6=8-20+4+6=\mathbf{-2}$

Wait: $f(2)=8-20+4+6=-2$. So correct index is 2 (option C: $-2$).

The magnitude of change in velocity is $|\Delta v| = 2v \sin(\frac{\theta}{2})$.

• $|\Delta v| = 2(10) \sin(60^{\circ}/2) = 20 \sin 30^{\circ}$
• $|\Delta v| = 20 \times \frac{1}{2} = 10\text{ m/s}$.