Answer:
Molar mass of vitamin K = 450.56\frac{g}{mol}[/tex]
Explanation:
The freezing point of camphor = 178.4 ⁰C
the Kf of camphor = 37.7°C/m
where : m = molality
the relation between freezing point depression and molality is
Depression in freezing point = Kf X molality
Where
Kf = cryoscopic constant of camphor
molality = moles of solute dissolved per kg of solvent.
putting values
2.69°C = 37.7°C/m X molality
molality = 0.0714 mol /kg

moles of vitamin K = 0.0714X0.025 = 0.00178 mol
we know that moles are related to mass and molar mass of a substance as:

For vitamin K the mass is given = 0.802 grams
therefore molar mass = 
Volume = nRT/P
n = number of particles (moles)
R = universal gas constant (0.0821)
T = temperature (Kelvin)
P = pressure (atm)
(Assuming you have 1 mole of Helium in a chemical reaction) We would need to convert grams to moles: 12.0g He x 1 mol He/4 molar mass of He = 3 mol He
Convert Celsius to Kelvin: 100*C + 273.15 = 373.15 K
Now we can set up the equation for volume: (3mol)(0.0821)(373.15)/1.2atm = 76.6 L of Helium gas
When ice melts, the physicals state changes from solid to liquid. The energy or the heat required (q) required to change a unit mass (m) of a substance from solid to liquid is known as the enthalpy or heat of fusion (ΔHf). The variables; q, m and ΔHf are related as:
q = m * ΔHf
the mass of ice m = 65 g
the heat of fusion of water at 0C = ΔHf = 334 J/g
Therefore: q = 65 g * 334 J/g = 21710 J
Now:
4.184 J = 1 cal
which implies that: 21710 J = 1 cal * 21710 J/4.184 J = 5188.8 cal
Hence the heat required is 5188.8 cal or 5.2 Kcal (approx)
Answer:
The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.
Explanation:
According to the Arrhenius equation,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate of reaction at 
= rate of reaction at 
= activation energy of the reaction
R = gas constant = 8.314 J/K mol


![\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7B300%2C000%20J%2Fmol%7D%7B2.303%5Ctimes%208.314%20J%2FK%20mol%7D%5B%5Cfrac%7B1%7D%7B798.15%20K%7D-%5Cfrac%7B1%7D%7B898.15%20K%7D%5D)


The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.