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
2.24 liters is the volume of the gas if pressure is increased to 1000 Torr.
Explanation:
Data given:
Initial volume of the gas V1 = 2.6 liters
Initial pressure of the gas P1 = 860 Torr 1.13 atm
final pressure on the gas P2 = 1000 Torr 1.315 atm
final volume of the gas after pressure change V2 =?
From the data given above, the law used is :
Boyles Law equation:
P1V1 = P2V2
V2 = P1V1/P2
= 1.13 X 2.6/ 1.31
= 2.24 Liters
If the pressure is increased to 1000 Torr or 1.315 atm the volume changes to 2.24 liters. Initially the volume was 2.6 litres and the pressure was 860 torr.
Answer:
The half-life of Material 1 and Material 2 are equal.
step by step explanation;
Material 1 disintegrates to half its mass three times in 21.6 s, to go from 100g
to 12.5g. That is,
100g - 50g - 25g - 12.5g
Material 2 disintegrates to half its mass three times in 21.6 s, to go from 200g to 25g. That is,
200g - 50g - 25g - 12.5g.
This means that regardless of their initial masses involved, material 1 and material 2 have equal half-life.
Their half-life is 21.6 ÷ 3 = 7.2 sec
Answer:
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