You're oxidising agent will be reduced. In the molecule MnO4-, Mn has and oxidation number of +9 and in MnO2 it is +4, therefore its oxidation number has reduced, therefore Mn is reduced, therefore it is the oxidant/oxidising agent.
Answer:
The answer to your question is b. CaCrO₄
Explanation:
Data
H₂CrO₄ reacts with Ca(OH)₂
Process
H₂CrO₄ is an acid and Ca(OH)₂ is a base then this is a neutralization reaction.
H₂CrO₄ is an acid and Ca(OH)₂ is a base.
In a neutralization reaction, the products will always be water and a binary or ternary salt.
In the reaction, the salt will be Calcium chromate
Balanced chemical reaction
H₂CrO₄ + Ca(OH)₂ ⇒ CaCrO₄ + 2H₂O
Moles Cu+2 = M * V
= 0.05 L * 0.011 m
= 0.00055 moles
when the molar ratio of Cu2+: EDTA = 1:1 so moles od EDTa also =0.00055 moles
and when the Molarity of EDTa = 0.0630 M
∴ Volume of EDTA = moles / Molarity
= 0.00055 / 0.0630
= 0.0087 L = 8.7 L