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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Answer:
2.5 × 10² ppm
Explanation:
Step 1: Given data
- Mass of the sample: 200. g
Step 2: Convert 0.050 g to μg
We will use the conversion factor 1 g = 10⁶ μg.
0.050 g × 10⁶ μg/1 g = 5.0 × 10⁴ μg
Step 3: Calculate the concentration of NaCl in ppm
The concentration of NaCl in ppm is equal to the micrograms of NaCl per gram of the sample.
5.0 × 10⁴ μg NaCl/200. g = 2.5 × 10² ppm
Answer:
Glucose, found in the food animals eat, is broken down during the process of cellular respiration into an energy source called ATP. When excess ATP and glucose are present, the liver converts them into a molecule called glycogen, which is stored for later use.
The final temperature : 78.925°C
<h3>Further explanation </h3>
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
Energy releases = 130 kcal = 130 x 4.18 kJ=543.4 kJ
The final temperature :
Final temperature :
ΔT=final-initial
51.925°c=final-27°c
final = 51.925+27=78.925°C
