<span>Answer: 0.00649M
The question is incomplete,
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<span>You are told that the first ionization of the sulfuric acid is complete and the second ionization of the sulfuric acid has a constant Ka₂ = 0.012
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With that you can solve the question following these steps"
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<span>1) First ionization:
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<span>
H₂SO₄(aq) --> H⁺ (aq) + HSO₄⁻ (aq)
Under the fully ionization assumption the concentration of HSO4- is the same of the acid = 0.01 M
2) Second ionization
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<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
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<span>Do the mass balance:
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<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
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<span /><span /><span> 0.01 M - x x x
</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
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=> Ka₂ = (x²) / (0.01 - x) = 0.012
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<span>3) Solve the equation:
</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
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x² + 0.012x - 0.0012 = 0
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<span>Using the quadratic formula: x = 0.00649
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<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>
Answer:
(b) fully filled valence s orbitals
Explanation:
Electron configuration of Be: 1s22s2
2s2 is fully filled
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
Answer:
32.8 C
Explanation:
- Use combined gas law formula and rearrange.
- Hope that helped! Please let me know if you need further explanation.
