You are given two beakers, distilled water, two hot plates, two thermometers and salt. These materials are enough in order to test the effect of salt in the boiling point water. To do this, you set up two beakers. In one of the beakers, you add pure distilled water and nothing else. For the other beaker, you put a solution of salt and water. You place these beakers on separate hot plates and place inside the beakers the thermometers. You heat these substances until they boil and then you measure the boiling points of the substances. You would observe that the boiling point of the solution would have a higher boiling point than the pure liquid.
Answer:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
Explanation:
For the reaction:
H₂C₂O₄(g) → CO₂(g) + HCOOH(g)
At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:
H₂C₂O₄(g) = P₀ - x
CO₂(g) = x
HCOOH(g) = x
P at t=20000 is:
P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x
For 1st point:
x = 92,8-65,8 = 27
Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8
2nd point:
x = 130-92,1 = 37,9
H₂C₂O₄(g): 92,1 - 37,9 = 54,2
3rd point:
x = 157-111 = 46
H₂C₂O₄(g): 111-46 = 65
Now, as the rate law is :
v = k P[H₂C₂O₄]
Based on integrated rate law, k is:
(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k
1st point:
k = 2,64x10⁻⁵
2nd point:
k = 2,65x10⁻⁵
3rd point:
k = 2,68x10⁻⁵
The averrage of this values is:
k = 2,66x10⁻⁵
That means law is:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
I hope it helps!
<span>Colligative properties are dependent upon the number of molecules or ions present in solution. Therefore, 1 mole of Na2SO4 will produce 3 moles of ions and so it will have 3 times as much of an effect as 1 mole of sugar, which is not an electrolyte and can't dissociate to an appreciable extent.</span>
Answer: Thus concentration of 
in 
is 0.011 and in 
is 0.814
Explanation:
To calculate the concentration of , we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is
We are given:
Putting values in above equation, we get:
The concentration in is 
Thus concentration of is 
and 
