Question

  The standardization of KMnO₄ solution, if the magnetic stir bar was dropped into the flask, and some FAS solution splashed out, how would the calculation for the molarity of KMNO₄ be affected?    The calculated molarity of KMnO₄ would be:

		Lower.   Less FAS would need less MnO4-, therefore, the reduced volume would result in a smaller M.
		Unchanged.
		Lower.   More FAS would need less MnO4-, therefore, the increased volume would result in a smaller M.
		Higher.  Less FAS would need less MnO4-, therefore, the reduced volume would result in a larger M.
		Higher.  More FAS would need less MnO4-, therefore, the increased volume would result in a smaller M.

The following redox reaction, determine what is oxidized and is also the reducing agent:

3 Cu (s) + 8 HNO₃ (aq) → 3 Cu²⁺ (aq) + 2 NO (g) + 6 NO₃^- (aq) + 4 H₂O (l)

		H2O
		NO
		Cu
		HNO3
		None of the above

2)

he following redox reaction, determine what is reduced and is also the oxidizing agent:

3 SO₂ (aq) + Cr₂O₇^2- (aq) + 2 H⁺ (aq) → 3 SO₄^2- (aq) + 2 Cr³⁺ (aq) + H₂O (l)

		H2O
		SO42-
		SO2
		Cr2O72-
		None of the above

2 MnO₄^- (aq) + 5 H₂O₂ (aq) + 6 H⁺(aq) → 2 Mn²⁺(aq) + 5 O₂(g) + 8 H₂O(l)

		3.4%
		24%
		100%
		7.9%
		12%

Please help, this is a new investigation and I am having trouble understanding the concepts. Please show work :)

Answer 1

1)

Higher.  Less FAS would need less MnO4-, therefore, the reduced volume would result in a larger M.

M1V1=M2V2

volume reduced. So, molaity will be high to make MV constant

2) Cu . It is oxidised to Cu+2 . The compound which is oxidised acts as a reducing agent

3) Cr2O7-2 . Cr changed its oxidaion state from +6 to +3. The compound which is reduced acts as a oxidising agent

4) only equation given. No question