Answer:
32.7 kilograms of aluminium oxide will be produced.
Explanation:
Mass of aluminum = 17.3 kg = 17300 g (1 kg = 1000 g )
Moles of aluminium = 
According to reaction, 2 moles of aluminum gives 1 mole of aluminum oxide,then 640.74 moles of aluminum will give:
of aluminum oxide
Mass of 320.37 moles of aluminum oxides:
320.37 mol × 102 g/mol = 32,677.74 g = 32.67774 kg ≈ 32.7 kg
32.7 kilograms of aluminium oxide will be produced.
Answer:
the ion will have additional 2 electrons so it will be 18
Nitrate
Hope this helps :))
Answer:
V₂ = 1.92 L
Explanation:
Given data:
Initial volume = 0.500 L
Initial pressure =2911 mmHg (2911/760 = 3.83 atm)
Initial temperature = 0 °C (0 +273 = 273 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
by putting values,
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm
V₂ = 522.795 atm .L. K / 273 K.atm
V₂ = 1.92 L
