Answer:
0.35 atm
Explanation:
It seems the question is incomplete. But an internet search shows me these values for the question:
" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."
Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>
First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:
- 137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene
- 111 g heptane ÷ 100 g/mol = 1.11 moles heptane
Total number of moles = 1.63 + 1.11 = 2.74 moles
The<u> mole fraction of thiophene</u> is:
Finally, the <u>partial pressure of thiophene vapor is</u>:
Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene
- Partial Pressure = 0.59 * 0.60 atm
Answer:
-1
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
<u>Δn = (2)-(2+1) = -1 </u>
Thus, Kp is:
<span>density = mass / volume
given the quotient , we have density and mass, volume can be easily calculated as:
volume = mass / p =15.5 g / 0.789 g/cm^3
=~ 20 cm^3 (dimension ally constant)</span>
Answer:
Boiling point for the solution is 100.237°C
Explanation:
We must apply colligative property of boiling point elevation
T° boiling solution - T° boiling pure solvent = Kb . m
m = molalilty (a given data)
Kb = Ebulloscopic constant (a given data)
We know that water boils at 100°C so let's replace the information in the formula.
T° boiling solution - 100°C = 0.512 °C/m . 0.464 m
T° boiliing solution = 0.512 °C/m . 0.464 m + 100°C → 100.237 °C
Either he’s busy, or he is just trying to ignore you.
Need more info tho
