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.
It is, 1,986.56.
Hope this helps.
<h3>Further explanation</h3>
1.Atomic Number (Z) = Mass Number (A) - Number of Neutrons
neutrons = mass number-atomic number
Atomic mass Cl-37= 17
Mass number Cl-37=37
Neutrons = 37-17=20
2. Mass atom X = mass isotope 1 . % + mass isotope 2.%...
3. The energy in one photon can be formulated as
f = c / λ, so :
Energy is directly proportional to frequency and inversely proportional to the wavelength
So, as the frequency of photon increases, the energy of photon increases
4. Based on answer number 3 :
A. The wavelength becomes longer, and the energy decreases
1) Chemical reaction: AgNO₃ + HCl → AgCl + HNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(HCl) = 1 : 1.
0,00675 mol : n(HCl) = 1 : 1.
n(HCl) = 0,00675 mol.
V(HCl) = n(HCl) ÷ c(HCl).
V(HCl) = 0,00675 mol ÷ 0,130 mol/L.
V(HCl) = 0,0519 L = 51,92 ml.
2) 1) Chemical reaction: AgNO₃ + KCl → AgCl + KNO₃.
V(AgNO₃) = 30,0 mL = 0,03 L.
c(AgNO₃) = 0,225 mol/L.
n(AgNO₃) = 0,03 L · 0,225 mol/L.
n(AgNO₃) = 0,00675 mol.
From chemical reaction: n(AgNO₃) : n(KCl) = 1 : 1.
0,00675 mol : n(KCl) = 1 : 1.
n(KCl) = 0,00675 mol.
m(KCl) = n(KCl) · M(KCl).
m(KCl) = 0,00675 mol · 74,55 g/mol.
m(KCl) = 0,503 g.
n - amount of substance.
M - molar mass.
