The thing you MUST do FIRST is look for any H's, O's, or F's in the equation
1)any element just by itself not in a compound, their oxidation number is 0
ex: H2's oxidation number is 0
ex: Ag: oxidation number is 0 if its just something like Ag + BLA = LALA
2) the oxidation number of H is always +1, unless its just by itself (see #1)
3) the oxidation number of O is always -2, unless its just by itself (see #1)
4) the oxidation number of F is always -1, unless its just by itself (see#1)
ok so after you have written those oxidation numbers in rules 1-4 over each H, F, or O atom in the compound, you can look at the elements that we havent talked about yet
for example::::
N2O4
the oxidation number of O is -2.
since there are 4 O's, the charge is -8. now remember that N2O4 has to be neutral so the N2 must have a charge of +8
+8 divided by 2 = +4
N has an oxidation number of +4.
more rules:
5) the sum of oxidation numbers in a compound add up to 0 (when multiplied by the subscripts!!!) (see above example)
6) the sum of oxidation numbers in a polyatomic ion is the charge (for example, PO4 has a charge of (-3) so
oxidation # of O = -2. (there are 4 O's = -8 charge on that side ) P must have an oxidation number of 5. (-8+5= -3), and -3 is the total charge of the polyatomic ion
The correct response would be A. The slowest step that determines the speed(rate) of the overall reaction.
Answer: The pH of phenol solution is 4.7
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydronium ion concentration.
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
Given : ![[H_3O^+]=1.8\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1.8%5Ctimes%2010%5E%7B-5%7DM)
Putting in the values we get
![pH=-\log[1.8\times 10^{-5}M]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B1.8%5Ctimes%2010%5E%7B-5%7DM%5D)

Thus pH of phenol solution is 4.7
Answer:

Explanation:
We can use the Combined Gas Laws to solve this problem

Data
p₁ = 571.2 Torr; p₂ = 400 Torr
V₁ = 3.5 L; V₂ = ?
T₁ = 21.5 °C; T₂ = 6.8 °C
Calculations
(a) Convert the temperatures to kelvins
T₁ = (21.55 + 273.15) K = 294.70 K
T₂ = (6.8 + 273.15) K = 279.95 K
(b) Calculate the new volume

Answer: Well then I guess Sara is having a good breakfast
Explanation: what was the point of this question?