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.
Answer:
half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive ...
Explanation:
braniest
Each section would be 230 centimeters or 2.30 meters long.
<h3>Sectional division</h3>
The original wood pole is 12 meters long.
12 meters = 12 x 100 = 1200 centimeters
It is to be cut into 5 equal sections: 1200/5 = 240 centimeters
But each cut wastes 10 cm of the pole.
240 - 10 = 230 centimeters.
Thus, each section would be 230 centimeters or 2.30 meters long.
More on sectional divisions can be found here: brainly.com/question/15381501
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Answer:
![Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BC_6H_5O%5E-%5D%5BH%5E%2B%5D%7D%7B%5BC_6H_5OH%5D%7D)
Explanation:
Hello,
In this case, weak acids are characterized by the fact they do not dissociate completely, it means they do not divide into the conjugated base and acid at all, a percent only, which is quantified via equilibrium. In such a way, the chemical equation representing such incomplete dissociation is said to be:

Thus, we can write the law of mass action, which consider the equilibrium concentrations of all the involved species, which is also known as the acid dissociation constant which accounts for the capacity the acid has to yield hydronium ions:
![K=Ka=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}](https://tex.z-dn.net/?f=K%3DKa%3D%5Cfrac%7B%5BC_6H_5O%5E-%5D%5BH%5E%2B%5D%7D%7B%5BC_6H_5OH%5D%7D)
Best regards.