Chemistry

(a) $\frac12\times6.023\times10^{23}$

At NTP, 22.4 L of gas is 1 mole, so 11.2 L is 0.5 mol, which contains $0.5\times6.023\times10^{23}$ molecules of $O_2$. This is the option that matches the 0.5 mol of gas. Strictly, 11.2 L of $O_2$ holds $6.023\times10^{23}$ oxygen atoms (2 atoms per molecule), a value that is not listed.

(c) 6

Let the oxidation number of S be $x$: $2(+1)+x+4(-2)=0$, so $x=+6$.

(d) Mol wt/1

In basic medium, $MnO_4^-+e^-\to MnO_4^{2-}$, a change of 1 electron per formula unit. So the equivalent weight is molecular weight/1.

(a) acidic

Moles of $H^+$ from $H_2SO_4$: $100\times0.01\times2=2$ mmol.

Moles of $H^+$ from HCl: $50\times0.1=5$ mmol.

Total $H^+=7$ mmol.

Moles of $OH^-$ from KOH: $500\times0.012=6$ mmol.

Since $7>6$, 1 mmol of $H^+$ is left over, so the solution is acidic.

(c) 5.39 kg

Energy for a week: $12000\times7=84000$ kJ.

$\text{moles of glucose}=\frac{84000}{2808}=29.91$

$\text{mass}=29.91\times180\approx5385\text{ g}\approx5.39\text{ kg}$