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Chemistry

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Here are the answers to the first five questions:

Question One (1)

a)
i- Protons are subatomic particles found in the nucleus of an atom, carrying a positive elementary electric charge (+1e+1e). Their discovery is attributed to Ernest Rutherford in 1919, who observed that alpha particles colliding with nitrogen atoms produced hydrogen nuclei, which he identified as protons.

ii- Neutrons are subatomic particles found in the nucleus of an atom, having no net electric charge (neutral) and a mass slightly greater than that of a proton. They were discovered by James Chadwick in 1932, who observed that beryllium bombarded with alpha particles emitted a highly penetrating, uncharged radiation that could eject protons from paraffin.

b)
To determine the empirical formula, we assume a 100g sample.

  1. Calculate the mass of each element.

    Mass of Carbon (C) = 75 g
    Mass of Hydrogen (H) = 25 g

  2. Calculate the number of moles for each element.

    Molar mass of C = 12 g/mol
    Molar mass of H = 1 g/mol

    Moles of C=75 g12 g/mol=6.25 mol\text{Moles of C} = \frac{75 \text{ g}}{12 \text{ g/mol}} = 6.25 \text{ mol} Moles of H=25 g1 g/mol=25 mol\text{Moles of H} = \frac{25 \text{ g}}{1 \text{ g/mol}} = 25 \text{ mol}
  3. Determine the simplest mole ratio by dividing by the smallest number of moles.

    Smallest number of moles = 6.25 mol

    Ratio for C=6.256.25=1\text{Ratio for C} = \frac{6.25}{6.25} = 1 Ratio for H=256.25=4\text{Ratio for H} = \frac{25}{6.25} = 4
  4. Write the empirical formula.

    The empirical formula is CH4\boxed{\text{CH}_4}.

    Question Two (2)

    a)
    An empirical formula represents the simplest whole-number ratio of atoms of each element present in a compound. It shows the relative number of atoms, but not necessarily the actual number of atoms in a molecule.

    b)
    i- Four examples of non-metallic radicals are:

    • Sulfate (SO42−\text{SO}_4^{2-})
    • Nitrate (NO3−\text{NO}_3^{-})
    • Carbonate (CO32−\text{CO}_3^{2-})
    • Phosphate (PO43−\text{PO}_4^{3-})

    ii- Two natural phenomena that support the kinetic theory of gas are:

    • Diffusion: The spontaneous mixing of gases due to the random motion of their particles.
    • Brownian motion: The random movement of microscopic particles suspended in a fluid (liquid or gas), resulting from their collision with the fast-moving atoms or molecules in the fluid.

    Question Three (3)

    a)
    Degenerate orbitals are atomic orbitals that have the same energy level within a subshell. For example, the three p orbitals (px,py,pzp_x, p_y, p_z) in a given principal energy level are degenerate.

    b)
    i- Electronic configuration:
    a- Scandium (Sc, atomic number 21): 1s22s22p63s23p64s23d1\boxed{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1} or [Ar]4s23d1\boxed{[\text{Ar}] 4s^2 3d^1}
    b- Nickel (Ni, atomic number 28): 1s22s22p63s23p64s23d8\boxed{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8} or [Ar]4s23d8\boxed{[\text{Ar}] 4s^2 3d^8}

    ii- Two physical properties of water are:

    • It is a colorless, odorless, and tasteless liquid at room temperature.
    • It has a relatively high boiling point (100∘C100^\circ\text{C} at standard pressure) and freezing point (0∘C0^\circ\text{C} at standard pressure).

    Question Four (4)

    a)

  5. Use the given pOH\text{pOH} value.

    pOH=12.0\text{pOH} = 12.0

  6. Calculate the concentration of hydroxide ions using the formula.

    [OH−]=10−pOH[\text{OH}^-] = 10^{-\text{pOH}} [OH−]=10−12.0[\text{OH}^-] = 10^{-12.0} [OH−]=1.0×10−12 M[\text{OH}^-] = \boxed{1.0 \times 10^{-12} \text{ M}}

    b)
    i- Sodium chloride (NaCl\text{NaCl}) is formed through electrovalent bonding (ionic bonding) by the transfer of electrons between a sodium atom and a chlorine atom. A sodium atom (Na) has one valence electron, which it readily loses to achieve a stable octet, forming a positively charged sodium ion (Na+\text{Na}^+). A chlorine atom (Cl) has seven valence electrons and readily gains one electron to achieve a stable octet, forming a negatively charged chloride ion (Cl−\text{Cl}^-). The oppositely charged ions (Na+\text{Na}^+ and Cl−\text{Cl}^-) are then held together by strong electrostatic forces of attraction, forming the ionic compound sodium chloride.

    ii- Two properties of electrovalent (ionic) compounds are:

    • They typically have high melting and boiling points due to the strong electrostatic forces between ions.
    • They are generally soluble in polar solvents like water and can conduct electricity when molten or dissolved in water.

    Question Five (5)

    a)
    i- The solubility of a salt refers to the maximum amount of that salt (solute) that can dissolve in a given amount of solvent at a specific temperature to form a saturated solution.

    ii- Two examples of insoluble salts are:

    • Silver chloride (AgCl\text{AgCl})
    • Barium sulfate (BaSO4\text{BaSO}_4)

    b)

  7. Calculate the molar mass of KNO3\text{KNO}_3.

    Molar mass of K = 39 g/mol
    Molar mass of N = 14 g/mol
    Molar mass of O = 16 g/mol

    Molar mass of KNO3=39+14+(3×16)=39+14+48=101 g/mol\text{Molar mass of KNO}_3 = 39 + 14 + (3 \times 16) = 39 + 14 + 48 = 101 \text{ g/mol}
  8. Calculate the number of moles of KNO3\text{KNO}_3.

    Moles of KNO3=MassMolar mass=30.5 g101 g/mol≈0.302 mol\text{Moles of KNO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{30.5 \text{ g}}{101 \text{ g/mol}} \approx 0.302 \text{ mol}
  9. Convert the volume of water from cm3\text{cm}^3 to dm3\text{dm}^3.

    Volume of water=30 cm3=301000 dm3=0.030 dm3\text{Volume of water} = 30 \text{ cm}^3 = \frac{30}{1000} \text{ dm}^3 = 0.030 \text{ dm}^3
  10. Calculate the solubility in moldm−3\text{moldm}^{-3}.

    Solubility=Moles of soluteVolume of solvent=0.302 mol0.030 dm3≈10.07 moldm−3\text{Solubility} = \frac{\text{Moles of solute}}{\text{Volume of solvent}} = \frac{0.302 \text{ mol}}{0.030 \text{ dm}^3} \approx 10.07 \text{ moldm}^{-3}

    The solubility of KNO3\text{KNO}_3 is approximately 10.1 moldm−3\boxed{10.1 \text{ moldm}^{-3}}.

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