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1.
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g) …… ΔHrxn

ΔHrxn
= 3 × ΔHf[CO₂(g)] + 4 × ΔHf[H₂O(g)] - ΔHf[C₃H₈(g)] - 5 × ΔHf[O₂(g)]
= 3 × (-393.5) + 4 × (-241.8) - (-103.9) - 5 × (0.00) kJ
= -2043.8 kJ


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2.
For 1 mole of CH₄O(g) at STP :
Mass = (12.0 + 1.0×4+ 16.0) g = 32.0 g
Volume = 22.4 L

Density of CH₄O(g) at STP = (32.0 g) / (22.4 L) = 1.43 g/L


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3.
A.
For 1 mole of benzene (C₆H₆) :
Mass = (12.0×6 + 1.0×6) g = 78.0 g
No. of molecules = 6.02 × 10²³

Mass of one benzene molecule = (78.0 g) / (6.02 × 10²³) = 1.30 × 10⁻²² g

B.
Mass of one mole of benzene molecules = (12.0×6 + 1.0×6) g = 78.0 g

C.
Molar mass of C₆H₆ = (12.0×6 + 1.0×6) g/mol = 78.0 g/mol
No. of moles of C₆H₆ in 156-g sample = (156 g) / (78.0 g/mol) = 2.00 mol

D.
Avogadro constant = 6.02 × 10²³ /mol
No. of moles of C₆H₆ in 156-g sample = 2.00 mol
No. of molecules of C₆H₆ in 156-g sample = (2.00 mol) × (6.02 × 10²³ /mol) = 1.204 × 10²⁴ (molecules)

E.
No. of C atoms in one molecule of C₆H₆ = 6

F.
No. of H atoms in one molecule of C₆H₆ = 6
From Part D, no. of molecules of C₆H₆ in 156-g sample = 1.204 × 10²⁴ (molecules)
Hence, no. of H atoms in 156-g C₆H₆ sample = (1.204 × 10²⁴) × 6 = 7.224 × 10²⁴ (atoms)