Answer:
The heat required to vaporize 102.3 grams of H₂O(l) is 231198 J.
Explanation:
The heat required to vaporize 102.3 g of H₂O(l) can be calculated as follows:
Where:
q: is the heat
ΔHv: is the heat of vaporization of water = 2260 J/g
m: is the mass = 102.3
Therefore, the heat required to vaporize 102.3 grams of H₂O(l) is 231198 J.
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The sequence of coefficient that should be placed in the blank space to balance ii as follows
the sequence is 2 infront of Al 3 infront of O2 and 2 infront of Al2O3
that is 2Al +3O2 = 2Al2O3
from equation above 2 moles of Al reacted with 3 moles of O2 to form 2 moles of Al2O3
Answer:
Sodium metaborate: sodium+boron+oxygen= NaBO 2.
The formula can be written also as Na 2O·B
Explanation:
Answer:
2.17 mol
Explanation:
Given data:
Number of moles of potassium = 4.34 mol
Moles of hydrogen gas produced at STP = ?
Solution:
Chemical equation:
2K + 2H₂O → 2KOH + H₂
Now we will compare the moles of hydrogen and potassium.
K : H₂
2 : 1
4.34 : 1/2×4.34 = 2.17 mol
So, from 4.34 moles of potassium 2.17 moles of hydrogen are produced at STP.