Answer:
false
Explanation:
not too sure sorry if im wrong
Answer:
The answer to your question is 8.21 g of H₂O
Explanation:
Data
mas of water = ?
mass of hydrogen = 4.6 g
mass of oxygen = 7.3 g
Balanced chemical reaction
2H₂ + O₂ ⇒ 2H₂O
Process
1.- Calculate the atomic mass of the reactants
Hydrogen = 4 x 1 = 4 g
Oxygen = 16 x 2 = 32 g
2.- Calculate the limiting reactant
Theoretical yield = H₂/O₂ = 4 / 32 = 0.125
Experimental yield = H₂/ O₂ = 4.6/7.3 = 0.630
From the results, we conclude that the limiting reactant is Oxygen because the experimental yield was higher than the theoretical yield.
3.- Calculate the mass of water
32 g of O₂ ---------------- 36 g of water
7.3 g of O₂ --------------- x
x = (7.3 x 36) / 32
x = 262.8 / 32
x = 8.21 g of H₂O
Answer:
Sodium peroxide can be prepared on a large scale by the reaction of metallicsodium with oxygen at 130–200 °C, a process that generates sodium oxide, which in a separate stage absorbs oxygen: 4 Na + O2 → 2 Na2O. The ozone oxidizes the sodium to form sodium peroxide.
Boyle’s Law P1V1 = P2V2
P1 = 0.80 atm V1 = 1.8 L
P2 = 1.0 atm V2 = ??
(.8 atm)(1.8 L) = (1.0 atm)(V2)
1.44 atm x L = 1 atm V2
