Answer:
Explanation:
Hello there!
In this case, since the modelling of titration problems can be approached via the Henderson-Hasselbach equation to set up a relationship between pH, pKa and the concentration of the acid and its conjugate base, we can write:
![pH=pKa+log(\frac{[NO_2^-]}{[HNO_2]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BNO_2%5E-%5D%7D%7B%5BHNO_2%5D%7D%20%29)
Whereas the pH is given as 3.14 and the concentrations are the same, that is why the pH would be equal to the pKa as the logarithm gets 0 (log(1)=0); thus, we can calculate the Ka via:
Best regards!
True, Energy cannot be created or destroyed therefore it remains constant
Answer:
709 g
Step-by-step explanation:
a) Balanced equation
Normally, we would need a balanced chemical equation.
However, we can get by with a partial equation, as log as carbon atoms are balanced.
We know we will need an equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 30.07 236.74
C₂H₆ + … ⟶ C₂Cl₆ + …
m/g: 90.0
(i) Calculate the moles of C₂H₆
n = 90.0 g C₂H₆ × (1 mol C₂H₆ /30.07 g C₂H₆)
= 2.993 mol C₂H₆
(ii) Calculate the moles of C₂Cl₆
The molar ratio is (1 mol C₂Cl₆/1 mol C₂H₆)
n = 2.993 mol C₂H₆ × (1 mol C₂Cl₆/1 mol C₂H₆)
= 2.993 mol C₂Cl₆
(iii) Calculate the mass of C₂Cl₆
m = 2.993 mol C₂Cl₆ × (236.74 g C₂Cl₆/1 mol C₂Cl₆)
m = 709 g C₂Cl₆
The reaction produces 709 g C₂Cl₆.
Solution
Given mass of FeBr2 = 85.5 g
Molar mass of FeBr2 = 215.65
No. of Moles of FeBr2 = Given Mass / Molar mass = 85.5 / 215.65
No. of Moles = 0.3964
Given Volume is 450 mL i.e. 0.450 L
Now Concentration = 0.3964/0.450
Hence Concentration = 0.881 mol/L
