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2 years ago

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1 answer:

8_murik_8 [283]2 years ago

7 0

There should be a sufficient amount of the selected isotope in the rock.

The half-life of the isotope must be long enough to capture the age of the rock.

Explanation:

Sully must consider two main aspect before selecting her choice isotope for dating.

There must be sufficient amount of the selected isotope in the rock.

The half - life of the isotope must be long enough to capture the age of the rock.

Radiometric dating gives a rock an absolute numerical age.
The half-life of an isotope is time take for half of a radioactive element to decay.
If the half-life of an isotope is very short, all the parent nuclide would have turned to daughter nuclides.
Also, we must have sufficient amount of both the daughter and parent isotope in the selected rock.

learn more:

Radiometric dating brainly.com/question/7022607

#learnwithBrainly

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In an aqueous solution of a certain acid the acid is 0.10% dissociated and the pH is 4.06. Calculate the acid dissociation const

Kruka [31]

Answer : The acid dissociation constant Ka of the acid is, $8.7\times 10^{-5}$

Explanation :

First we have to calculate the concentration of hydrogen ion.

$pH=-\log [H^+]$

Given: pH = 4.06

$4.06=-\log [H^+]$

$[H^+]=8.71\times 10^{-5}M$

The dissociation of acid reaction is:

$HA\rightarrow H^++A^-$

Initial conc. c 0 0

At eqm. c-cα cα cα

Given:

Degree of dissociation = α = 0.10 % = 0.001

$[H^+]=c\alpha$

$8.71\times 10^{-5}=c\times 0.001$

$c=0.0871M$

The expression of dissociation constant of acid is:

$K_a=\frac{[H^+][A^-]}{[HA]}$

$K_a=\frac{(c\times \alpha)\times (c\times \alpha)}{(c-c\alpha)}$

Now put all the given values in this expression, we get:

$K_a=\frac{(0.0871\times 0.001)\times (0.0871\times 0.001)}{(0.0871-0.0871\times 0.001)}$

$K_a=8.7\times 10^{-5}$

Thus, the acid dissociation constant Ka of the acid is, $8.7\times 10^{-5}$

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