Answer:
Δ S = 93.8 J/mol-K
Explanation:
Given,
Boiling point of chloroform = 61.7 °C
= 273 + 61.7 = 334.7 K.
Enthalapy of vapourization = 31.4 kJ/mol.
Using Gibbs free energy equation
Δ G = Δ H - T (ΔS)
at equilibrium (when the liquid is boiling), Δ G = 0
so, 0 = ΔH - T (Δ S)
T (Δ S) = Δ H
and ΔS = ΔH / T
Δ S = (31400 J/mol.) / 334.7 K
Δ S = 93.8 J/mol-K
Answer:
the energy of the products is less than the energy of the reactants.
Explanation:
the the change is enthalpy is negative, and heat is released to the surroundings.
<span>conductor because it conducts the electrons</span><span />
Answer:
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Explanation:
Data Given
M1 = 6.00 M
M2 = 2.5 M
V1 = 250 mL
V2 = ?
Solution:
As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.
Now
first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution
For this Purpose we use the following formula
M1V1=M2V2
Put values from given data in the formula
6 x V1 = 2.5 x 250
Rearrange the equation
V1 = 2.5 x 250 /6
V1 = 104 mL
So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M aqueous solution of ammonium hydroxide
But we have to prepare 250 mL of the solution.
so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.
in this question you have to tell about the amount of water that is 146 mL
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
<h3><u>Answer</u>;</h3>
1.0875 x 10-2 atm
<h3><u>Explanation;</u></h3>
2O3(g) → 3O2(g)
rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t
The average rate of disappearance of ozone ... is found to
be 7.25 × 10–3 atm over a certain interval of time.
This means (ignoring time)
∆[O3]/∆t = -7.25 × 10^–3 atm
(it is disappearing, thus the negative sign)
rate = -(1/2)∆[O3]/∆t
rate = -(1/2)*(-7.25 × 10^–3 atm)
= 3.625 × 10^–3 atm
Now use the other part of the expression:
rate = +(1/3)∆[O2)∆t
3.625 × 10–3 atm = +(1/3)∆[O2)/t
∆[O2)/∆t = (3)*(3.625× 10^–3 atm)
= 1.0875 x 10-2 atm over the same time interval
