Answer:
67.5% ≅ 67.6%
Explanation:
Given data:
Mass of water = 17.0 g
Mass of oxygen produced (actual yield)= 10.2 g
Percent yield of oxygen = ?
Solution:
Chemical equation:
2H₂O → 2H₂ + O₂
Number of moles of water:
Number of moles = mass/ molar mass
Number of moles = 17.0 g/ 18.016 g/mol
Number of moles = 0.944 mol
Now we will compare the moles of oxygen with water to know the theoretical yield of oxygen.
H₂O : O₂
2 : 1
0.944 : 1/2×0.944 = 0.472 mol
Mass of oxygen:
Mass = number of moles× molar mass
Mass = 0.472 mol × 32 g/mol
Mass = 15.104 g
Percent yield:
Percent yield = [Actual yield / theoretical yield] × 100
Percent yield = [ 10.2 g/ 15.104 g] × 100
Percent yield = 0.675 × 100
Percent yield = 67.5%
Reaction on positive electrode (cathode):
PbSO₄₍s₎ + 2H₂O₍l₎ → 2e⁻ + PbO₂₍s₎ + 4H⁺₍aq₎ + SO₄²⁻₍aq₎.
s - solid.
l - liquid.
aq - dissolve in water.
PbSO4 is converted to Pb at one electrode (anode) and to PbO₂ at the other (cathode). Lead battery can be recharged, during recharging, an external source of energy is used.
Answer:
where are those two images which you have sent
Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.
<h3>What is the Ki for the inhibitor?</h3>
The Ki of an inhibitor is known as the inhibition constant.
The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.
Using the Michaelis-Menten equation for inhibition:
Making Ki subject of the formula:
where:
- Kma is the apparent Km due to inhibitor
- Km is the Km of the enzyme-catalyzed reaction
- [I] is the concentration of the inhibitor
Solving for Ki:
where
[I] = 26.7 μM
Km = 1.0
Kma = (150% × 1 ) + 1 = 2.5
Ki = 26.7 μM/{(2.5/1) - 1)
Ki = 53.4 μM
Therefore, the Inhibition constant, Ki of the inhibitor is 53.4 μM.
Learn more about enzyme inhibition at: brainly.com/question/13618533
