After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml
<h3>Subtraction of Numbers</h3>
Given Data
- Volume of Hexane dispensed = 40ml
Let us compute the amount of excess hexane/ the volume that will remain
Remainder = The difference in volume dispensed and the volume needed
Remainder = 40-5
Remainder = 35 ml
The remainder is 35ml
Learn more about subtraction of numbers here:
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The statement above is FALSE.
Unlabeled atom joined to carbon atoms which are not directly part of a ring structure are assumed to be CARBON ATOMS. In a ring structure, an unlabeled atom at the angle where two lines joined together is always assumed to be a carbon atom<span />
The expected radius of a nucleus having 82 protons and 125 neutrons would be 5.2 fm.
<h3>
What is a nucleus in the atom?</h3>
The nuclei are incredibly tiny and dense. They are 10 thousand times smaller than an atom and have more than 99.9% of their mass. Protons, which have a positive charge, and neutrons, which have no electrical charge, make up the nucleus. Alternately, protons and neutrons make up an atom's nucleus. While neutrons have no charge but weigh the same as protons, protons have a positive charge that is equivalent to the orbiting electrons. Based on the 1909 Geiger-Marsden gold foil experiment, Ernest Rutherford identified the atomic nucleus in 1911, which is the compact, dense region made up of protons and neutrons at the heart of an atom.
To learn more about the nucleus, visit:
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Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>