Answer:
v = 37.9 ml
Explanation:
Given data:
Mass of compound = 1.56 kg
Density = 41.2 g/ml
Volume of compound = ?
Solution:
First of all we will convert the mass into g.
1.56 ×1000 = 1560 g
Formula:
D=m/v
D= density
m=mass
V=volume
v = m/d
v = 1560 g / 41.2 g/ml
v = 37.9 ml
Answer:
101,37°C
Explanation:
Boiling point elevation is one of the colligative properties of matter. The formula is:
ΔT = kb×m <em>(1)</em>
Where:
ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-
kb is ebulloscopic constant (0,52°C/m)
And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):
203g × (1mol /62,07g) = <em>3,27moles of ethlyene glycol</em>
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Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m
Replacing these values in (1):
X - 100°C = 0,52°C/m×2,64m
X - 100°C = 1,37°C
<em>X = 101,37°C</em>
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I hope it helps!
When a water vapor condenses, heat is being released from the process. This heat is called latent heat of vaporization since the phase change happens without any change in the temperature. This value is constant per mole of a substance as a function of pressure and temperature. For this problem, we are given the heat of vaporization at a certain T and P. We use this value to calculate the total heat released from the process. We calculate as follows:
Total heat released: 32.4 g ( 1 mol / 18.02 g ) (40.67 kJ / mol) = 73.12 kJ
Therefore, 73.12 kJ of heat is released from the condensation of 32.4 g of water vapor.
Answer:
lesser the molar mass of the gas higher the no. of moles included in a certain mass sample. ie at STP more volume is required for the gas having less molar mass.
He has the smallest molar mass.
Therefore bag of He is the biggest.
