The given question is incomplete. The complete question is:
Suppose a current of 0.920 A is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for 47.0 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell.
Answer: 0.0484 g
Explanation:
where Q= quantity of electricity in coloumbs
I = current in amperes = 0.920 A
t= time in seconds = 47.0 sec

96500 Coloumb of electricity electrolyzes 1 mole of Ag
43.24 C of electricity deposits =
of Ag
Thus the mass of pure silver deposited on a metal object made into the cathode of the cell is 0.0484 g
Answer:
1 M
Explanation:
The molarity of a solution, M, is a measure of the concentration of that solution and it refers to the number of moles of solute (mol) per liter (L) of solution. The molarity (M) can be calculated using the formula:
M = number of moles (n) /volume (V)
In this question, a 500 ml aqueous solution of Na3PO4 was prepared using 82g of the solute.
Molar mass of Na3PO4 = 23(3) + 15 + 16(4)
= 69 + 31 + 64
= 164g/mol
Mole = mass/molar mass
mole = 82/164
mole = 0.5 mol
Volume in Litres (L) = 500 ml ÷ 1000 = 0.500L
Therefore, Molarity (M) = 0.5/0.500
Molarity = 1 M or 1 mol/L
They all have the same number of electrons.
Answer: 1.
2. 3 moles of
: 2 moles of 
3. 0.33 moles of
: 0.92 moles of 
4.
is the limiting reagent and
is the excess reagent.
5. Theoretical yield of
is 29.3 g
Explanation:
To calculate the moles :

The balanced chemical equation is:
According to stoichiometry :
3 moles of
require = 2 moles of
Thus 0.33 moles of
will require=
of
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 3 moles of
give = 2 moles of
Thus 0.33 moles of
give =
of
Theoretical yield of
Thus 29.3 g of aluminium chloride is formed.
Carbon carbon triple bonds