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jekas [21]
3 years ago
13

Rubidium is comprised of two isotopes, one of which has a natural abundance of 72% and contains 48 neutrons in the nucleus. give

n that the calculated atomic mass for rubidium is 85.5, how many neutrons are contained in the nucleus of an atom of the second isotope?
Chemistry
1 answer:
SOVA2 [1]3 years ago
6 0
Since there are only two isotopes, then that means the other isotope has an abundance of: 100 - 72 = 28%. Let y be the mass of the first isotope, and x be the mass of the second. The equation would be

85.5 = 0.72y + 0.28x

Now, rubidium has 37 protons, and each proton has a mass of 1.00727647 amu. When neutral, it must also have 37 electrons in which each electron weighs 0.000548597 amu. Let n be the number of neutrons, in which each neutron weighs <span>1.008664 amu. The solution is as follows:

y = 37(</span>1.00727647) + 37(0.000548597) + 48(1.008664)
y = 85.7054 amu

Then, x would be:
85.5 = 0.72(85.7054) + 0.28x
x = 84.972 amu
So,
x = 84.972 = 37(1.00727647) + 37(0.000548597) + n(1.008664)
Solving for n,
n = 47.27 ~ 47

<em>Therefore, there are 47 neutrons in the second isotope.</em>
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Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer
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Answer:

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

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Now, if the total concentration is

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[HA] + [A−] 0.5754

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= 0.0635 M

−−−−−−−−

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and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

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nHA = 0.0635 molL × 0.050L =

0.003175 mols

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So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

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