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List common laboratory equipment, explain distinguishing CO32- from HCO3-, define titration, explain why KMnO4 is a common oxidizing agent, and calculate heat change.

Chemistry

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Answer

4.64 J\text{4.64 J}

Step-by-step solution

2 steps

  1. Test tube, beaker, funnel, Bunsen burner.

  2. A suitable reagent for distinguishing aqueous solution of CO32−\text{CO}_3^{2-} and HCO3−\text{HCO}_3^{-} is magnesium chloride (MgCl2\text{MgCl}_2).

  3. Titration is a common method of quantitative analysis used to determine the concentration of an unknown substance in a solution.

  4. Three reasons why KMnO4\text{KMnO}_4 is the most widely used oxidizing agent in redox titrations:

    • It is a strong oxidizing agent.
    • It is self-indicating (its intense purple color disappears at the endpoint, so no external indicator is usually needed).
    • It is relatively inexpensive.

  5. Step 1: Identify the given values and the formula for heat change. Given: Change in temperature, ΔT=9.2 °C=9.2 K\Delta T = 9.2 \text{ °C} = 9.2 \text{ K} Mass of solution, m=120 grams=0.120 kgm = 120 \text{ grams} = 0.120 \text{ kg} Specific heat capacity, c=4.2 JKg−1K−1c = 4.2 \text{ JKg}^{-1}\text{K}^{-1} The formula for heat change (ΔH\Delta H or qq) is:

    ΔH=mcΔT\Delta H = mc\Delta T

  1. Substitute the values into the formula.

    $$
\Delta H = (0.120 \text{ kg}) \times (4.2 \text{ JKg}^{-1}\text{K}^{-1}) \times (9.2 \text{ K})
$$

  2. Calculate the result.

    $$
\Delta H = 0.48 \times 9.2 \text{ J}
$$
$$
\Delta H = 4.6368 \text{ J}
$$

The change in enthalpy ($\Delta H$) is $\boxed{\text{4.64 J}}$.
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