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QUESTION 4 8 points Save Answer Calculations for the amount of ammonium nitrate used during Lab 1 must be completed before th
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Answer #1

Formula to get solution :- moles = grams / molar mass

1. Mass of NH4NO3 = 0.0500 mole × 80.043 grams /1mole

= 4.0022 grams

2 Mass of NH4NO3 = 0.0375 mole × 80.043gram / 1mole

= 3.002 grams

3. Mass of NH4NO3 = 0.0250 mole × 80.043 gram / 1mole

= 2.0011 grams

4. Each sample tested twice =4.0022× 2+3.002 ×2 + 2.0011× 2

= 8.0044 + 6.004 + 4.0022

total amount of salt = 18.1 grams

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