<span>0.333 ounce * 0.75 * 28.35 grams/ounce = 7.080 grams gold
7.080 grams Au / 196.967g/mol Au = 0.0359 mol Au
0.0359 mol Au * (6.023 * 10^23) atoms/mol = 2.24 * 10^22 Atoms Au</span>
Answer:

Explanation:
Hello,
In this case, given the acid, we can suppose a simple dissociation as:

Which occurs in aqueous phase, therefore, the law of mass action is written by:
![Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
That in terms of the change
due to the reaction's extent we can write:

But we prefer to compute the Kb due to its exceptional weakness:

Next, the acid dissociation in the presence of the base we have:
![Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BOH%5E-%5D%5BHA%5D%7D%7B%5BA%5E-%5D%7D%3D1x10%5E%7B6%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.1-x%7D)
Whose solution is
which equals the concentration of hydroxyl in the solution, thus we compute the pOH:
![pOH=-log([OH^-])=-log(0.0999)=1](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.0999%29%3D1)
Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:

Regards.
Answer : The number of moles of solute
is, 0.0788 moles.
Explanation : Given,
Molarity = 0.225 M
Volume of solution = 0.350 L
Formula used:

Now put all the given values in this formula, we get:


Therefore, the number of moles of solute
is, 0.0788 moles.
B. It has a negative charge and much less mass than a proton.