Answer:
13.5g of AgNO3 will be needed
Explanation:
Silver nitrate, AgNO3 contains 1 mole of silver, Ag, per mole of nitrate. To solve this problem we need to convert the mass of Ag to moles. Thee moles = Moles of AgNO3 we need. With the molar mass of AgNO3 we can find the needed mass:
<em>Moles Ag-Molar mass: 107.8682g/mol-</em>
8.6g * (1mol / 107.8682g) = 0.0797 moles Ag = Moles AgNO3
<em>Mass AgNO3 -Molar mass: 169.87g/mol-</em>
0.0797 moles Ag * (169.87g/mol) =
<h3>13.5g of AgNO3 will be needed</h3>
<span>Oxidation is the loss of electrons and corresponds to an increase in oxidation state. A reduction is the gain of electrons and corresponds to a decrease in oxidation state. Balancing redox reactions can be more complicated than balancing other types of reactions because both the mass and charge must be balanced. Redox reactions occurring in aqueous solutions can be balanced by using a special procedure called the half-reaction method of balancing. In this procedure, the overall equation is broken down into two half-reactions: one for oxidation and the other for reduction. The half-reactions are balanced individually and then added together so that the number of electrons generated in the oxidation half-reaction is the same as the number of electrons consumed in the reduction half-reaction.</span>
So let's convert this amount of mL to grams:

Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):

Then we need to convert moles to atoms using Avogadro's number:
![\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms](https://tex.z-dn.net/?f=%20%5Cfrac%7B6.022%2A10%5E%7B23%7Datoms%7D%7B1mole%7D%20%2A%5B8.135%2A10%5E%7B-2%7Dmol%5D%3D4.90%2A10%5E%7B22%7Datoms%20)
So now we know that in 1.2 mL of liquid mercury, there are
present.