Below are the answers to the questions provided:
Question 1.1 [2 marks]
To determine the molecular formula from the empirical formula, we first need to calculate the molar mass of the empirical formula Mg₃Si₂H₃O₈.
- Molar mass of Mg = 24.31 g/mol (3 Mg gives 3 × 24.31 = 72.93 g/mol)
- Molar mass of Si = 28.09 g/mol (2 Si gives 2 × 28.09 = 56.18 g/mol)
- Molar mass of H = 1.01 g/mol (3 H gives 3 × 1.01 = 3.03 g/mol)
- Molar mass of O = 16.00 g/mol (8 O gives 8 × 16.00 = 128.00 g/mol)
Calculating the total: \[ 72.93 + 56.18 + 3.03 + 128.00 = 260.14 \text{ g/mol} \]
Next, we find the ratio of the molar mass of the compound (520.8 g/mol) to the molar mass of the empirical formula (260.14 g/mol): \[ \text{Ratio} = \frac{520.8}{260.14} \approx 2 \]
Thus, the molecular formula is: \[ (Mg₃Si₂H₃O₈) \times 2 = Mg₆Si₄H₆O₁₆ \]
Answer: B. Mg6Si4H6O16
Question 1.2 [2 marks]
Bromothymol blue is an indicator that changes color at a pH range typically between 6.0 (yellow) and 7.6 (blue). It is commonly used in titrations involving strong acids and strong bases.
Answer: D. Strong base and strong acid
Question 1.3 [2 marks]
In an exothermic reaction, the energy is released, and the potential energy of the products is lower than that of the reactants. Therefore, ΔH (the change in enthalpy) is typically negative.
Answer: B. ΔH is negative and the products have less potential energy than the reactants.
Question 1.4 [2 marks]
A standard solution is defined as a solution of known concentration, which may contain one mole of solute per dm³ but isn't limited to that single definition.
Answer: C. is a solution of which the concentration is precisely known.
Question 1.5 [2 marks]
In a potential energy diagram, the activation energy for the forward reaction is represented by the change in potential energy from the reactants level to the peak of the curve (the transition state). Therefore, from the description of the options:
- \( y3 \) is likely the potential energy of the activated complex (transition state)
- \( y2 \) is the potential energy of the reactants
The activation energy would therefore be calculated as: \[ \text{Activation energy} = y3 - y2 \]
Answer: A. Activation energy = y3 - y2
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