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:
rate= k[A]²[B]²[C]
Explanation:
When concentration of A is increased two times ,keeping other's concentration constant , rate of reaction becomes 4 times .
So rate is proportional to [A]²
When concentration of B is increased two times , keeping other's concentration constant,rate of reaction becomes 4 times.
So rate is proportional to [B]²
When concentration of C is increased two times , keeping other's concentration constant, rate of reaction becomes 2 times.
So rate is proportional to [C]
So rate= k[A]²[B]²[C]
Transpiration is the progression of <em>water </em>inside a plant! So, the molecule representing transpiration is going to be good ol' H2O! =)
