Amu valie on the top and abundsnce percent in decimals on the bottom for each isotope
Answer:
= -457.9 kJ and reaction is product favored.
Explanation:
The given reaction is associated with 2 moles of 
Standard free energy change of the reaction (
) is given as:
, where T represents temperature in kelvin scale
So, 
So, for the reaction of 1.57 moles of
, 
As,
is negative therefore reaction is product favored under standard condition.
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Answer : The percentage reduction in intensity is 79.80 %
Explanation :
Using Beer-Lambert's law :



where,
A = absorbance of solution
C = concentration of solution = 
l = path length = 2.5 mm = 0.25 cm
= incident light
= transmitted light
= molar absorptivity coefficient = 
Now put all the given values in the above formula, we get:



If we consider
= 100
then, 
Here 'I' intensity of transmitted light = 20.198
Thus, the intensity of absorbed light
= 100 - 20.198 = 79.80
Now we have to calculate the percentage reduction in intensity.


Therefore, the percentage reduction in intensity is 79.80 %
Answer:
The attractive force between them decreases
Explanation:
This is because they become localised.