Answer:
A) pH of Buffer solution = 4.59
B) pH after 5.0 ml of 2.0 M NaOH have been added to 400 ml of the original buffer solution = 4.65
Explanation:
This is the Henderson-Hasselbalch Equation:
![pH = pKa + log\frac{[conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
to calculate the pH of the following Buffer solutions.
i just learnt this recently but i need to look through my notes again.
Explanation believe the answer is A. Average.
2KClO₃ → 2KCl + 3O₂
mole ratio of KClO₃ to O₂ is 2 : 3
∴ if moles of O₂ = 5 mol
then moles of KClO₃ =
= 3.33 mol
Mass of KClO₃ needed = mol of KClO₃ × molar mass of KClO₃
= 3.33 mol × ((39 × 1) + (35.5 × 1) + (16 × 3) g/mol
= 407.93 g
Answer:
Explanation:
Here we have to use stoichiometry.
First of all, we have to calculate the mass of 100% of yield:
1.7 g ------- 98%
X -------- 100%
X = 1.73 g (approximately)
Second, we have to calculate the mass of N2 that is necessary to react to produce the mass of 1.73g of NH3. To do that, we have to use the Molar mass of N2 and NH3 and don't forget the stoichiometric relationship between them.
Molar Mass N2 : 14x2 = 28 g/mol
Molar Mass NH3: 14 + 3 = 17 g/mol
28g (N2) ------- 17x2 (NH3)
X ------------ 1.73 g
X = 1.42 g (approximately)
