We read your photo as:

Give two examples of iron ores. State three adverse effects of chemical products.

Chemistry

ScanSolve AI Answer

Step-by-step solution

9 steps

2. (a) (i) Two examples of iron ores are:

  • Hematite (Fe2O3\text{Fe}_2\text{O}_3)
  • Magnetite (Fe3O4\text{Fe}_3\text{O}_4)

2. (a) (ii) Three adverse effects of chemical products are:

  • Environmental pollution: They can contaminate air, water, and soil, harming ecosystems.
  • Health hazards: Exposure can cause various health issues, including respiratory problems, skin irritation, and long-term illnesses like cancer.
  • Damage to infrastructure: Corrosive chemicals can degrade materials and structures.

2. (b) (i) The alkali metals Na, Li, K in decreasing order of their melting points are:

  • Li (180.5∘C180.5^\circ\text{C}) > Na (97.8∘C97.8^\circ\text{C}) > K (63.5∘C63.5^\circ\text{C})

2. (b) (ii) The three rules/principles of electron distribution in an atom are:

  • Aufbau principle
  • Pauli exclusion principle
  • Hund's rule of maximum multiplicity

2. (c) (i)

  • An oxidizing agent is a substance that gains electrons and causes the oxidation of another substance.
  • A reducing agent is a substance that loses electrons and causes the reduction of another substance.

2. (c) (ii) The ions formed when the following compounds dissociate are:
I. KOH: K+(aq)+OH−(aq)\text{K}^+ (\text{aq}) + \text{OH}^- (\text{aq})
II. Pb(NO3)2\text{Pb(NO}_3)_2: Pb2+(aq)+2NO3−(aq)\text{Pb}^{2+} (\text{aq}) + 2\text{NO}_3^- (\text{aq})
III. Na2CO3\text{Na}_2\text{CO}_3: 2Na+(aq)+CO32−(aq)2\text{Na}^+ (\text{aq}) + \text{CO}_3^{2-} (\text{aq})

2. (d) (i)
I. An anode is the electrode where oxidation (loss of electrons) occurs during an electrochemical reaction.
II. A cathode is the electrode where reduction (gain of electrons) occurs during an electrochemical reaction.

2. (d) (ii)

  1. Calculate the total charge passed.

    The current I=1.0 AI = 1.0 \text{ A} and time t=15 minutes=15×60 s=900 st = 15 \text{ minutes} = 15 \times 60 \text{ s} = 900 \text{ s}.

    Q=I×t=1.0 A×900 s=900 CQ = I \times t = 1.0 \text{ A} \times 900 \text{ s} = 900 \text{ C}

  2. Determine the moles of electrons.

    Using Faraday's constant F=96,500 C/molF = 96,500 \text{ C/mol}:

    Moles of electrons=QF=900 C96500 C/mol≈0.009326 mol\text{Moles of electrons} = \frac{Q}{F} = \frac{900 \text{ C}}{96500 \text{ C/mol}} \approx 0.009326 \text{ mol}

  3. Write the half-reaction for hydrogen production at the cathode.

    During the electrolysis of dilute tetraoxosulphate (VI) acid (H2SO4\text{H}_2\text{SO}_4), hydrogen gas is produced at the cathode:

    2H+(aq)+2e−→H2(g)2\text{H}^+ (\text{aq}) + 2\text{e}^- \to \text{H}_2 (\text{g})

    From the stoichiometry, 2 moles of electrons produce 1 mole of H2\text{H}_2 gas.

  4. Calculate the moles of hydrogen gas produced.

    Moles of H2=12×Moles of electrons=12×0.009326 mol≈0.004663 mol\text{Moles of } \text{H}_2 = \frac{1}{2} \times \text{Moles of electrons} = \frac{1}{2} \times 0.009326 \text{ mol} \approx 0.004663 \text{ mol}

  5. Calculate the volume of hydrogen gas at s.t.p.

    At s.t.p., 1 mole of any gas occupies 22.4 dm322.4 \text{ dm}^3.

    Volume of H2=Moles of H2×22.4 dm3/mol=0.004663 mol×22.4 dm3/mol≈0.10445 dm3\text{Volume of } \text{H}_2 = \text{Moles of } \text{H}_2 \times 22.4 \text{ dm}^3/\text{mol} = 0.004663 \text{ mol} \times 22.4 \text{ dm}^3/\text{mol} \approx 0.10445 \text{ dm}^3

    The volume of hydrogen produced is 0.104 dm3\boxed{\text{0.104 dm}^3}.

