Answer:
Hg(NO3)2 + Na2CO3 --> 2NaNO3 + HgCO3
<h2>Answer:</h2>
5.65moles
<h2>Explanations:</h2>
The formula for calculating the number of moles the compound contain is given as:

Given the following parameters
Mass of Ag = 700grams
Determine the molar mass of AgO
Molar mass = 107.87 + 16
Molar mass = 123.87g/mol
Determine the moles of AgO

Hence the moles of AgO present is 5.65moles
It is B, and also for a moment I didn't understand that 4.69 x 10^22. I almost did this whole problem wrong.
Answer:
Products are AgBr and KNO3
Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18