Answer:
pH = 8.314
Explanation:
equil: S S 3S
∴ Ksp = [ Y+ ] * [ OH- ]³ = 6.0 E-24
⇒ 6.0 E-24 = ( S )*( 3S )³
⇒ 6.0 E-24 = 27S∧4
⇒ 2.22 E-25 = S∧4
⇒ ( 2.22 E-25 )∧(1/4) = S
⇒ S = 6.866 E-7 M
⇒ [ OH- ] = 3*S =2.06 E-6 M
⇒ pOH = - Log [ OH- ]
⇒ pOH = - Log ( 2.06 E-6 )
⇒ pOH = 5.686
∴ pH = 14 - pOH
⇒ pH = 8.314
A: Trial 1, because the average rate of the reaction is lower.
The rate of reaction is the speed with which reactants are converted into products. It is also the rate at which reactants disappear and products appear. The higher the rate of reaction, the greater the amount of product formed in a reaction.
If we look at the graph, we will realize that trial 1 produces a lesser amount of product than trial 2. This implies that the average rate of the reaction in trial 1 is lower than in trial 2.
Lower average rate of reaction implies lower concentration of the reactants since the rate of reaction depends on the concentration of reactants.
Hence trial 1 has a lower concentration of reactants because the average rate of the reaction is lower.
Answer:
a=28600J; b=90.6 J/K; c=402 torr
Explanation:
(a) considering the data given
Vapour pressure P1 =0 at Temperature T1 = 42.43˚C,
Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)
Using the Clausius-Clapeyron Equation
ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
In 760/140 = ΔH/8.314 J/mol/K × (1/315.58K -- 1/273.15K)
ΔH vap= +28.6 kJ/mol or 28600J
(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.
Since ΔG at boiling point is zero,
ΔS =(ΔH°vap/Τb)
ΔS = 28600 J/315.58 K
= 90.6 J/K
(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
ln P298 K/1 atm = 28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)
P298 K = 0.529 atm
= 402 torr
Answer:
0.208mole of CO2
Explanation:
First, let us calculate the number of mole of HC3H3O2 present.
Molarity of HC3H3O2 = 0.833 mol/L
Volume = 25 mL = 25/100 = 0.25L
Mole =?
Mole = Molarity x Volume
Mole = 0.833 x 0.25
Mole of HC3H3O2 = 0.208mole
Now, we can easily find the number of mole of CO2 produce by doing the following:
NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2
From the equation,
1mole of HC2H3O2 produced 1 mole of CO2.
Therefore, 0.208mole of HC2H3O2 will also produce 0.208mole of CO2
Answer:
Option C. 1.60x10^26 molecules
Explanation:
Avogadro's hypothesis proved that that 1 mole of any substance contains 6.02x10^23 molecules.
From the above, we understood that 1 mole of CCl4 contains 6.02x10^23 molecules.
If 1 mole of CCl4 contains 6.02x10^23 molecules,
then, 265 moles of CCl4 will contain = 265 x 6.02x10^23 = 1.60x10^26 molecules
From the calculation made above, 265 moles of CCl4 contains 1.60x10^26 molecules.
