Answer:
Percentage abundance of B 10 is = 20 %
Percentage abundance of B 11 is = 80 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope, B 10:
% = x %
Mass = 10.0129 u
For second isotope, B 11:
% = 100 - x
Mass = 11.0093 u
Given, Average Mass = 10.81 u
Thus,

Solving for x, we get that:
x = 20 %
Thus percentage abundance of B 10 is = 20 %
Percentage abundance of B 11 is = 100 - 20 % = 80 %
Explanation:
The different subatomic particles in an atom are the:
- Protons which are the positively charged particles.
- Electrons which are the negatively charged particles.
- Neutrons which do not carry any charges.
Protons and neutrons are located in the nucleus of an atom which is the tiny center of the atom.
Electrons orbits around the nucleus and fill the rest of the volume of atom.
In order to increase the strength of his electromagnet and pick up more thumbtacks Braddy should use two batteries instead of one. Using two batteries will increase the amount of current in the coils which will induce a higher electromagnetic field in the nail.
Another option would be to increase the number of coils of wire.
Fe - 3
O - 6
H - 6
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Answer:
6,45mmol/L of NaOH you need to add to reach this pH.
Explanation:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ <em>pka = 4,74</em>
Henderson-Hasselbalch equation for acetate buffer is:
5,0 = 4,74 + log₁₀![\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
Solving:
1,82 =
<em>(1)</em>
As total concentration of acetate buffer is:
10 mM = [CH₃COOH] + [CH₃COO⁻] <em>(2)</em>
Replacing (2) in (1)
<em>[CH₃COOH] = 3,55 mM</em>
And
<em>[CH₃COO⁻] = 6,45 mM</em>
Knowing that:
<em>CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O</em>
Having in the first 10mmol/L of CH₃COOH, you need to add <em>6,45 mmol/L of NaOH. </em>to obtain in the last 6,45mmol/L of CH₃COO⁻ and 3,55mmol/L of CH₃<em>COOH </em>.
I hope it helps!