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Yo sup??
Let the percentage of K-39 be x
then the percentage of K-40 is 100-(x+0.01)
We know that the net weight should be 39.5. Therefore we can say
(39*x+40*(100-(x+0.01))+38*0.01)/100=39.5
(since we are taking it in percent)
39*x+40*(100-(x+0.01))+38*0.01=3950
39x+4000-40x-0.4+0.38=3950
2x=49.98
x=24.99
=25 (approx)
Therefore K-39 is 25% in nature and K-40 is 75% in nature.
Hope this helps.
This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:
Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰
Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻
Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.
In such a way, the correct choice is C.
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Answer:
37 mmol of acetate need to add to this solution.
Explanation:
Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-
![pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%28acetic%20acid%29%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7Bmmol%20of%20CH_%7B3%7DCOOH%20%7D%5D)
Here pH is 5.31,
(acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.
Plug in all the values in the above equation:
![5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]](https://tex.z-dn.net/?f=5.31%3D4.74%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7B10%7D%5D)
or, mmol of
= 37
So 37 mmol of acetate need to add to this solution.