You dissolve 15.71 g of NH4NO3 in 150.0 mL of solution (call this solution A). You take 20.0 mL of solution A and add water until the total volume is 75.0 mL (call this solution B). You take 15.0 mL of solution B and add 25.0 mL of water to it (call this solution C). You mix 10.0 mL of solution B and 10.0 mL of solution C (call this solution D). What is the concentration of ammonium nitrate in solution D

[NH₄NO₃] at D → 0.279 M

Explanation:

This exercise involves a series of dilutions one after the other.

First of all, we calcualte ammonium nitrate's concentration at A.

15.71 g . 1 mol/ 80 g = 0.196 mol / 0.150 mL = 1.31 M

At B → 1.31 M . 20 mL/ 75 mL = 0.349  M

At C →  0.349 M . 15 mL / 25 mL = 0.209 M

[NH₄NO₃] at B = 0.349 M

[NH₄NO₃] at C = 0.209 M

So let's calculate the new moles

In 1 mL of B we have 0.349 mmoles

In 10 mL of B we have 3.49 mmoles

In 1 mL of C we have 0.209 mmoles

In 10 mL of C we have 2.09 mmoles

Volume of D = 10 ml + 10ml = 20 mL

Total mmmoles = 3.49 mmoles + 2.09 mmoles = 5.58 mmoles

[NH₄NO₃] at D = 5.58 mmoles / 20mL → 0.279 M

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

im not suree sorry if wrong