EJC 2022 GCE A-Level H1 Chemistry Paper 2 Suggested Solution

EUNOIA JUNIOR COLLEGE

Bond angle is 107º; Shape is trigonal pyramidal. Electron pairs around the central atom arrange themselves to be as far away from each other as possible to minimise inter-electronic repulsion. Order of repulsion: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair There are 3 σ bond pairs and 1 lone pair around the central nitrogen in NH3, hence giving a trigonal pyramidal shape with a bond angle slightly smaller than 109.5º. Each N–H bond is polar. The dipoles due to the three N–H bonds do not cancel each other in the trigonal pyramidal arrangement. Hence ammonia is a polar molecule. 130º. This is because the unpaired electron on N will occupy less space than the two electron pairs, giving an angle larger than the ideal 120º for a trigonal planar.

Across Period 3 from Na to Cl, first electron affinity involves adding one electron to the valence shell of the gaseous atom of the element. Since nuclear charge increases across the Period due to an increase in the number of protons in the nucleus, while shielding remains essentially constant as the core electrons remains the same. Hence effective nuclear charge experienced by the incoming electron increases across the Period, resulting in stronger attraction between the nucleus and the incoming electron, releasing more energy, giving a higher magnitude of the first electron affinity. X–(g) + e– → X2–(g) Mg(g) → Mg2+(g) + 2e– amount of electrons removed 2= amount of Mg = 2 mol −19 23charge of 1 mole of electrons 10=− 1.6010  6.02 =− 96320 coulombs charge removed from 1 mole of Mg atoms= 2− 96320 =− 192640 coulombs –192640

2C18H35O2Na(aq) + Ca2+(aq) → (C18H35O2)2Ca(s) + 2Na+(aq) carbon and hydrogen Non-polar covalent bonds as carbon and hydrogen are non-metals with similar electronegativity.

Instantaneous dipole-induced dipole attraction since the hydrophobic part of the detergent molecule is hydrocarbon in nature, which is non-polar. Hydrogen bonding since the hydrophilic parts of the detergent molecule is able to bond strongly with water, rinsing out with water. Ionic bonding since the cationic surfactant molecules bond to the hair but do not rinse off, the interaction with hair must be stronger than its ion-dipole interaction with water. The hair strands must have been negatively charged. The positive hydrophilic head of the cationic surfactant forms strong electrostatic forces of attraction with the hair strands, hence cannot be rinsed off. On the other hand, repulsion between the hair strands and the negative hydrophilic head of the anionic surfactants facilitates their rinsing off the hair.

When the eggshell powder is not longer visible or is seen to have completely dissolved and there is no more effervescence or bubbling observed. This is because the CaSO4 formed being insoluble, will coat the surface of the eggshell powder particles, preventing further reaction of the CaCO3 inside the particle.

The pH when the strong acid, hydrochloric acid, is just neutralised by the strong base, aqueous sodium hydroxide, is neutral at pH 7. Hence an indicator with a pH range of colour change including pH 7, that is, bromothymol blue, is more suitable. 30.0500= 1.0210 mol−20.40amount of NaOH used= 1000 NaOH + HCl → NaCl + H2O amount of HCl in 25.0 cm of diluted solution= amount of NaOH used=1.0210 mol 3− 2250 1.021025.0 1.02 10 mol−==  2 0.5 00 0.0= 500 mol1000 amount of HCl reacted with eggshell= 0.0500−1.0210 0.0398 mol− = 3amount of HCl in 250 cm of diluted solution 100initial amount of HC presentl= 3 1amount of CaCO present in eggshell  amount of HCl reacted 0.0398 2 1 2 = == 0.0199 mol )3  mass of CaCO in the eggshell= 0.0199 (40.1+12.0+ 16.0 3 = 1.992 g 3 1.992percentage by mass of CaCO in eggshell 1003.62=  = 55.0% Since the percentage of calcium carbonate in the eggshell is less than 88%, the farmer needs to change the feed.

Upon increasing the pressure, by Le Chatelier's Principle, the system will attempt to counter this change by decreasing the pressure of CO2, shifting the position of equilibrium to the left side, with no gaseous particles. CaO(s) + H2O(l) → Ca(OH)2(aq)

C12H22O elimination condensation

Restricted rotation about the carbon-carbon double bond. Add orange aqueous bromine. Decolourisation of the orange aqueous bromine.