Answer:
pH = 12.33
Explanation:
Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.
The titration reaction is
HA + KOH ---------------------------- A⁻ + H₂O + K⁺
number of moles of HA : 118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA
number of moles of OH : 115.4 mL/1000ml/L x 0.400 mol/L = 0.046 mol A⁻
therefore the weak acid will be completely consumed and what we have is the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.
n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH
pOH = - log (KOH)
M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M
pOH = - log (0.0021) = 1.66
pH = 14 - 1.96 = 12.33
Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.
Answer:
1.64 moles O₂
Explanation:
Part A:
Remember 1 mole of particles = 6.02 x 10²³ particles
So, the question becomes, how many '6.02 x 10²³'s are there in 9.88 x 10²³ molecules of O₂?
This implies a division of given number of particles by 6.02 x 10²³ particles/mole.
∴moles O₂ = 9.88 x 10²³ molecules O₂ / 6.02 x 10²³ molecules O₂ · mole⁻¹ = 1.64 mole O₂
_______________
Part B needs an equation (usually a combustion of a hydrocarbon).
I'm pretty sure it's yards
Answer:
7.5 moles of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Finally, we shall determine the number of mole of O₂ produced by the decomposition of 5 moles of KClO₃. This can be obtained as follow:
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Therefore, 5 moles of KClO₃ will decompose to produce = (5 × 3)/ 2 = 7.5 moles of O₂.
Thus, 7.5 moles of O₂ were obtained from the reaction.