<span>Answer:
The HCl and KOH will react until one or the other is gone. As you have a larger volume of an equal concentration of HCl, the KOH will go first.
moles HCl = 0.04000 L * 0.100 M = 0.00400 moles
moles KOH = 0.02500 L * 0.100 M = 0.00250 moles
moles HCl left = 0.00400 - 0.00250 = 0.00150 moles
Your total volume is now 65.00 mL, so the [HCl] = 0.00150 moles / 0.06500 L = 0.0231 M = [H+]
pH = -log [H+] = -log (0.0231) = 1.64</span>
Electronic Configuration of elements in a period is same because If you see the electronic Configuration of elements in a period you will notice that the valence shell electrons for all elements are present in the same Shell. For example, in first period consisting of Hydrogen and Helium, both the elements' valence electrons are present in the same Shell.
Electronic Configuration of Hydrogen,
1s^1
Electronic Configuration of Helium,
1s^2
Both elements' valance electrons are present in the 1st shell
(This is just a small example to understand the concept because other periods are long but the first period is short that's why I gave the example of the first period)
2Fe + 3Cl₂ ---> 2FeCl₃
4.4mol of Fe, you have a 2:3 ratio of Fe to Cl₂ so divide 4.4/2 = 2.2 and multiply by three 2.2 x 3 = 6.6mol of Cl₂
hope that helps :)
Answer:
No effect.
Explanation:
Hello,
In this case, considering the widely studied Le Chatelier's principle, we can realize that the factors affecting equilibrium are concentration, temperature and pressure and volume if the reaction is in gaseous phase and with non-zero change in the number of moles. In such a way, by adding a catalyst to given reaction will have no effect on the equilibrium direction.
Best regards.
