Answer:
The answer to your question is P2 = 0.78 atm
Explanation:
Data
Temperature 1 = T1 = 263°K Temperature 2 = T2 = 298°K
Volume 1 = V1 = 24 L Volume 2 = V2 = 35 L
Pressure 1 = P1 = 1 Pressure 2 = P2 = ?
Process
1.- To solve this problem use the Combined gas law
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
-Substitution
P2 = (1)(24)(298) / (263)(35)
-Simplification
P2 = 7152 / 9205
-Result
P2 = 0.777
or P2 = 0.78 atm
The answer is: II.The endpoint is recorded when the solution is dark red in color rather than light pink.
The endpoint is the point at which the indicator changes colour in a colourimetric titration and that is point when titration must stop.
Phenolphthalein is colorless in acidic solutions and pink in basic solutions. If this indicator change color to dark red, more base is added and endpoint is not accurate.
If the the acid is spilled before titration, that does not make endpoint wrong and molar mass can be calculated.
In this example we can take acetic acid as carboxylic acid; basic salt sodium acetate CH₃COONa is formed from the reaction between weak acid (in this example acetic acid CH₃COOH) and strong base (in this example sodium acetate NaOH).
Balanced chemical reaction of acetic acid and sodium hydroxide:
CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l).
Neutralization is is reaction in which an acid (in this example vinegar or acetic acid CH₃COOH) and a base react quantitatively with each other.
Answer:
Explanation:
moles of acetic acid = 500 x 10⁻³ x .1 M
= 5 X 10⁻³ M
.005 M
Moles of NaOH = .1 M
Moles of sodium acetate formed = .005 M
Moles of NaOH left = .095 M
pOH = 4.8 + log .005 / .095
= 4.8 -1.27875
= 3.52125
pH = 14 - 3.52125
= 10.48
Answer:
ΔU = −55.45 kJ
Explanation:
From first law of thermodynamics in chemistry, we have;
ΔU = Q + W
where;
ΔU is change in internal energy
Q is the net heat transfer
W is the net work done
We are given;
Q = 74.6 kJ
But Q will be negative since heat is released
Thus;
ΔU = -74.6 kJ + W
We are given;
Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²
Volume before reaction; Vi = 8.2 L = 0.0082 m³
Volume after reaction; V_f = 2.8 L = 0.0028 m³
Now,
W = -P(V_f - V_i)
W = - 3546375(0.0028 - 0.0082)
W = 19.15 KJ
Thus;
ΔU = Q + W
ΔU = -74.6 kJ + 19.15 KJ =
ΔU = −55.45 kJ
