Answer:
The percent isotopic abundance of Ir-193 is 60.85 %
The percent isotopic abundance of Ir-191 is 39.15 %
Explanation:
we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193
First of all we will set the fraction for both isotopes
X for the isotopes having mass 193
1-x for isotopes having mass 191
The average atomic mass is 192.217
we will use the following equation,
193x + 191(1-x) = 192.217
193x + 191 - 191x = 192.217
193x- 191x = 192.217 - 191
2x = 1.217
x= 1.217/2
x= 0.6085
0.6085 × 100 = 60.85 %
60.85% is abundance of Ir-193 because we solve the fraction x.
now we will calculate the abundance of Ir-191.
(1-x)
1-0.6085 =0.3915
0.3915× 100= 39.15 %
Answer:
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Explanation:
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37.1×10−8M is the concentration of pb 2 ions in a solution prepared by adding 5. 00 g of lead(ii) iodide to 500. ml of 0. 150 m ki? [ k sp(pbi 2) =1. 4 × 10 –8] by common ion effect.
A phenomenon known as the "common ion effect" allows for the modification of a salt's molar solubility by adding another salt in which one ion dissociates in solution and maintains equilibrium with the undissociated salt.
The term "common-ion effect" describes the reduction in solubility of an ionic precipitate caused by the addition of a soluble molecule that shares an ion with the precipitate to the solution. Le principle for the equilibrium response of ionic association/dissociation leads to this behavior.
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brainly.com/question/13194653
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Explanation:
Reaction equation for this reaction is as follows.
It is given that 
= 0.0118.
According to the ICE table,
Initial: 0.86 0.86 0 0
Change: -x -x +x +x
Equilibrium: 0.86 - x 0.86 - x x x
Hence, value of will be calculated as follows.
0.0118 = 
x = 0.084 atm
Thus, we can conclude that 
is 0.084 atm.
Here first we will identify the number of significant figures in the given two dimensions
2.3mm : there are two significant numbers
8.00mm: there are three significant numbers
so we will report the answer to minimum number of significant number , which is two
the answer is 18.4
We will report it to be 18 mm^2
