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:
density=1.43 g/L
Explanation:
Since the density formula is density = mass / volume, we need to find out the mass of the gas and the volume is that of the container.
The mass of the gas is 130.0318 g-129.6375 g=0.3943 g
The gas volume is 276mL*(1L/1000mL) 0.276 L
density = mass / volume=0.3943g/0.276L
density =1.43g/L
There would be 79 electrons present in each atom of gold. I hope this helps :)
Answer:
M=0.380 M.
Explanation:
Hello there!
In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:
Now, we compute the total moles of bromide:
Then, the total volume in liters:
Therefore, the concentration of total bromide is:
Best regards!
