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tatiyna
3 years ago
8

Potassium is isotopic and has RAM of 39.5 work out the percentage abundance of each isotope in a given sample of potassium which

is found to contain 3 isotopes of K-39, K-40 and K-38 with the abundance of K-38 being 0.01%
Chemistry
1 answer:
Troyanec [42]3 years ago
3 0

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.

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Option (B) is the right answer.

Hydronium ion (H_{3}O^{+}) concentration decreases by the <u>factor of 100</u>, if the pH of a solution increases from 2.0 to 4.0.

<h3>What is pH?</h3>

The hydrogen ion concentration in water is expressed by pH. Specific to aqueous solutions, pH is the <u>negative logarithm</u> of the hydrogen ion (H+) concentration (mol/L) : pH = -log_{10}[H_{3}O^{+}  ]

Acidic solutions are those with a pH under 7, and basic solutions are those with a pH over 7. At this temperature, solutions with a pH of 7 are neutral (e.g.<u> pure water</u>). The pH neutrality <u>relies on temperature, falling below 7 if the temperature rises above 25 °C</u>.

<h3>Given: </h3>

pH1( initial pH) = 2.0

pH2( initial pH) = 4.0

[H3O+] =  initial hydronium concentration

[H3O+]* = final hydronium concentration

<h3>Formula used : </h3>

pH = -log_{10}[H_{3}O^{+}]

<h3>Solution: </h3>

pH = - log_{10}[H_{3}O^{+}] \\\\= > 10^{-pH} = [H_{3}O^{+}] \\\\similarly, \\\\10^{-pH} = [H_{3}O^{+}]*\\\\ = > 10^{-4} = [H_{3}O^{+}]*\\\\Now, \frac{[H_{3}O^{+}]*}{[H_{3}O^{+}]}  = 10^{-2}

Thus , the concentration of hydronium ion decreases by 100.

To learn more about pH :

brainly.com/question/15289741

#SPJ4

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