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:
the four main spheres of the earth are geosphere, hydrosphere, atmosphere and biosphere
Explanation:
geosphere consists of all rocks on Earth
atmosphere which are the gases that surrounds the earth
hydrosphere which is all the water on the earth
biosphere which are the living things on the earth
I disagree with the answer pick of D. If you have a neutral pH 7 solution and you proceed to add a base even with an relatively insignificantly low Kb your solution would still be more basic then acidic. The answer should be b which is true that the base only ionizes slightly in aqueous solution. This is also truer to the definition of what Kb represents.
Answer:
add x to 7 and divide by 3
Explanation:
easier formula
Answer:
T2 = 51.6°C
Explanation:
Given:
P1 = 1.01 atm
T1 = 25°C + 273 = 298K
P2 = 1.10 atm
T2 = ?
P1/T1 = P2/T2
Solving for T2,
T2 = (P2/P1)T1
= (1.10 atm/1.01 atm)(298K)
= 324.6 K
= 51.6°C
where Tc = Tk - 273