Suppose that a solid cube of some material floats on the surface of an oil puddle at a service station with 75.1 % of its volume
1 answer:
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Answer:
Explanation:
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In this case, for the given reaction at equilibrium:
We can write the law of mass action as:
That in terms of the change due to the reaction extent we can write:
Nevertheless, for the carbon monoxide, we can directly compute as shown below:
Finally, we can compute the equilibrium constant:
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Answer:
The empirical formula is =
Explanation:
Given that:- Mass of nickel = 2.241 g
Mass of the oxide formed = 2.852 g
Mass of the oxygen reacted = Mass of the oxide formed - Mass of nickel = 2.852 g - 2.241 g = 0.611 g
Molar mass of nickel = 58.6934 g/mol
Moles of nickel = = 0.03818 mol
Molar mass of oxygen = 15.999 g/mol
Moles of nickel = = 0.03818 mol
Taking the simplest ratio for Ni and O as:
0.03818 : 0.03818 = 1 : 1
<u>The empirical formula is =
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Answer : The percent abundance of the heaviest isotope is, 78 %
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
As we are given that,
Average atomic mass = 48.68 amu
Mass of heaviest-weight isotope = 49.00 amu
Let the percentage abundance of heaviest-weight isotope = x %
Fractional abundance of heaviest-weight isotope =
Mass of lightest-weight isotope = 47.00 amu
Percentage abundance of lightest-weight isotope = 10 %
Fractional abundance of lightest-weight isotope =
Mass of middle-weight isotope = 48.00 amu
Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %
Fractional abundance of middle-weight isotope =
Now put all the given values in above formula, we get:
Therefore, the percent abundance of the heaviest isotope is, 78 %