Since Janice was given a mixture of alcohol and water, her teacher suggested that she use temperature to separate the two substances. The property demonstrated by the experiment is D. boiling. The boiling point refers to the temperature at which the liquid phase of the substance will turn into vapor. Water is known to boil at 100°C at atmospheric pressure while alcohols are generally known to have a boiling point lower than that of water. In this experiment, knowing that the two substances had a significant difference in boiling temperature was crucial to be able to separate them into their pure substances.
Answer:
2CH2Cl2(g) Doublearrow CH4(g) + CCl4(g)
0.205 moles of CH2Cl2 is introduced. Let by the time an equilibrium is reached x moles each of CH4 and CCl4 are formed => remaining moles of CH2Cl2 are 0.205-x
i.e at equilibrium the concentration on CH2Cl2 is (0.205-2x) mol/L, CH4 is x mol/L, CCl4 is x mol/L
Now the equilibrium constant equation : K = [CH4][CCl4]/[CH2Cl2]^2 ([.] - stands for concentration of the term inside the bracket)
10.5 = x*x/(0.205-2x)^2
=> 10.5(4x^2-0.82x+0.042) = x^2
=>42x^2-8.61x+0.441=x^2
=>41x^2-8.61x+0.441 = 0
This is a Quadratic in x, solving for the roots, we get x = 0.0886 , x = 0.121
The second solution for x will lead 0.205-2x to become negative, so is an infeasible solution.
Therefore equilibrium concentrations of the products and reactants correspond to x=0.0886 and they are , [CH2Cl2] = 0.205-2*0.0886 =0.0278 mol/L , [CH4] = 0.0886 mol/L , [CCl4] = 0.0886 mol/L
Answer:
According to coulomb's law their potential energy will increase and the particles will repel each other.
Explanation:
As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.
So, using ideal gas equation we will calculate the number of moles as follows.
PV = nRT
or, n = 
= 
= 0.0052 mol
Also, No. of moles = 
0.0052 mol = 
mass = 0.0104 g
As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at
= 24 mm Hg
=
atm
= 0.03158 atm
Now, P = 
= 0.963 atm
Hence, n =
= 0.0056 mol
So, mass of
= 0.0056 mol × 2
= 0.01013 g (actual yield)
Therefore, calculate the percentage yield as follows.
Percent yield = 
=
= 97.49%
Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.