Answer:
neutral
Explanation:
pH range goes from 0-14, since 7 is in between 0 and 14, it is neutral.
Answer:
- <u>C₂H₄</u> (option number 4)
Explanation:
A hydrocarbon with a <em>double bond</em> in its carbon skeleton is an alkene and has the general form:
-
.
This is, the number of hydrogen atoms is twice the number of carbon atoms.
On the other hand, alkanes have only single bonds, and the compounds with a triple bond in its carbon skeleton are alkynes.
Review each choice:
1) <u>C₃H₈:</u>
- In this case, the number of hydrogen atoms is 2×3 + 2 = 6 + 2 = 8, which is corresponds to an alkane, not an alkene.
2)<u> C₂H₆</u>
- For this, the number of hydrogen atoms is 2 × 2 + 2 = 4 + 2 = 6. Again an alkane, not alkene.
3) <u>CH₄</u>
- Hydrogen atoms: 1 × 2 + 2 = 4 ⇒ an alkane
4) <u>C₂H₄ </u>
- Hydrogen atoms: 2 × 2 = 4. This is precisely the relation for an alkene, so this is the hydrocarbon that has a double bond in its carbon skeleton.
- The chemical formula may be writen as CH₂ = CH₂, to show the double bond.
So, this is the correct answer.
5) <u>C₂H₂</u>
- Hydrogen atoms: 2 × 2 - 2 = 4 - 2 = 2. This relation of carbon and hydrogen atoms corresponds to a compound with triple bond, i.e an alkyne: CH≡CH.
That would be a depression..
Answer:

Explanation:
Hello there!
In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:
![HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=HA%5Crightleftharpoons%20H%5E%2B%2BA%5E-%5C%5C%5C%5CKa%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:
![Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7Bx%5E2%7D%7B%5BHA%5D_0-x%7D%20%3D10%5E%7B-4.74%7D%3D1.82x10%5E%7B-5%7D)
Thus, it is possible to find x given the pH as shown below:

So that we can calculate the initial concentration of the acid:
![\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B%281.82x10%5E%7B-5%7D%29%5E2%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1.82x10%5E%7B-5%7D%5C%5C%5C%5C%5Cfrac%7B1.82x10%5E%7B-5%7D%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1%5C%5C%5C%5C)
![[HA]_0=3.64x10^{-5}M](https://tex.z-dn.net/?f=%5BHA%5D_0%3D3.64x10%5E%7B-5%7DM)
Therefore, the percent dissociation turns out to be:
![\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%](https://tex.z-dn.net/?f=%5C%25%20diss%3D%5Cfrac%7Bx%7D%7B%5BHA%5D_0%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%3D%5Cfrac%7B1.82x10%5E%7B-5%7DM%7D%7B3.64x10%5E%7B-5%7DM%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%20%3D%2050%5C%25)
Best regards!