Answer:
pH = 4.09
Explanation:
molarity of oxalic acid in the solution
= 0.1 x 25 / (25 + 35)
= 0.0417 M
molarity of NaOH in the solution
= 0.1 x 35 / (25 +35)
= 0.0583 M
H2C2O4 + NaOH -------------------> NaHC2O4 + H2O
0.0417 0.0583 0 0
0 0.0166 0.0417
now second acid -base titration
NaHC2O4 + NaOH -------------------> Na2C2O4 + H2O
0.0417 0.0166 0 0
0.0251 0 0.0166 ---
now
pH = pKa2 + log [Na2C2O4 / NaHC2O4]
pH = 4.27 + log (0.0166 / 0.0251)
pH = 4.09
The balanced chemical equation is given as:
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.
Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3
% yield = .775 = actual yield / 264.66
actual yield = 205.11 g CH3CH2OCH2CH3
This is a combination reaction. Look at the 2 elements on left and a compound on the right.
Answer:
Ag 0 is the reducing agent.
Explanation:
Reducing -> gaining electrons
Oxidizing -> losing electrons
Ag lost electrons (became more positive) since it went from a 0 charge to a +1 charge. Therefore it was oxidized. Ag+ is the oxidized product. Reactants that create an oxidized product are called reducing agents. This would make Ag 0 the reducing agent in this reaction.