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Answer:
a) volume of ammonium iodide required =349 mL
b) the moles of lead iodide formed = 0.0436 mol
Explanation:
The reaction is:

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.
Let us calculate the moles of lead nitrate taken in the solution.
Moles=molarityX volume (L)
Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol
the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol
The volume of ammonium iodide required will be:

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol
Answer:
rate= k[A]²[B]²[C]
Explanation:
When concentration of A is increased two times ,keeping other's concentration constant , rate of reaction becomes 4 times .
So rate is proportional to [A]²
When concentration of B is increased two times , keeping other's concentration constant,rate of reaction becomes 4 times.
So rate is proportional to [B]²
When concentration of C is increased two times , keeping other's concentration constant, rate of reaction becomes 2 times.
So rate is proportional to [C]
So rate= k[A]²[B]²[C]
F (Fluorine) is in column (group/family) VIIA, or the "halogens". When you see the halogens (Fluorine, Chlorine, Bromine, and Iodine) in combination with a metal, each halogen atom present will carry a -1 charge. We can see that the atom has no charge, so the metal must cancel out the negative charges brought by the two fluorine atoms.
(Charge on m) + 2*(charge on fluorine) = 0
(Charge on m) + 2*(-1) = 0
(Charge on m) - 2 = 0
Charge on m ion = +2
<u>Answer:</u> The standard potential of the cell is 0.77 V
<u>Explanation:</u>
We know that:

The substance having highest positive
reduction potential will always get reduced and will undergo reduction reaction.
The half reaction follows:
<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u>
( × 2)
To calculate the
of the reaction, we use the equation:

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation follows:

Hence, the standard potential of the cell is 0.77 V