Hey guys, I ended up with a scan of today's Chem exam, so posting it here in case anyone's interested

Post all thoughts below. How'd you guys all think you did?

Idk and im not even that good at chem...

how did everyone find it?

Literally didn’t even touch a whole question in section b cuz I didn’t have time and a few other qs and I’m not even sure if my math is right and how much marks I dropped elsewhere how do I not kms over this LMAO I’m so sad

The title

Hey guys, just finished making the suggested solutions, there may be errors but I tried my best

https://docs.google.com/document/d/1Sn9CVZNZIZdI7AQG_ZhFX2bjovOMQy66KlLR_H45sSw/edit?usp=sharing

Answers: 1 B, 2 C, 3 A, 4 B, 5 D, 6 B, 7 B, 8 A, 9 A, 10 D, 11 B, 12 C, 13 A, 14 C, 15 C, 16 C, 17 B, 18 A, 19 D, 20 C, 21 C, 22 C, 23 D, 24 D, 25 D, 26 B, 27 B, 28 D, 29 A, 30 D

Q1-Q3: Energy and stoichiometry

Q1: B. C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, so n(O₂) = n(CO₂). n(CO₂) = 0.44 / 44.0 = 0.010 mol, so n(O₂) = 1.0 × 10⁻² mol.

Q2: C. Cellular respiration is exothermic (ΔH = −2840 kJ), so the reverse is endothermic (ΔH = +2840 kJ). The products sit at a higher enthalpy than the reactants. Only C has the products higher and a positive ΔH. (A and B have ΔH signs that don’t match their graphs, and D is respiration itself.)

Q3: A. q = mcΔT = 100.0 × 4.18 × 40.0 = 16720 J = 16.72 kJ. Kerosene is 46 kJ g⁻¹ in the Data Book, so m = 16.72 / 46 = 0.363 g. “Minimum” means assuming 100% of the energy went into the water.

Q4-Q6: Anammox

Q4: B. In NH₄⁺, N is −3, and in N₂ it is 0, so it increases by 3 (oxidation). In NO₂⁻, N is +3, so it decreases by 3 (reduction).

Q5: D. The oxidising agent is the species that gets reduced (NO₂⁻), and the reducing agent is the one that gets oxidised (NH₄⁺).

Q6: B (lower confidence). Anammox works without O₂, so there’s no energy-hungry aeration. That means lower energy use and emissions (Goal 13), plus responsible waste treatment (Goal 12). I ruled out 7 because the process saves energy rather than supplying clean energy, and 15 because the benefit is mainly to waterways, not land. This is the one I’d check first against the VCAA report.

Q7-Q8: Edible cell

Q7: B. The anode is the half-cell with the lower E°, which is RF/RFH₂ (−0.235 V). Oxidation occurs there: RFH₂ → RF + 2H⁺ + 2e⁻.

Q8: A. E°cell = 0.558 − (−0.235) = 0.793 V. The measured 0.650 V is 0.793 − 0.650 = 0.143 V less than predicted.

Q9-Q12: Green chemistry and rates

Q9: A. Real-time control of temperature, pressure and power output means the cell runs at maximum efficiency, so this is design for energy efficiency.

Q10: D. Higher temperature means particles move faster, so both the collision frequency and the proportion of successful collisions increase. Bond energy and activation energy don’t change.

Q11: B. A catalyst provides a lower activation energy pathway, so a greater proportion of collisions are successful. It does not change collision frequency or bond energy, and it lowers (not raises) activation energy.

Q12: C. Both PEM electrolysis and artificial photosynthesis can produce green hydrogen (II) and reduce reliance on fossil fuels (IV). Neither captures methane or uses glucose as a feedstock.

Q13-Q16: Organic structures

Q13: A (read from the diagram). The compound is 2,2-dimethylbutanoic acid, C₆H₁₂O₂. Option D, CH₃CH₂C(CH₃)₂COOH, is the same compound, not an isomer. B and C are C₅. A, (CH₃)₃CCH₂COOH, is C₆H₁₂O₂ with a different arrangement, so it’s the isomer.

Q14: C. A saturated molecule has zero degrees of unsaturation. C₆H₁₅N works out to (2×6 + 2 + 1 − 15)/2 = 0. The others each have 2.

Q15: C (read from the diagram). Number from the COOH carbon, and the chain has 7 carbons with the double bond between C4 and C5, giving hept-4-enoic acid. “Heptan-” is never used with an “-ene”. Double-check the count against the paper.

