Answer : The value of equilibrium constant (Kc) is, 0.0154
Explanation :
The given chemical reaction is:

Initial conc.
0 0
At eqm.
x x
As we are given:
Concentration of
at equilibrium = 
That means,

The expression for equilibrium constant is:
![K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BSO_2%5D%5BCl_2%5D%7D%7B%5BSO_2Cl_2%5D%7D)
Now put all the given values in this expression, we get:



Thus, the value of equilibrium constant (Kc) is, 0.0154
Answer:
The answer of this question is molecule
Answer
0.9516 grams / mL (50.00 has 4 sig digs.)
Remark
You have a couple of extraneous numbers there. You don't care about anything except the mass of the flask + water/alcohol mixture (88.219 grams). and the mass of the flask (40.638 grams)
Formulas
- mass water/alcohol mixture = mass of the flask with fluid - mass flask
- density = mass / volume
Solution
mass water/alcohol mixture = 88.219 - 40.638 = 47.581
- Volume = 50 mL
- Density = mass / Volume
- Density = 47.581/50
- Density = 0.95162 There are 4 sig digs so the answer should be
- 0.9516
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