Answer:
2,577°F
Also,
Boron: 3,769°F (2,076°C)
Neon: -415.5°F (-248.6°C)
Beryllium: 2,349°F (1,287°C)
<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
</span>
<span>
With that you can solve the question following these steps"
</span>
<span>1) First ionization:
</span>
<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
</span>
<span>Do the mass balance:
</span>
<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
</span>
<span /><span /><span> 0.01 M - x x x
</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
<span /><span>
=> Ka₂ = (x²) / (0.01 - x) = 0.012
</span><span />
<span>3) Solve the equation:
</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
<span /><span>
x² + 0.012x - 0.0012 = 0
</span><span />
<span>Using the quadratic formula: x = 0.00649
</span><span />
<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>
Answer:
I know someone that has the answer
Explanation:
I know someone that has the answer
The element that gains electrons, becomes reduced.
While the one which loses electrons, becomes oxidized.
In this equation,
CH₃OH + Cr₂O₇²⁻---- --> CH₂O + Cr³⁺.
By balancing the equation, we will get:
3CH₃OH + Cr₂O₇²⁻ + 8H⁺ --> 3CH₂O + 2Cr³⁺ + 7H₂O
Here the oxidation state of Cr changes from +6 to +3 that is it is being reduced thus serving as a oxidizing agent while other element retain their charges.
Here Cr₂O₇²⁻ is reduced while CH₃OH is oxidized.
So Cr₂O₇²⁻ serves as a oxidizing agent, while CH₃OH serves as reducing agent .
The molecules vibrate faster and space between atoms increase
