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1 answer:
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Explanation:
It is known that for high concentration of , reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.
Now, = 0 and the general reaction equation is as follows.
3.00 M n = 2 30 mM
E =
=
= 0.038 V
Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.
Answer:
0.444 mol/L
Explanation:
First step is to find the number of moles of oxalic acid.
n(oxalic acid) =
Now use the molar ratio to find how many moles of NaOH would be required to neutralize of oxalic acid.
n(oxalic acid): n(potassium hydroxide)
1 : 2 (we get this from the balanced equation)
: x
x = 0.0111 mol
Now to calculate what concentration of KOH that would be in 25 mL of water: