The work is in the picture, post-lab questions. show all work or i will not get any !

1. Cu(s) + 2AgNO₃(aq) ⟶ Cu(NO₃)₂(aq) + 2Ag(s)

2. 50.92 g

3. $8.09

Step-by-step explanation:

1. Balanced equation

Cu(s) + 2AgNO₃(aq) ⟶ Cu(NO₃)₂(aq) + 2Ag(s)

2. Mass of Ag

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:   63.55                                        107.87

            Cu + 2AgNO₃ ⟶ Cu(NO₃)₂ + 2Ag

m/g:    15.00

(a) Calculate the moles of Cu  

Moles Cu = 15.00 × 1/63.55  

Moles Cu = 0.2360 mol Cu  

(b) Calculate the moles of Ag  

The molar ratio is 2 mol Ag:1 mol H₂O  

Moles of Ag = 0.2360 × 2/1  

Moles of Ag = 0.4721 mol Ag  

(c) Calculate the mass of Ag  

Mass of Ag = 0.4721 × 107.87  

Mass of Ag = 50.92 g Ag

3. Value of Ag

Mass of Ag = 50.92 × 0.0353/1

Mass of Ag = 1.797 oz

Value of Ag = 1.797 ×4.50/1

Value of Ag = $8.09

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

i'm pretty sure its b i might be wrong though!