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Over [174]
3 years ago
14

Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °

C, the vapor pressure of CCl4 is 54.0 kPa, and its enthalpy of vaporization is 33.05 kJ/mol. Use this information to estimate the normal boiling point for CCl4.
Chemistry
1 answer:
Mila [183]3 years ago
3 0
This problem is to use the Claussius-Clapeyron Equation, which is:

ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]

Where p2 and p1 and vapor pressure at estates 2 and 1

ΔH is the enthalpy of vaporization

R is the universal constant of gases = 8.314 J / mol*K

T2 and T1 are the temperatures at the estates 2 and 1.

The  normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa

Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol 

=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]

=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x =  0.000157 + 1/330.95 = 0.003179

=> x = 314.6 K => 314.6 - 273.15 = 41.5°C

Answer: 41.5 °C 
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How many moles are in 20 grams of O₂ gas?
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Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo
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Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

CH_4+2O_2\rightarrow CO_2+2H_2O

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

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Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

left\ over=0g

Regards.

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