<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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<span>
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:
Potassium (K) = (2,8,8,1)
This is the element that has three(3) filled energy level and one valence electron
Answer:
1) Endothermic.
2)
3)
Explanation:
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1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.
2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:
3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:

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