Answer me number 1-5

Mathematics

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  1. Complete Question 1

    ...no effect on litmus paper. This is because ammonia does not ionize in methylbenzene to produce hydroxide ions (OH−OH^-), which are responsible for basic properties.

  2. Answer Question 2

    a) Calculate the number of moles of H2SO4H_2SO_4.
    Volume of H2SO4=25.0 cm3=0.0250 dm3H_2SO_4 = 25.0 \text{ cm}^3 = 0.0250 \text{ dm}^3.
    Concentration of H2SO4=0.100 mol dm−3H_2SO_4 = 0.100 \text{ mol dm}^{-3}.

    Moles of H2SO4=Concentration×Volume\text{Moles of } H_2SO_4 = \text{Concentration} \times \text{Volume} Moles of H2SO4=0.100 mol dm−3×0.0250 dm3\text{Moles of } H_2SO_4 = 0.100 \text{ mol dm}^{-3} \times 0.0250 \text{ dm}^3 Moles of H2SO4=0.00250 mol\text{Moles of } H_2SO_4 = \boxed{\text{0.00250 mol}}

    b) Calculate the number of moles of NaOHNaOH that reacted.
    The balanced chemical equation is:

    H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)H_2SO_4(aq) + 2NaOH(aq) \to Na_2SO_4(aq) + 2H_2O(l)

    From the equation, 1 mole of H2SO41 \text{ mole of } H_2SO_4 reacts with 2 moles of NaOH2 \text{ moles of } NaOH.

    Moles of NaOH=2×Moles of H2SO4\text{Moles of } NaOH = 2 \times \text{Moles of } H_2SO_4 Moles of NaOH=2×0.00250 mol\text{Moles of } NaOH = 2 \times 0.00250 \text{ mol} Moles of NaOH=0.00500 mol\text{Moles of } NaOH = \boxed{\text{0.00500 mol}}

    c) Calculate the concentration of the NaOHNaOH solution.
    Volume of NaOH=20.0 cm3=0.0200 dm3NaOH = 20.0 \text{ cm}^3 = 0.0200 \text{ dm}^3.

    Concentration of NaOH=Moles of NaOHVolume of NaOH\text{Concentration of } NaOH = \frac{\text{Moles of } NaOH}{\text{Volume of } NaOH} Concentration of NaOH=0.00500 mol0.0200 dm3\text{Concentration of } NaOH = \frac{0.00500 \text{ mol}}{0.0200 \text{ dm}^3} Concentration of NaOH=0.250 mol dm−3\text{Concentration of } NaOH = \boxed{\text{0.250 mol dm}^{-3}}
  3. Answer Question 3

    a) The type of reaction that occurs when H2SO4H_2SO_4 reacts with NaOHNaOH is a neutralization reaction.

    b) A common indicator used in this titration (strong acid-strong base) is phenolphthalein.

  4. Answer Question 4

    a) If phenolphthalein is used, the colour of the indicator in the conical flask (containing H2SO4H_2SO_4) at the start of the titration is colorless.

    b) If phenolphthalein is used, the colour of the indicator in the conical flask at the end-point of the titration is pink.

  5. Answer Question 5

    a) Apparatus A is a burette.

    b) Apparatus B is a pipette.

    c) Apparatus C is a conical flask (or Erlenmeyer flask).

    d) Apparatus D is a retort stand (with a clamp).

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