Capture all of the smoke and weight it. it will weigh exactly the same before and after you burn it but will just be CO2 and H2O gas.
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
Answer:
2Li(s) + ⅛S₈(s, rhombic) + 2O₂(g) → Li₂SO₄(s)
Explanation:
A thermochemical equation must show the formation of 1 mol of a substance from its elements in their most stable state,.
The only equation that meets those conditions is the last one.
A and B are wrong , because they show Li₂SO₄ as a reactant, not a product.
C is wrong because Li⁺ and SO₄²⁻ are not elements.
D is wrong because it shows the formation of 8 mol of Li₂SO₄.
Answer:
The chemical elements are arranged in order of increasing atomic number.
Explanation:
Hydrogen = 1
Helium = 2
Lithium = 3
Answer:
136.63 °C
Explanation:
ΔTb=Tb solution - Tb pure
Where; Tb pure = 133.60°C
molar mass of solute = 121.14 g/mol
number of moles of solute; 52.2g/121.14 g/mol = 0.431 moles
molality = 0.431 moles/350 * 10^-3 = 1.23 molal
Then;
ΔTb = Kb * m * i
Kb = 2.46°C kg mol^-1
m = 1.23 molal
i = 1
ΔTb = 2.46 * 1.23 * 1
ΔTb = 3.03 °C
Hence;
Tb solution = ΔTb + Tb pure
Tb solution = 3.03 °C + 133.60°C
Tb solution = 136.63 °C
