Solve this chemistry questions

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). 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. Step 1: Calculate the mass of each element. Mass of Carbon (C) = 75 g Mass of Hydrogen (H) = 25 g Step 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 Moles of H = 25 g1 g/mol = 25 mol Step 3: Determine the simplest mole ratio by dividing by the smallest number of moles. Smallest number of moles = 6.25 mol Ratio for C = (6.25)/(6.25) = 1 Ratio for H = (25)/(6.25) = 4 Step 4: Write the empirical formula. The empirical formula is 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 (SO_4^2-) • Nitrate (NO_3^-) • Carbonate (CO_3^2-) • Phosphate (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 (p_x, p_y, p_z) in a given principal energy level are degenerate. b) i- Electronic configuration: a- Scandium (Sc, atomic number 21): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1 or [Ar] 4s^2 3d^1 b- Nickel (Ni, atomic number 28): 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8 or [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^ at standard pressure) and freezing point (0^ at standard pressure). Question Four (4) a) Step 1: Use the given pOH value. pOH = 12.0 Step 2: Calculate the concentration of hydroxide ions using the formula. [OH^-] = 10^-pOH [OH^-] = 10^-12.0 [OH^-] = 1.0 × 10^-12 M b) i- Sodium chloride (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^+). 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^-). The oppositely charged ions (Na^+ and 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) • Barium sulfate (BaSO_4) b) Step 1: Calculate the molar mass of 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 KNO_3 = 39 + 14 + (3 × 16) = 39 + 14 + 48 = 101 g/mol Step 2: Calculate the number of moles of KNO_3. Moles of KNO_3 = MassMolar mass = 30.5 g101 g/mol ≈ 0.302 mol Step 3: Convert the volume of water from cm^3 to dm^3. Volume of water = 30 cm^3 = (30)/(1000) dm^3 = 0.030 dm^3 Step 4: Calculate the solubility in moldm^-3. Solubility = Moles of soluteVolume of solvent = 0.302 mol0.030 dm^3 ≈ 10.07 moldm^-3 The solubility of KNO_3 is approximately 10.1 moldm^-3. What's next? Send 'em! 📸