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Ionization energy is said to be the amount of energy required to remove a valence electron from the atom of an element to become a gaseous specie. Ionization energy decrease as you move down the group and increase across the period in the periodic table. Consequently, the elements under consideration are in the same period, period 3, ranging from Aluminium to Silicon to Phosphorus to Sulphur. i.e Al < Si < P < S.

C\(_3\)H\(_8\) + 5 O\(_2\) → 3CO\(_2\) + 4H\(_2\)O From the equation: 1 mole : 3 moles If 0.1 mole reacted, then 0.3 mole will be formed. ⇒ 0.3 mole of CO\(_2\) will be produced. Recall that Number of moles of gases = \(\frac{Volume}{Molar Volume of gas at stp}\) n = \(\frac{V}{M.V}\) 0.3 = \(\frac{V}{22.4}\) V = 0.3 x 22.4 V = 6.72 dm\(^3\)

Relative Atomic Mass of Boron = [19.9% \(^{10}\)B} + [80.1% \(^{11}\)B] = [ 19.9% of 10] + [80.1% of 11] = [ \(\frac{19.9}{100}\times{10}\) ] + [ \(\frac{80.1}{100}\times{11}\)] = 1.99 + 8.811 = 10.801

In methane (CH\(_4\)), there are four shared pairs of electrons. A single carbon atom in the center shares one pair of electrons with each of the four hydrogen atoms, resulting in four shared pairs.

Isotopes differ in the number of neutrons in their nucleus, but they have the same number of protons (which determines the number of electrons). Since chemical properties are primarily determined by the electron configuration, isotopes of the same element behave similarly in chemical reactions.

Milk of magnesia is alkaline because it has a pH of 10.5, which means that it would neutralize any acidic substance encountered. Hence, it is an excellent antacid. It neutralizes excess stomach acid or hydrochloric acid. The hydroxide ions of the milk combine with the hydrogen ions in the stomach acid to reduce excess acidity in the stomach.

Given: M\(_r\) of MgSO\(_4\).7H\(_2\)O = 246 M\(_r\) of Oxygen = 16 Percentage by mass of Oxygen = \(\frac{Total mass of Oxygen}{Total mass of MgSO_4.7H_2O}\times 100\) % by mass of Oxygen = \(\frac{11moles of oxygen}{MgSO_4.7H_2O}\times 100\) (4moles of oxygen + 7 moles of oxygen from 7H\(_2\)O ) = \(\frac{11 X 16}{246}\times 100\) = \(\frac{176}{246}\times 100\) = 71.544% = 71.5%

A catalyst alters the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy, meaning it makes it easier for reactants to reach the transition state and form products, thus significantly increasing the reaction rate without being consumed in the process.

2C\(_2\)H\(_6\)\(_(g)\) + 7O\(_2\)\(_(g)\) → 6H\(_2\)O\(_(g)\) + 4CO\(_2\)\(_(g)\) 2 moles 6 moles 400cm\(^3\) X cm\(^3\) X = \(\frac{{400}\times{6}}{2}\) X = 1200cm\(^3\)

A saturated solution at 30^\circC will normally produce crystals when it is cooled below the saturation point for that specific solute at a lower temperature; this means that crystals will typically begin to form as the solution cools down from 30^\circC, with the exact temperature depending on the solubility curve of the dissolved substance.

Although fats are complex molecules that are made up of fatty acids and glycerol, they are not polymers. Polymers are large molecules made up of many smaller units called monomers that are chemically bonded together. Rubber, starch and proteins are polymers because they have several units of isoprene, glucose and amino acids respectively, chemically bonded together.

The collision theory assumes that the colliding molecules have sufficient energy to be able to overcome the intermolecular forces of attraction and the activation energy barriers, but it does not require that the collision be perfectly elastic. i.e no loss of energy.

While a strong acid dissociates or ionizes completely in water, weak acids only partially ionizes in water.

Esterification reaction is the reaction between an alkanol (alcohol) and an alkanoic acid (carboxylic acid) in the presence of an acid catalyst forms an ester (alkanoate) and water. To produce alkanoates (esters) from alkanols and alkanoic acids, you need to perform esterification, which requires a concentrated acid catalyst (like sulphuric acid) and heat.

Given: m = 22g M.V = 22.4dm\(^3\) V = ? First, calculate the number of moles n, by dividing mass of CO\(_2\) by the molar mass of CO\(_2\), since n = \(\frac{mass}{Molar mass}\) Molar mass of CO\(_2\) = 12 + (2 x 16) = 44g/mol So, n = \(\frac{mass}{Molar mass}\) n = \(\frac{22}{44}\) = 0.5 mole Also recall that number of moles of gases, n = \(\frac{Volume}{Molar Volume}\) n = \(\frac{V}{M.V}\) 0.5 = \(\frac{V}{22.4}\) V = 0.5 X 22.4 V = 11.2 dm\(^3\) Alternatively, Given the mass of CO\(_2\) = 22g Molar mass of CO\(_2\) = 12 + (16X2) = 44g number of moles, n = \(\frac{m}{M}\) = \(\frac{22}{44}\) = 0.5 mole. If 1 mole of CO\(_2\) = 22.4dm\(^3\) 0.5 mole of CO\(_2\) = X dm\(^3\) X = 0.5 x 22.4 = 11.2 dm\(^3\)