Answer:
addition polymerization
Explanation:
In addition polymerization, the monomers are simply joined to each other to form a polymer having the same empirical formula as the monomer but of higher relative molecular mass. The monomers in addition polymerization are usually simple unsaturated molecules such as alkenes.
We can deduce the reaction to be an addition polymerization because of the the attachment of n to both the unsaturated monomer and the saturated polymer without the loss of any small molecule. If it was a condensation polymerization, there would have been an accompanying loss of a small molecule such as water.
It would be 35.8 Calories or calories. Not sure about that part. Hope this helps though.
The bigger the atomic radius the easier it is to oxidise the atom. Remember that an atom is oxidized by the loss of an electron.
Explanation:
The bigger the atomic radius the further away the valence electron are from the attractive force of the atomic nucleus. This means that the energy required to remove an electron from the valence shell is easier compared to an atom with a smaller atomic radius. This is because you need to overcome the attractive force of the nucleus on the electron for you to oxidize the atom.
Learn More:
For more on oxidation energy check out;
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probably none becuase it going staright unless its going down a hill
Answer:
New pH = 3.84
Explanation:
First of all we may think that if the buffer has pH 3.98 and we're adding H⁺, pH's buffer will be lower, as the [H⁺] is been increased.
Let's determine the moles of each compound:
0.23 M . 1.3L = 0.299 moles of NaBz
0.38 M . 1.3L = 0.494 moles of HBz
We add 0.058 of HCl, which is the same as 0.058 moles of H⁻
HCl → H⁺ + Cl⁻
As we add the moles of protons, these are going to react to the Bz⁻
In the buffer system we have these dissociations:
NaBz → Na⁺ + Bz⁻
HBz → H⁺ + Bz⁻
So, as we add protons, we have a new equilibrium:
Bz⁻ + H⁺ ⇄ HBz
In 0.299 0.058 0.494
Eq 0.241 - 0.552
Protons are substracted to benzoate, so the [HBz] is now higher than before. We calculate the new pH, with the Henderson Hasselbach equation
pH = pKa + log (Bz⁻/HBz)
pH = 4.20 + log (0.241 / 0.552) → 3.84
