Question 1:

Step 1: Volume given
227 cm³ of ammonia gas at STP.

Use 22.7 dm³ as the gas molar volume, which equals 22700 cm³:

Step 2: Moles

Step 3: Molecules

This matches option C exactly.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 2:

We know:

• Three identical containers → same volume, temperature, pressure.

• Ideal gas law: PV=nRT PV = nRT → same number of moles n n in each container.

• So same number of molecules in each container.

Step 1: Check each option

A. All three containers contain the same number of atoms

• X (H₂): 2 atoms per molecule → 2natoms

• Y (CO₂): 3 atoms per molecule → 3natoms

• Z (C₂H₆): 8 atoms per molecule → 8natoms
  → Different number of atoms → False.

B. Container X contains the most hydrogen atoms

• X: H₂ → 2n H atoms

• Y: CO₂ → 0 H atoms

• Z: C₂H₆ → 6n H atoms
  → Z has more H atoms than X → False.

C. Container Z contains the most molecules

• Same moles → same molecules in each → False.

D. Container Y has the largest mass
Mass = moles × molar mass. Molar masses:

• X (H₂): 2 g/mol

• Y (CO₂): 44 g/mol

• Z (C₂H₆): 30 g/mol
  Same n → largest molar mass gives largest mass → CO₂ (44) wins → True.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 3:

Step 1: Find moles of NO₂

n = 22.7 45.4​ = 2.00 mol

Step 2: Number of molecules

Step 3: Atoms per molecule

NO₂ has 1 N + 2 O = 3 atoms per molecule.

Step 4: Match with options

That is 3.61 × 10²⁴ → Option D.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 4:

We use the mole ratio from the balanced equation and assume all gases are measured at the same temperature and pressure (so volumes are proportional to moles).

Step 1: Reaction

Mole ratio:
2 volumes C₄H₁₀ : 13 volumes O₂ : 8 volumes CO₂

Step 2: Scale from 2 volumes to 100 cm³ of butane

If 2 volumes → 13 volumes O₂, then 1 volume → 6.5 volumes O₂.

So 100 cm³ C₄H₁₀ requires:

Step 3: CO₂ produced

2 volumes C₄H₁₀ → 8 volumes CO₂, so 1 volume → 4 volumes CO₂.

100 cm³ C₄H₁₀ produces:

Step 4: Match with table

O₂ = 650 cm³, CO₂ = 400 cm³ → Option D.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 5:

At STP, 1 mole of gas occupies 22.7 dm³ = 22700 cm³.

Step 1: Find moles from volume

Step 2: Simplify

Step 3: Molar mass

Step 4: Match with options

That’s 28 g mol⁻¹ → Option B.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 6:

Step 1: Balanced equation

Water is liquid, so only gases contribute to final volume.

Step 2: Determine limiting reagent
Volumes at same T,P → volume ratio = mole ratio.

Stoichiometry: 1 vol CH₄ reacts with 2 vol O₂.

Given: 200 cm³ CH₄ and 600 cm³ O₂.

O₂ required for 200 cm³ CH₄ = 200×2=400 cm³.
We have 600 cm³ O₂ → O₂ is in excess.
CH₄ is limiting.

Step 3: Reaction completion
200 cm³ CH₄ reacts with 400 cm³ O₂ → produces:

• CO₂: 200 cm³ (1:1 ratio with CH₄)

• Liquid H₂O: no gas volume.

Step 4: Remaining gases
O₂ left = 600 − 400 = 200 cm³.
CO₂ produced = 200 cm³.

Total gas volume = 200 (O₂) + 200 (CO₂) = 400 cm³.

Step 5: Match option
That’s B.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 7:

We’ll compare the number of atoms in each.

Option A: 454 cm³ CO₂ at STP
Molar volume = 22700 cm³/mol.
Moles = 454/22700=0.02 mol.
Molecules = 0.02×N_A
Atoms per molecule = 3 (C + 2O).
Total atoms = 0.02×N_A×3=0.06N_A

Option B: 908 cm³ He at STP
Moles = 908/22700=0.04 mol.
Atoms = 0.04×N_A (each molecule is 1 atom).
Total atoms = 0.04 N_A .

Option C: 0.16 g CH₄
Molar mass CH₄ = 16 g/mol.
Moles = 0.16/16=0.01 mol.
Atoms per molecule = 5 (C + 4H).
Total atoms = 0.01×N_A ×5=0.05N_A

Option D: 0.64 g SO₂
Molar mass SO₂ = 64 g/mol.
Moles = 0.64/64=0.01 mol.
Atoms per molecule = 3 (S + 2O).
Total atoms = 0.01×N_A×3=0.03N_A

Compare:
A: 0.06N_A
B: 0.04N_A
C: 0.05N_A
D: 0.03N_A​

Greatest is A.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 8:

Step 1: Reaction

All products are gases here.

Step 2: Volume ratios (same T,P)
From equation:
2 vol H₂S + 3 vol O₂ → 2 vol H₂O + 2 vol SO₂ → total 4 vol products.

Per 2 vol H₂S: products = 4 vol.
So 1 vol H₂S → 2 vol products.

Step 3: Limiting reagent
Given: 40 dm³ H₂S, 100 dm³ O₂.

O₂ needed for 40 dm³ H₂S:
From ratio 2 : 3 → 40 dm³ H₂S requires 40×3/2=60 dm³ O₂.
We have 100 dm³ O₂ → O₂ in excess.
H₂S is limiting.

Step 4: Products volume
From 40 dm³ H₂S:
Products total = 40×4/2=80 dm³.

Step 5: Oxygen remaining
O₂ used = 60 dm³.
Initial O₂ = 100 dm³.
Remaining O₂ = 100−60=40 dm³.

Step 6: Match table
Products = 80 dm³, O₂ remaining = 40 dm³ → Option C.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 9:

Step 1: General combustion reaction

All volumes measured at same T,P → volume ratio = mole ratio.

Step 2: From data
20 cm³ hydrocarbon → 120 cm³ CO₂.
Mole ratio: 1 mole hydrocarbon → x moles CO₂.
So:

120/20=x

x=6

Step 3: Water vapour
20 cm³ hydrocarbon → 120 cm³ H₂O vapour.
1 mole hydrocarbon → y/2 moles H₂O.
So:

120/20=y/2

y = 12

Step 4: Formula
C₆H₁₂ → matches option B.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 10:

Step 1: Reaction for oxygen production
From the equation:

4 moles of electrons produce 1 mole of O₂ gas.

Step 2: Moles of electrons given
Number of electrons = 2.40×10²²
Avogadro’s number NA=6.02×10²³ mol⁻¹
Moles of electrons = 2.40×10²² x 6.02×10²³≈0.03987

Step 3: Moles of O₂ produced
From stoichiometry:

Step 4: Volume at STP
Molar volume at STP = 22.7 dm³/mol=22700 cm³/mol

Volume of O₂ =0.0099675×22700≈226.2 cm³

≈ 227 cm³.

Step 5: Match option
That’s A.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.