    3. (a)
    An energy profile diagram illustrating the effect of a catalyst on an exothermic reaction:

    \begin{tikzpicture}[scale=0.8] % Axes \draw[->] (0,0) -- (8,0) node[right] {Reaction Progress}; \draw[->] (0,0) -- (0,6) node[above] {Energy}; % Reactants and Products \node at (1,5) (reactants) {Reactants}; \node at (7,1) (products) {Products}; % Uncatalyzed reaction \draw[thick, blue] (1,5) .. controls (3,7) and (5,7) .. (7,1); \draw[dashed] (1,5) -- (1,0); \draw[dashed] (7,1) -- (7,0); \draw[<->] (1.2,5.2) -- (3.8,6.8) node[midway, above right] {$\text{E}_{\text{a (uncatalyzed)}}$}; % Catalyzed reaction \draw[thick, red] (1,5) .. controls (3,6) and (5,6) .. (7,1); \draw[<->] (1.2,5.2) -- (3.8,5.8) node[midway, below right] {$\text{E}_{\text{a (catalyzed)}}$}; % Enthalpy change \draw[<->] (7.3,5) -- (7.3,1) node[midway, right] {$\Delta\text{H}$}; % Labels \node[blue] at (4,7.3) {Uncatalyzed Pathway}; \node[red] at (4,6.3) {Catalyzed Pathway}; \end{tikzpicture}

    In an exothermic reaction, the energy of the products is lower than that of the reactants. A catalyst provides an alternative reaction pathway with a lower activation energy (Ea\text{E}_{\text{a}}), thereby increasing the reaction rate. The overall enthalpy change (ΔH\Delta\text{H}) of the reaction remains unchanged.

    3. (b) (i) One difference between endothermic and exothermic reactions is:

    • Endothermic reactions absorb heat from the surroundings, resulting in a positive enthalpy change (ΔH>0\Delta\text{H} > 0).
    • Exothermic reactions release heat to the surroundings, resulting in a negative enthalpy change (ΔH<0\Delta\text{H} < 0).

    3. (b) (ii) Le Chatelier's principle states that if a change of condition (such as temperature, pressure, or concentration) is applied to a system in equilibrium, the system will shift in a direction that relieves the stress and re-establishes a new equilibrium.

    3. (b) (iii) Three factors that affect the equilibrium position of a chemical reaction are:

    • Concentration of reactants or products
    • Temperature
    • Pressure (for reactions involving gases)

    3. (c) (i)
    I. A saturated solution is a solution that contains the maximum amount of solute that can be dissolved in a given amount of solvent at a specific temperature and pressure.
    II. A molar solution is a solution that contains one mole of solute per liter (or dm3\text{dm}^3) of solution. Its concentration is expressed in moles per dm3\text{dm}^3 (mol dm−3\text{mol dm}^{-3} or M).

    3. (c) (ii)

  6. Determine the chemical formula and molar mass of sodium trioxocarbonate (IV).

    Sodium trioxocarbonate (IV) is Na2CO3\text{Na}_2\text{CO}_3.
    Given atomic masses: Na=23\text{Na} = 23, O=16\text{O} = 16, C=12\text{C} = 12.
    Molar mass of Na2CO3=(2×23)+12+(3×16)=46+12+48=106 g/mol\text{Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106 \text{ g/mol}.

  7. Convert the volume from cm3\text{cm}^3 to dm3\text{dm}^3.

    Given volume =250 cm3= 250 \text{ cm}^3.

    250 cm3=2501000 dm3=0.250 dm3250 \text{ cm}^3 = \frac{250}{1000} \text{ dm}^3 = 0.250 \text{ dm}^3

  8. Calculate the number of moles of Na2CO3\text{Na}_2\text{CO}_3 needed.

    Concentration =0.2 moldm−3= 0.2 \text{ moldm}^{-3}.

    Moles=Concentration×Volume=0.2 moldm−3×0.250 dm3=0.05 mol\text{Moles} = \text{Concentration} \times \text{Volume} = 0.2 \text{ moldm}^{-3} \times 0.250 \text{ dm}^3 = 0.05 \text{ mol}

  9. Calculate the mass of Na2CO3\text{Na}_2\text{CO}_3 needed.

    Mass=Moles×Molar Mass=0.05 mol×106 g/mol=5.3 g\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.05 \text{ mol} \times 106 \text{ g/mol} = 5.3 \text{ g}

    The mass of sodium trioxocarbonate (IV) needed is 5.3 g\boxed{\text{5.3 g}}.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Chemistry Questions

QExplain three methods to prevent rusting of iron and determine the oxidation state of iron in…QWhat are two environmental impacts of chemical misuse?QA) Give the names of the ores used in the production of iron and aluminium.QHighlight four adverse effects of production activities on the environment and community.QDiscuss the functions, sources, deficiency diseases, and symptoms of key minerals like iron, zinc…
See all Chemistry questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp