We have that the the liquid is
- C_2H_5OH (ethanol
- And at a condition of H_2SO4 as catalyst and temp 170
From the question we are told
- A student wished to prepare <em>ethylene </em>gas by <em>dehydration </em>of ethanol at 140oC using sulfuric acid as the <em>dehydrating </em>agent.
- A low-boiling liquid was obtained instead of ethylene.
- What was the liquid, and how might the reaction conditions be changed to give ethylene
<h3>
Ethylene formation</h3>
Generally the equation is
2C_2H_5OH------CH3CH_2O-CH_2CH_3+H_20
Therefore
with ethanol at 140oC
The product is diethyl ethen
The reaction at 170 ethylene will give
C_2H_5OH-------CH_2=CH_2+H_2O( at a condition of H_2SO4 as catalyst and temp 170)
Therefore
The the liquid is
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brainly.com/question/20117360
Answer : The oxidizing element is N and reducing element is O.
is act as an oxidizing agent as well as reducing agent.
Explanation :
An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.
Reducing agent is the agent which has ability to reduce other or lower in oxidation number.
The given reaction is :
act as an oxidizing agent.
The oxidation number of N in is calculated as:
(+1)+(x)+3(-2) = 0
x = +5
And the oxidation number of N in 
is calculated as:
(+1)+(x)+2(-2) = 0
x = +3
From the oxidation number method, we conclude that the oxidation number reduced this means itself get reduced to and it can act as an oxidizing agent.
act as a reducing agent.
The oxidation number of O in is calculated as:
(+1)+(+5)+3(x) = 0
x = -2
The oxidation number of O in 
is Zero (o).
Now, we conclude that the oxidation number increases this means itself get oxidized to and it can act as reducing agent.
The answer is
2Ag+(aq) + SO4-2(aq) → Ag2SO4(s)
<span>NO3- and K+ ions are spectators </span>
Answer:
d
Explanation:
sugar molecules are being broken down
