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
In order to find molarity, you must first find the number of moles that was dissolved.
Now, Moles = Mass ÷ Molar Mass
⇒ Moles of NaCl = 2.922 g ÷ 58.44 g/mol
= 0.05 moles
∴ the Molarity of the NaCl is 0.05 M [Option 1]
Answer:the amount of matter the object contains
Explanation:
C + H2O -> H2 + CO
n(C) = 15.9/12 = 1.325 (mol)
=> n(H2) = 1.325 mol
We have:
PV = nRT
=> V = (nRT)/P
(R = 22.4/273 = 0.082)
V = (1.325 x 0.082 x 360)/1 = 39.114 (L)