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.
Answer:
small- even though they are huge suns themselves.
Explanation:
Answer:
132.17 g
Explanation:
The reaction given , in the question is -
CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S ( g )
From the reaction , 4 mole of S is required for the production of 1 mole of CS₂ .
since ,
Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂
Since ,
the Molecular weight of CS₂ = 76
Given , mass of CS₂ = 72.57 g
Moles of CS₂ = 72.57 / 76 = 0.95 mol
Since ,
The yield is 92.0 % .
Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles
Mass of S required = 4.13 * 32 = 132.17 g .
Answer: 90 percent water, electrolytes such as potassium and phosphorus, and chemicals called urea and uric acid
Explanation:
find mass of oxygen by subtracting mass of phosphorus from the total mass
divide the masses by the molar mass to get moles
divide moles by the smallest amount of moles
multiply by 2 to get a nice number
P4O5