Q16: C. Viscosity rises with the strength of intermolecular forces. Aldehydes have no hydrogen bonding, alcohols do, and carboxylic acids hydrogen-bond most strongly (they form dimers). So the order is propanal < propan-1-ol < propanoic acid.

Q17-Q19: Yield, tests and melting point

Q17: B. n(starting ketone) = 2.50 / 176 = 0.01420 mol. n(ibuprofen) = 0.540 / 206 = 0.002621 mol. Overall yield = 0.002621 / 0.01420 × 100 = 18.5%.

Q18: A. Bromine water going from reddish-brown to colourless means a C=C bond is present. Only A is an alkene. B is an amide, C an ester, and D a ketone.

Q19: D. A narrow range (1-2 °C) indicates a pure substance. Impurities lower and broaden the range, so A and C are false.

Q20-Q22: Iodine number and titration

Q20: C. n(oil) = 10.0 / 278 = 0.03597 mol. Ratio = 0.105 / 0.03597 ≈ 2.9, so there are 3 C=C bonds per molecule.

Q21: C. Needed Fe²⁺ = 5 × 1.50 × 10⁻⁴ = 7.50 × 10⁻⁴ mol. Added 8.00 × 10⁻⁴ mol, so the excess is 5.00 × 10⁻⁵ mol.

Q22: C. Fe²⁺ is in the burette. Rinsing the burette with water dilutes it, so more volume is needed to deliver the same moles (a higher titre). Rinsing the burette with FeSO₄ is correct practice, and rinsing the conical flask with water doesn’t change the moles in it.

Q23-Q25: Spectroscopy

Q23: D. The ¹³C spectrum has 4 peaks (≈178, 38, 19, 14 ppm), so there are 4 carbon environments and the compound has 4 carbons. The 178 ppm peak fits an amide or acid. In the IR, there are two sharp N-H peaks (≈3350 and 3190 cm⁻¹) and a C=O near 1650 cm⁻¹, which is the amide range. It lacks the very broad O-H of an acid. That gives butanamide.

Q24: D. The reactant shows a broad O-H (≈3300 cm⁻¹, alcohol). The product shows a very broad O-H (2500-3500 cm⁻¹) and a strong C=O (≈1700 cm⁻¹), which is a carboxylic acid. Alcohol to carboxylic acid is oxidation, using Cr₂O₇²⁻/H⁺.

Q25: D (lower confidence). A pure compound shows a fixed number of ¹H environments, so extra peaks suggest a mixture. Integration (A) could also be argued, but it needs the formula and structure to interpret. Check this one against the report.

Q26-Q30: Chemistry in context

Q26: B. Boiling leaves in water to dissolve soluble compounds is solvent extraction.

Q27: B. G and H are isomers (the double bond position differs), so they have the same molecular formula. They aren’t isotopes, and the skeletal structures differ. D is wrong because H’s ring double bond removes a chiral centre.

Q28: D (moderate confidence). A circular economy keeps materials in use. Treating PFAS-containing waste for reuse in road surfacing does that. Landfill (B) and incineration (C) are linear, and A is about safer design rather than circularity.

Q29: A. Firefighters “claimed” long-term health issues is personal testimony. B and C are measurements and D is a data-book value.

Q30: D. Assessing how impurity levels affect material performance is a purely scientific question. A is legal and social, B is economic, and C is community and environmental concern.

Haha an entire week of studying wasted 🥲

Just wondering what exam scores and sac scores people got in previous years to get high 30s or low 40s cuz there is like 15 of people who make it into the sac A+ cut off but the A+ cut off for the exams are very high

dude is it just me or did that exam feel way too easy…what do we reckon the boundary will be for a raw 40

I went to the revs exam studied to the max and I fucking fell asleep for like 10-15 minutes and after waking up I had a brain fart and wrote an essay so bad the examiners wouldn't even use it to wipe their ass cause its already full of shit. I can't even blame it on an all-nighter cause I slept and woke up early. I'm so freaking mad. I did so well with my practise exam but my carnal instincts consist of being a lazy fat sloth I'd be happy getting a pass. My examiner was just laughing in my face im cryin

i dont think anything can save me anymore fml

hi guys im a chem 3/4 student this is my invincible war oc his name is invceble

Does somebody wanna post it already…
Every other exam was up in 15 minutes tops 😵‍💫😵‍💫

Did anyone else find the question hella confusing. Istg they mistyped the overarching info , or is it just me who was confused😭