Answer: The vapor pressure of water at 298 K is 3.565kPa.
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
where,
= initial pressure at 298 K = ?
= final pressure at 373 K = 101.3 kPa
= enthalpy of vaporisation = 41.1 kJ/mol = 41100 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 298 K
= final temperature = 373 K
Now put all the given values in this formula, we get
![\log (\frac{101.3}{P_1})=\frac{41100}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{373K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B101.3%7D%7BP_1%7D%29%3D%5Cfrac%7B41100%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B298K%7D-%5Cfrac%7B1%7D%7B373K%7D%5D)
Therefore, the vapor pressure of water at 298 K is 3.565kPa.
Answer:
mass of the butter is 2.107 kg
Explanation:
specific gravity = destiny of the substance / density of
the water at 4°C
0.86 = density of the butter / 1000
where density of the water is 1000 kg/m^3 at 4°C
density of the butter = 860kg/m^3
now,
density of the butter = mass of the butter / volume
860 = mass of the butter / 2.45 × 10^-3
( 1 L = 10^ -3 m^3)
mass of the butter = 2.107 kg
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Answer:
d) 8.01 E23 atoms
Explanation:
∴ mass C = 16 g
∴ molar mass C = 12.0107 g/mol
⇒ mol C = (16 g)*(mol/12.0107 g) = 1.332 mol
⇒ atoms C = (1.332 mol)*(6.022 E23 atoms/mol)
⇒ atoms C = 8.02 E23 atoms
