You cant draw here if i were u i would look it up
Aluminium Hydroxide on decomposition produces Al₂O₃ and Water vapors.
<span> 2 Al(OH)</span>₃ → Al₂O₃ + 3 H₂O
According to equation at STP,
67.2 L (3 moles) of H₂O is produced by = 78 g of Al(OH)₃
So,
65.0 L of H₂O will be produced by = X g of Al(OH)₃
Solving for X,
X = (65.0 L × 78 g) ÷ 67.2 L
X =
75.44 g of Al(OH)₂Result: 75.44 g of Al(OH)₂ is needed to decompose in order to produce 65.0 L of water at STP in stoichiometry
Answer: 2-Benzyloxyethanol is produced reaction in the fig. below.
Explanation:
Answer:
1.6 × 10²³ molecules of N2O5
Explanation:
To find the amount of molecules from the amount of moles of a substance we simply multiply the amount of moles by Avagadro's number.
0.27mol × 6.02×10²³ = 1.6 × 10²³ molecules of N2O5
Answer:
The element is Dysprosium.