3. What volume (mL) of 0.663M potassium hydroxide aqueous solution is needed to

ID: 792859 • Letter: 3

Question

3. What volume (mL) of 0.663M potassium hydroxide aqueous solution is needed to neutralize [?this means to

completely react with] 842 mL of 0.554M nitric acid?

4. 6. For each of the following, assign oxidation numbers to each element in each species, and label with arrows

the oxidation and reduction; write the oxidizing agent and the reducing agent. Do not use hand-in this

question paper; re-write these equations on your own paper. Follow the example:

NO2 --> NO3^- + NO

N= +4 O= -2 N= +5 O=-2 N= +2 O= -2

Oxidizing agent = NO2 and Reducing agent = NO2

1. Mn^2+ + BiO3^ -> MnO4^- + Bi^3+

2. ClO3^- + Cl^- -> Cl2 + ClO2

3. PH3 + I2 -> H3PO2^- + I^-

4. ClO2 -> ClO2^- + ClO3^-

5. Zn + NO3^- -> Zn(OH)4^2- + NH3

6. Al + OH^- -> AlO2^- + H2

3) 842 mL x 0.554M = 0.663M x V

V= 703.57ml

4)
1)Mn2+ to Mn7+ means ita getting oxidise

Bi5+ to Bi3+ gettind reduced

2) Cl1- to Cl0 means getting oxidised

Cl5+ to Cl4+ means getting reduced as well

3) P3- to P0 getiing oxidise

I0 to I1- getting reduced

4) Cl+4 to Cl3+ and Cl5+ oxidation and redcution as well

5) Zn0 to Zn2+ means getting oxidised

N5+ to N3- getting receduced

6) Al0 to Al3+ getting oxidiesd

H1+ to H0 means getting reduced

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