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Water decomposes when electrolyzed to produce hydrogen and oxygen gas. If 2.5 grams of water were decomposed 1.04 grams of oxygen will be formed.
BCA table:
2O ⇒
+
B 0.13 0 + 0
C -0.13 0.065 + 0.065
A 0 0.065
Explanation:
Balanced equation for water decomposition into hydrogen and oxygen gases
2O ⇒
+
B 0.13 0 + 0
C -0.13 0.065 + 0.065
A 0 0.065
Number of moles of water =
mass = 2.5 grams
atomic mass= 18 grams
number of moles can be known by putting the values in the formula,
n =
= 0.13 moles
2 moles of water gives one mole of oxygen on decomposition
so, 0.13 moles of water will give x moles of oxygen on decompsition
=
x = 0.065 moles of oxygen will be formed.
moles to gram will be calculated as
mass =number of moles x atomic mass
= 0.065 x 16
= 1.04 grams of oxygen.
Answer:
The concentration of the solution is 5.8168 × mol.
Explanation:
Here, we want to calculate the concentration of the solution.
The unit of this is mol/dm^3
So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3
The number of moles = mass/molar mass
molar mass of AlCO3 = 27 + 12 + 3(16) = 27 + 12 + 48 = 87g/mol
Number of moles = 33.4/87 = 0.384 moles
This 0.384 moles is present in 660 L
x moles will be present in 1 dm^3
Recall 1 dm^3 = 1L
x * 660 = 0.384 * 1
x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3