Answer:
The specie which is oxidized is:- 
The specie which is reduced is:- 
Explanation:
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
For the given chemical reaction:
The half cell reactions for the above reaction follows:
Oxidation half reaction: 
Reduction half reaction: 
Thus, the specie which is oxidized is:-
The specie which is reduced is:-
Answer:
Explanation:
Carbon dioxide is the answer I think.
Hope it helps...
Answer:
V₂ = 107.84 L
Explanation:
Given data:
Initial volume = 100 L
Initial pressure = 80 KPa (80/101 =0.79 atm)
Initial temperature = 200 K
Final temperature =273 K
Final volume = ?
Final pressure = 1 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁T₂ /T₁P₂
V₂ = 0.79 atm × 100 L × 273 K / 200 K × 1 atm
V₂ =21567 atm.L.K /200 K.atm
V₂ = 107.84 L
File:///Users/user/Downloads/Chemistry%20Exam%20Study%20Guide%20Fall%202012.pdf
Idk if this will help you, cuz it says what question are going to be in ur final
Answer:
Product: ethyl L-valinate
Explanation:
If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> (
) and a <em>carboxylic acid</em> group (
). Additionally, we have an <u>alcohol </u>(
) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (
).
When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (
). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base () removes the hydrogen from the C=O bond to produce ethyl L-valinate
See figure 1
I hope it helps!
