Combustion equation of n-hexane:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles
LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1%
b)
1.1 volume percent required for LFL
1.1% x 1
= 0.0011 m³ of n-hexane required
DNA
sorry if it’s wrong, but that answer seems reasonable
Answer:
244.76
Explanation:
The weight of grams by 1.82 moles of lithium carbonate would be: 134.481438 So we need to use this equation to find mass by grams m × g = ms Where m is for moles, g for grams, and ms for mass. So now we need use this equation: 1.82 × 134.481438 = ? 1.82 × 134.481438 = 244.75621716 244.75621716, rounded-up (to the nearest-tenths place) is 244.76. So now you have it! The mass of 1.82 moles of lithium carbonate is 244.76!
