Calculate first the number of moles of ethylene glycol by dividing the mass by the molar mass.
n = (6.21 g ethylene glycol) / 62.1 g/mol
n = 0.1 mol
Then, calculate the molality by dividing the number of moles by the mass of water (in kg).
m = 0.1 mol/ (0.025 kg) = 4m
Then, use the equation,
Tb,f = Tb,i + (kb)(m)
Substituting the known values,
Tb,f = 100°C + (0.512°C.kg/mol)(4 mol/kg)
<em>Tb,f = 102.048°C</em>
Answer:
2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.
Explanation:
Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g
1 kg = 1000 g
Molar mass of fat = M
M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]
Moles of fat =
According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give ;
of water
Mass of 148.31 moles of water ;
148.31 mol × 18 g/mol = 2,669.58 g
2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.
Answer:
volume of
Explanation:
Firstly balance the given chemical equation,
From the given balance equation it is clearly that,
2 mole of Li gives 1 mole of H2 gas
⇔
⇔
⇔
hence
3 mole of Li will give 1.5 mole H2 gas
therefore volume of gas produced from 3 mole Li at
volume of H2=33.6 litre