Answer:
1.51367e+10 inches
Explanation:
1 mile = 63360
63360 x 238900 = 15136704000
Hope this helped!
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
Heavy rainfall because that’s a natural thing that happens and can never stop
Answer:
2.3 * 10^-5
Explanation:
Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.
Hence;
Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols
From the question;
1.2 * 10^-3 mols dissolves in 250 mL
x moles dissolves in 1000mL
x = 1.2 * 10^-3 mols * 1000/250
x= 4.8 * 10^-3 moldm^-3
CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)
Hence Ksp = [Ca^2+] [C2O4^2-]
Where;
[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3
Ksp = (4.8 * 10^-3)^2
Ksp = 2.3 * 10^-5
