give me right ans!!! 2!!!!!no bad ans!!!fuc % ;) Fundamental constants e 3.0x10*
ID: 1076624 • Letter: G
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give me right ans!!!
2!!!!!no bad ans!!!fuc
% ;) Fundamental constants e 3.0x10* m/s, e 1.6101 C N 6.02x10 k = 1.38x 1 O-23 J/K,h=6.626x10%.. mor R-831AS K M. A sample o 1 2 3 4 5 6 781O 11 12 3 64°C.. The Find q, wA he enthalpy of cene is 28.8 kJ/mol and its melting point is Calculate its ideal solubility (g/kgjin benzene als00K MWof anthraceneens mo nsider the equilibrium N2O4(g) 2NO2(g), which can be easily studied in the laboratory through a measurement of the vapor density of theequlibrium mixture The degree of dissociation (a) of dinitrogen tetroxide is a function of the pressure )Show that if the mixture remains in equilibrium as the pressure is changed, heExplanation / Answer
You have to use the next equation
ln xb = Fusion enthalpy / R * (1 / Tf - 1/T) , T is temperature xb is the mole fraction
ln xb = 28 800 / 8.314 * (1 / 490 - 1 / 300) = 3464.0365 * (-0.00129) = -4.466
xb = exp (-4.477) = 0.011
mole fraction is
xb = moles of anthracene / (moles of anthracene + moles of benzene)
lets calculate the moles in 1 kg of benzene , 1000 grams / 78 = 12.82 moles of benzene, molar mass of benzene is 78 g/gmol.
now lets calculate moles of anthracene
moles of anthracene / (moles of anthracene + moles of benzene) = 0.011
let na be the moles of anthracene
na / (na + 12.82) = 0.011
na = 0.011 * (na + 12.82)
na = 0.011na + 0.1457
0.989 na = 0.1457 moles
na = 0.1473 moles of anthracene
grams = moles * molar mass
multiply this by molar mass of anthracene which is given , 178 g/gmol
grams of anthracene = 0.1473 * 178 = 26.221 grams of anthracene in 1 kg of benzene
solubility of anthracene in benzene is 26.23 grams / Kg
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