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:b
Explanation: it’s in the picture just go counterclockwise
Answer:
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Explanation:
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Answer:
1.51 X 10^23 ions
Explanation:
The number of ions in 17.1 gm of aluminum sulphate Al2 (SO4)3 =….. [Molar mass of Al2 (SO4)3 = 342 gm]
in one molecule of Al2(SO4)3 there are 5 ions 2 aluminum and 3 sulfate ions
in 2 molecules there are 2X5= 10 ions
in 10 molecules there are 10X5 = 50 ions
molar mass of Al2(SO4)3 = (2 X 26.98) +( 3 X 32.1) + (3 X 4 X 16.0 ) =342.gms = 17.1/342 =0.0500 moles
1 mole =6.02 X 10^23 molecules ( see Avogadros number)
0.0500 moles = 0.0500 X 6.02 X 10^23 molecules =
0.301 X 10^23 molecules = 3.01 X 10^22 molecules
We determined that each molecule of Al2(SO4)3 has 5 ions
so 3.01 X10^22 molecules have 5 X 3.01 X 10^22 ions =
15.05 X 10^22 ions = 1.51 X 10^23 ions
