Explanation:
Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:
Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.
Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.
However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.
Elimination of the hydrogen from the methyl group is easier.
Thus, the major product will A
Answer:
How the incident happened
Any chemicals involved in an incident
Any other hazards present in the lab
Explanation:
Above are the types of information that are necessary to communicate with emergency responders. The emergency responders ask the first question that how the incident happened. After that they ask that is there any harmful chemicals are present in the laboratory or what types of chemicals present in the laboratory. These questions were asked by the emergency responders in order to give the patient a suitable treatment.
Answer:
Hey hi
Explanation:
Can you pls tell me which language is this.... Pls really sorry... I wanna help you
Answer: Final temperature = 206∘C
Explanation:
Heat Energy is given as
q= mCΔT
ehere
q= Heat energy = 87.4J
m= mass=1.25g
C=specific heat c= 0.386Jg∘C) ,
ΔT = Change in temperate of which the final temperature= 25.0∘C
q= mCΔT
ΔT = q/mC
ΔT = 87.4/ 1.25 X 0.386=181.14∘C
But,
T final- T initial = ΔT
T final = T initial + ΔT
T final = 25.0∘C +181.14∘C=206.14∘C rounded to 206∘C
a. 0.137
b. 0.0274
c. 1.5892 g
d. 0.1781
e. 5.6992 g
<h3>Further explanation</h3>
Given
Reaction
2 C4H10 + 13O2 -------> 8CO2 + 10H2O
2.46 g of water
Required
moles and mass
Solution
a. moles of water :
2.46 g : 18 g/mol = 0.137
b. moles of butane :
= 2/10 x mol water
= 2/10 x 0.137
= 0.0274
c. mass of butane :
= 0.0274 x 58 g/mol
= 1.5892 g
d. moles of oxygen :
= 13/2 x mol butane
= 13/2 x 0.0274
= 0.1781
e. mass of oxygen :
= 0.1781 x 32 g/mol
= 5.6992 g