Answer: The freezing-point depression constant (Kf) of nitrobenzene is
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte nitrobenzene)
= freezing point constant = ?
m= molality
Weight of solvent = 20 g = 0.02 kg
mass of solute (ethanol) = 1.0 g
Molar mass of ethanol = 46 g/mol
Thus freezing-point depression constant (Kf) of nitrobenzene is
Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
</span>Look up the freezing point of benzene, then delta T = f.p.=3.5 = Kf*m
<span>Solve for molality. </span>
<span>molality = moles/kg solvent </span>
<span>Solve for moles. </span>
<span>moles acetic acid = grams acetic acid/molar mass acetic acid </span>
<span>Solve for molar mass. You would expect to find 60 for the molar mass CH3COOH.</span>
It means it will only allow some to go through but not all.
Firstly, the chemical equation between the calcium metal and water will be:
Ca(s) + 2 H₂O(l) → Ca(OH)₂(aq) + H₂(g)
We can see from the equation the bubbles of hydrogen gas which are formed during the reaction stick to the surface of the metal and hence calcium floats on water.
The other metal that will float on the water during the reaction is magnesium which have the same chemical behavior like calcium, we can illustrate that by the chemical equation:
Mg(s) + 2 H₂O(l) → Mg(OH)₂(aq) + H₂(g)
Answer:
C
Explanation:
Charles law states that volume of gas is directly proportional to temperature at constant pressure
V/T = k
where V - volume , T - temperature and k - constant
\frac{V1}{T1} = \frac{V2}{T2}
T1
V1
=
T2
V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
in the question it states that the temperature has been increased from 278 K to 231 K but it should actually be temperature is decreased from 278 K to 308 K
substituting the values in the equation
\frac{417cm^{3} }{278K} = \frac{V}{308 K}
278K
417cm
3
=
308K
V
V = 462 cm³
the answer should be C. 462 cm³