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
First off chlorine is not a metal so you can ignore that one.
Sodium and Rubidium are in group 1 of the periodic table and Magnesium is in group 2.
Group one metals are more reactive than group two because it is harder for the group two metals to lose their 2 valence (outer most) electrons.
As you go down group 1 there is an increase in the reactivity this is because as you go down there is an increase in the atomic radius which leads to more shielding. This weakens the electrostatic forces of attraction making it easier to lose the outermost electrons, therefore they are more reactive.
Given :
Human blood should have around 1.04 kg/L platelets.
A blood sample of 4.01 milliliters is collected from a patient to be analyzed for a platelet count.
To Find :
The expected mass in grams of platelets in the blood sample.
Solution :
1 L of human blood contains 1.04 kg of platelets.
So, amount of platelets is 1 ml blood is :
Mass of platelets in 4.01 ml blood is :
Hence, this is the required solution.
Answer:
0.823 M was the molarity of the KOH solution.
Explanation:
(Neutralization reaction)
To calculate the concentration of base , we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is KOH.
We are given:
Putting values in above equation, we get:
0.823 M was the molarity of the KOH solution.
