3 Will the motor car stop before reaching the wreck of the truck? Show all working in your answer.

here are the solutions to Question 5.

5.1 You are given $25 \text{ g}$ of sodium sulphate ($\text{Na}_2SO_4$).

5.1.1 Calculate the number of moles of sodium sulphate crystals.

Step 1: Calculate the molar mass of $\text{Na}_2SO_4$. Atomic masses: $\text{Na} = 22.99 \text{ g/mol}$, $\text{S} = 32.07 \text{ g/mol}$, $\text{O} = 16.00 \text{ g/mol}$. $M(Na_2SO_4) = (2 \times 22.99) + (1 \times 32.07) + (4 \times 16.00)$ $M(Na_2SO_4) = 45.98 + 32.07 + 64.00 = 142.05 g/mol$

Step 2: Calculate the number of moles using the given mass and molar mass. $n = \frac{m}{M}$ $n = \frac{25 g}{142.05 g/mol}$ $n \approx 0.176 mol$ The number of moles of sodium sulphate crystals is $\boxed{0.176 mol}$.

5.1.2 Calculate the number of sodium atoms present in $25 \text{ g}$ of sodium sulphate crystals.

Step 1: Determine the moles of sodium atoms from the moles of $\text{Na}_2SO_4$. From the formula $\text{Na}_2SO_4$, there are 2 moles of Na atoms for every 1 mole of $\text{Na}_2SO_4$. $Moles of Na atoms = 0.176 mol Na_2SO_4 \times \frac{2 mol Na}{1 mol Na_2SO_4}$ $Moles of Na atoms = 0.352 mol$

Step 2: Use Avogadro's number to find the number of sodium atoms. Avogadro's number ($N_A$) = $6.022 \times 10^{23} \text{ atoms/mol}$. $Number of Na atoms = n \times N_A$ $Number of Na atoms = 0.352 mol \times 6.022 \times 10^{23} atoms/mol$ $Number of Na atoms \approx 2.12 \times 10^{23} atoms$ The number of sodium atoms present is $\boxed{2.12 \times 10^{23} atoms}$.

5.2 A substance contains 40% Carbon, 6.67% Hydrogen, and 53.53% Oxygen by mass.

5.2.1 Define the term empirical formula.

The empirical formula is the simplest whole-number ratio of atoms in a compound.

5.2.2 Determine the empirical formula of the substance.

Step 1: Assume a $100 \text{ g}$ sample and convert percentages to mass. Mass of C = $40 \text{ g}$ Mass of H = $6.67 \text{ g}$ Mass of O = $53.53 \text{ g}$

Step 2: Convert mass of each element to moles. Atomic masses: $\text{C} = 12.01 \text{ g/mol}$, $\text{H} = 1.01 \text{ g/mol}$, $\text{O} = 16.00 \text{ g/mol}$. $Moles of C = \frac{40 g}{12.01 g/mol} \approx 3.33 mol$ $Moles of H = \frac{6.67 g}{1.01 g/mol} \approx 6.60 mol$ $Moles of O = \frac{53.53 g}{16.00 g/mol} \approx 3.35 mol$

Step 3: Divide by the smallest number of moles to find the simplest ratio. The smallest number of moles is approximately $3.33 \text{ mol}$ (for Carbon). $Ratio C = \frac{3.33}{3.33} = 1$ $Ratio H = \frac{6.60}{3.33} \approx 1.98 \approx 2$ $Ratio O = \frac{3.35}{3.33} \approx 1.006 \approx 1$ The empirical formula of the substance is $\boxed{CH_2O}$.

5.2.3 If the molecular mass of the substance is $60 g \cdot mol^{-1}$, determine its molecular formula.

Step 1: Calculate the empirical formula mass. $M(CH_2O) = (1 \times 12.01) + (2 \times 1.01) + (1 \times 16.00)$ $M(CH_2O) = 12.01 + 2.02 + 16.00 = 30.03 g/mol$

Step 2: Determine the ratio ($n$) between the molecular mass and the empirical formula mass. $n = \frac{Molecular mass}{Empirical formula mass}$ $n = \frac{60 g/mol}{30.03 g/mol} \approx 1.998 \approx 2$

Step 3: Multiply the subscripts in the empirical formula by $n$ to get the molecular formula. $Molecular formula = (CH_2O)_2 = C_2H_4O_2$ The molecular formula of the substance is $\boxed{C_2H_4O_2}$.