Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
Assume there is 100g of the substance at first
Answer:
A sample of an ideal gas has a volume of 2.21 L at 279 K and 1.01 atm. Calculate the pressure when the volume is 1.23 L and the temperature is 299 K.
You need to apply the ideal gas law PV=nRT
You have the pressure, P=1.01 atm
you have the volume, V = 2.21 L
The ideal gas constant R= 0.08205 L. atm/ mole.K at 273 K
find n = PV/RT = (1.01 atm x 2.21 L / 0.08205 L.atm/ mole.K x 273 K)
n= 0.1 mole, Now find the pressure for n=0.1 mole, T= 299K and
L=1.23 L
P=nRT/V= 0.1mole x 0.08205 (L.atm/ mole.K x 299 k)/ 1.23 L
= 1.994 atm
Explanation:
Answer:
light with a high enough intensity
Explanation:
