Answer:
![[OH^-]=9.24x10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D9.24x10%5E%7B-3%7DM)
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Explanation:
Hello,
In this case, since the ionization of methylamine is:
The equilibrium expression is:
![Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D)
And in terms of the reaction extent 
which is equal to the concentration of OH⁻ as well as that of CH₃NH₃⁺ via ice procedure we can write:
Whose solution for via quadratic equation is 9.24x10⁻³ M since the other solution is negative so it is avoided. Therefore, the concentration of OH⁻ is:
![[OH^-]=x=9.24x10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3Dx%3D9.24x10%5E%7B-3%7DM)
With which we can compute the pOH at first:
![pOH=-log([OH^-])=-log(9.24x10^{-3})=2.034](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%289.24x10%5E%7B-3%7D%29%3D2.034)
Then, since pH and pOH are related via:
The pH turns out:
Best regards.
Answer:
When the following redox equation is balanced with smallest whole number coefficients, the coefficient for the iodide ion will be __6__.
Explanation:
From the redox equation, we can see that NO₃⁻ is reduced to NO (from oxidation state +5 to +2), whereas I⁻ is oxidized to I₂ (from oxidation state -1 to 0). The half reactions are balanced with H⁺ (acidic solution), as follows:
Reduction : 2 x (NO₃⁻(aq) + 3 e- + 4 H⁺ → NO(g) + 2 H₂O)
Oxidation : 3 x (2 I⁻(aq) → I₂(s) + 2 e-)
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Total equation: 6 I⁻(aq) + 2 NO₃⁻(aq)+ 8 H⁺ → 3 I₂(s) + 2 NO(g) + 4 H₂O
That is the redox equation with the smallest whole number coefficients.
Accordin to this, the coefficient for the iodide ion (I⁻) is: 6.
Answer:
Because the Calvin cycle is dependent on the the product from light reaction.
Explanation:
The Calvin cycle is the light independent phase of photosynthesis during which carbon is fixed. This step requires energy generated during the light dependent phase of the photosynthesis.
Hence, if the light dependent reaction does not occur, the required energy to drive carbon fixation will be lacking and the Calvin cycle will not be able to proceed.
