Answer:
The nswer to the question is
The maximum fraction of the air in the room that could be displaced by the gaseous nitrogen is 0.548 or 54.8 %
Explanation:
To solve the question we note that
The density of the liquid nitrogen = 0.808g/mL and the volume is 195 L tank (vaporised)
Therefore since density = mass/volume we have
mass = Density × volume = 0.808 g/mL × 195 L × 1000 ml/L =157560 g
In gaseous form the liquid nitrogen density =1.15 g/L
That is density = mass/volume and volume = mass/density = 157560 g/(1.15g/L) or
volume = 137008.69565 L
The dimension of the room = 10 m × 10 m × 2.5 m = 250 m³ and
1 m³ is equivalent to 1000 L, therefore 250 m³ = 250 m³ × 1000 L/m³ = 250000L
Therefore fraction of the volume occupied by the gaseous nitrogen =
137008.69565 L/250000 L = 0.548
Therefore the gaseous nitrogen occpies 54.8% of the room
Data:
solute: ethylene glicol => not ionization
molar mass of ethylene glicol (from internet) = 62.07 g/mol
solute = 400 g
solvent = water = 4.00 kg
m =?
ΔTf = ?
Kf = 1.86 °C/mol
Formulas:
m: number of moles of solute / kg of solvent
ΔTf = Kf*m
number of moles of solute = mass in grams / molar mass
Solution
number of moles of solute = 400 g / 62.07 g/mol = 6.44 moles
m = 6.44 mol / 4 kg = 1.61 m <-------- molality (answer)
ΔTf = 1.86 °C / m * 1.61 m = 2.99 °C <---- lowering if freezing point (answer)
Answer:
x= 138.24 g
Explanation:
We use the avogradro's number
6.023 x 10^23 molecules -> 1 mol C2H8
26.02 x 10^23 molecules -> x
x= (26.02 x 10^23 molecules * 1 mol C2H8 )/6.023 x 10^23 molecules
x= 4.32 mol C2H8
1 mol C2H8 -> 32 g
4.32 mol C2H8 -> x
x= (4.32 mol C2H8 * 32 g)/ 1 mol C2H8
x= 138.24 g
