Answer:
Both reactions share a common intermediate and differ only in the leaving group
Explanation:
The elimination reaction of tertiary alkyl halides usually occur by E1 mechanism. In E1 mechanism, the substrate undergoes ionization leading to the loss of a leaving group and formation of a carbocation.
Loss of a proton from the carbocation completes the reaction mechanism yielding the desired alkene.
In the cases of t-butanol and t-butyl bromide, the mechanism is the same. The both reactions proceed by E1 mechanism. The leaving groups in each case are water and chloride ion respectively.
Answer:
The easiest way is to titrate a sample with a solution of a base.
Explanation:
Answer: 20L of H2O
Explanation:
C3H8 + 5O2 → 3CO2 + 4H2O
Recall 1mole of a gas contains 22.4L at stp
5moles of O2 contains = 5 x 22.4 = 112L
4moles of H2O contains = 4 x 22.4 = 89.6L
From the equation,
112L of O2 produced 89.6L H2O
There for 25L of O2 will produce XL of H2O i.e
XL of H2O = (25 x 89.6)/112 = 20L
Answer:
k= 1.925×10^-4 s^-1
1.2 ×10^20 atoms/s
Explanation:
From the information provided;
t1/2=Half life= 1.00 hour or 3600 seconds
Then;
t1/2= 0.693/k
Where k= rate constant
k= 0.693/t1/2 = 0.693/3600
k= 1.925×10^-4 s^-1
Since 1 mole of the nuclide contains 6.02×10^23 atoms
Rate of decay= rate constant × number of atoms
Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms
Rate of decay= 1.2 ×10^20 atoms/s
