Answer : The initial quantity of sodium metal used is 17.25 grams.
Solution : Given,
Volume of 
gas = 8.40 L
Molar mass of Na metal = 23 g/mole
The Net balanced chemical reaction is,
At STP, 22.4 L of volume is occupied by 1 mole of gas
so, 8.40 L of volume is occupied by = 
= 0.375 moles of gas
Now from the above reaction, we conclude that
1 mole of gas produced by the 2 moles of Na metal
0.375 moles of gas produced = 
of Na metal
The quantity of Na metal used = Moles of Na metal × Molar mass of Na metal = 0.75 moles × 23 g/mole = 17.25 grams
Therefore, the initial quantity of sodium metal used is 17.25 grams.
I don’t understand the question
<h3>
Answer:</h3>
482.596 grams
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Explanation:</h3>
The relationship between moles, molar mass and mass of a compound is given by;
In this case;
The compound in question is P₂O₅
Molar mass of P₂O₅ = 141.94 g/mol
Rearranging the formula;
Mass = Moles × molar mass
Therefore;
Mass of P₂O₅ = 3.4 mol × 141.94 g/mol
= 482.596 grams
Therefore, mass of P₂O₅ is 482.596 grams
P2 = 40 atm
Explanation:
Given:
P1 = 2 atm
V1 = 1200 L
V2 = 60 L
P2 = ?
Using Boyle's law and solving for P2,
P1V1 = P2V2
P2 = (V1/V2)P1
= (1200 L/60 L)(2 atm)
= 40 atm
