Answer:
B) is reduced.
Explanation:
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced and it is oxidizing agent.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized and it is reducing gent.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized.
The IUPAC name for the organic compound CH2(OH)-CH2-CH(C2H5)-OH is Pentan-1,3-diol
This compound above belong to class of organic compound called the alkanols
<h3>What are organic compounds?</h3>
Organic compounds are compounds which contains carbon and hydrogen only.
Some few classes of organic compounds includes the following:
- Alkanes
- Alkenes
- Alkynes
- Alkanols
- Alkanals
- Alkanones
- Esters
- Ketones and other
So therefore, the IUPAC name for the organic compound is CH2(OH)-CH2-CH(C2H5)-OH is Pentan-1,3-diol
Learn more about organic compounds:
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A challenge they could face is that the plants are not used to the climate in the new area . (sorry if it’s wrong)
Answer:
?
Explanation:
What are the statements? You've given the passage but not the statements
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M