Answer:
Distillation will generate the most cyclohexene.
Explanation:
Let us assume following attached reaction for the synthesis of cyclohexene from cyclohexanol which attains equilibrium after certain time.
As shown in figure the cyclohexanol upon treatment with phosphoric acid undergoes dehydration reaction (removal of water) and produces cyclohexene. On the other hand cyclohexene reacts with water (hydration reaction) and produces cyclohexanol.
Now, if this reaction is allowed in a single flask it will attain equilibrium and will not generate the cyclohexene in high quantity. On the other hand if we apply <em>Le Chatelier's principle</em> ( <u><em>removal of product moves the equilibrium in right direction</em></u>) and distillate cyclohexene (boiling the cyclohexene to convert it into vapors and then collect it after condensation) will move the reaction in forward direction and will allow us to generate cyclohexene in high amounts.
Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
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
For more information on Ethylene visit
brainly.com/question/20117360
The radius of the cation is much smaller than the corresponding neutral atom.(b) The radius of an anion is much larger than the corresponding neutral atom.Explanation:The size of the atom or ion is inversely proportional to the nuclear charge experienced by the electrons.(a)The size of the cation is smaller than the size of the corresponding neutral atom. This is because after removal of an electron from the highest principle energy level the nuclear charge experienced by the valence electrons increases resulting in the decrease in size.(b)The size of an anion is larger than the size of the corresponding neutral atom. In an anion, an extra electron is added to the highest principle energy level but the effective nuclear charge pulling the electrons towards the nucleus is still same. The net effective nuclear charge experienced by the electrons present in the outermost shell decrease. Moreover, due to the added electron, the repulsion between the electrons also increases resulting in the increase in size
Make since? i hope this helps
