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1) Reaction: 3Mg + N₂ → Mg₃N₂.
m(Mg) = 0,225 g
n(Mg) = 0,225 g ÷ 24,3 g/mol = 0,009 mol.
n(Mg) : n(N₂) = 3 : 1
n₁(N₂) = 0,003 mol.
n₂(N₂) = 0,5331 ÷ 28 = 0,019 mol.
n₃(N₂) = 0,019 mol - 0,003 mol = 0,016, m(N₂) = 0,016mol·28g/mol=0,4467g.
or simpler: m(N₂) = 0,225 g + 0,5331 - 0,3114 g = 0,4467 g.
2) Answer is: 6 <span>of fluorine atoms are combined with one uranium atom.
</span>m(U) = 209 g.
m(F) = 100 g.
n(U) = m(U) ÷ M(U)
n(U) = 209 g ÷ 238 g/mol.
n(U) = 0,878 mol.
n(F) = m(F) ÷ M(F)
n(F) = 5,263 mol
n(U) : n(F) = 0,878 mol : 5,263 mol /:0,878.
n(U) : n(F) = 1 : 6.
n - amount of substance
Answer:
1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).
2.6592 grams of oxygen remain in the flask.
Explanation:
Volume of the flask remains constant = V = 2.0 L
Initial pressure of the oxygen gas = 
Initial temperature of the oxygen gas = 
Final pressure of the oxygen gas = 
Final temperature of the oxygen gas = 
Using Gay Lussac's law:


1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).
Moles of oxygen gas = n
(ideal gas equation)

Mass of 0.08310 moles of oxygen gas:
0.08310 mol × 32 g/mol = 2.6592 g
2.6592 grams of oxygen remain in the flask.
The photos for the reaction arent here so ill insert them below
umm. what's your question? i'll answer it cause these r just the elements...