Answer: a) pH = 13.00 : basic
b)
: basic
c) pOH = 5.00 : basic
d)
: acidic
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.
a) pH = 13.00
As pH is more than 7, the solution is basic.
b) ![[H_3O^+]=1.0\times 10^{-12}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1.0%5Ctimes%2010%5E%7B-12%7D)
Putting in the values:
![pH=-\log[1.0\times 10^{-12}]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B1.0%5Ctimes%2010%5E%7B-12%7D%5D)

As pH is more than 7, the solution is basic.
c) pOH = 5.00


As pH is more than 7, the solution is basic.
d) ![[OH^-]=1.0\times 10^{-9}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.0%5Ctimes%2010%5E%7B-9%7D)
Putting in the values:
![pOH=-\log[1.0\times 10^{-9}]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B1.0%5Ctimes%2010%5E%7B-9%7D%5D)



As pH is less than 7, the solution is acidic.
The molar concentration will be greater than 0.01 M
.
Since more of the compound was measured out than what was calculated, you can think of the solution as being 'stronger' than what it was calculated to be. Since a 'stronger' concentration results in a number that is higher, the molarity of this solution is going to be greater than 0.01 M.
Not sure what you are asking. I have two possible answers though...
It could either be more negatively charged, or valence electrons.
The more away from the nucleus a electron is, the more negatively charged it is.
The electrons on the outermost electron shell is valence electrons.
Again, I don't know what you were asking, but one of these answers may be correct.
Explanation:
Charles' law gives the relationship between the volume and the temperature of the gas. Mathematically,
Volume ∝ Temperature
i.e. 
We have, V₁ = 1.6 L, T₁ = 278 K, T₂ = 253, V₂=?

So, the new volume is 1.45 L.
Answer:
(1) addition of HBr to 2-methyl-2-pentene
Explanation:
In this case, we will have the formation of a <u>carbocation</u> for each molecule. For molecule 1 we will have a <u>tertiary carbocation</u> and for molecule 2 we will have a <u>secondary carbocation</u>.
Therefore the <u>most stable carbocation</u> is the one produced by the 2-methyl-2-pentene. So, this molecule would react faster than 4-methyl-1-pentene. (See figure)