Answer:
mol times or devided by molar volume
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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:
.259 g
Explanation:
PV = nRT
n = PV / RT
= .986 x 0.144 / .082 x 293.6
= .005897 moles
= .005897 x 44 g
= .259 g
762 miles = 1226.32 kilometers
you multiply the miles by 1.609 to get the kilometers