To determine the amount of excess reactant left, we need to first identify which is the limiting and the excess reactant. We do as follows:
<span>A + 2B → AB2
2 mol A ( 2 mol B / 1 mol A) = 4 mol B needed to be consumed completely
5 mol B ( 1 mol A / 2 mol B) = 2.5 mol A needed to be consumed completely
Therefore, the limiting would be reactant A and the excess would be reactant B. The amount of the excess would be:
5 mol B - 4 mol B = 1 mol B <----------LAST OPTION</span>
MO Diagram of C₂⁻ is shown below,
Bond order is calculated as,
Bond Order = [# of e⁻s in BMO]-[#of e⁻s in ABMO] / 2
Where,
BMO = Bonding Molecular Orbital
ABMO = Anti-Bonding Molecular Orbital
Putting values,
Bond Order = [9]-[4] / 2
Bond Order = 5 / 2
Bond Order = 2.5
Answer:
pH = 10
The solution is basic.
Explanation:
A solution contains 1 × 10⁻⁴ M OH⁻ ions. First, we will calculate the pOH.
pOH = -log [OH⁻]
pOH = -log 1 × 10⁻⁴
pOH = 4
We can find the pH of the solution using the following expression.
pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 4 = 10
Since the pH > 7, the solution is basic.
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