<span>There are 1000 cm3 in 1 liters.
Hence 1 liter of the liquid would weigh:
1000 cm3 x (1.17 g/cm3) = 1170 gm
and there are 1000 gm in 1 kg, so we want enough liters to have a mass of
3.75 kg x 1000 gm/kg = 3750 gm
Hence, # of liters = desired mass / # of gm per liter
= 3750 gm / 1170 gm/liter
= 3.2051282 liters</span>
As,
5471 kJ heat is given by = 1 mole of Octane
Then,
5310 kJ heat will be given by = X moles of Octane
Solving for X,
X = (5310 kJ × 1 mol) ÷ 5471 kJ
X = 0.970 moles of Ocatne
So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
Moles = mass / M.mass
Or,
Mass = Moles × M.mass
Putting values,
Mass = 0.970 mol × 114.23 g/mol
Mass = 110.83 g of Octane
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