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ExtremeBDS [4]
3 years ago
11

Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank

items as equivalent, overlap them. resethelp hardesteasiest

Chemistry
1 answer:
lana [24]3 years ago
3 0
The rate of Formation of Carbocation mainly depends on two factors'

                    1)  Stability of Carbocation:
                                                              The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.

                     2) Ease of detaching of Leaving Group:
                                                                                   The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

                                            R-I > R-Cl > R-F

                                               B   >  C  >  A

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An alternative definition of electronegativity is Electronegativity= constant (I.E — E.A.) where I.E. is the ionization energy a
Kazeer [188]

Answer:

The electronengativity values of given elements is as follows.

Fluorine - 4

Chlorine -3

Bromine - 2.9

Iodine- 2.5

Explanation:

Electronegativity  =consant (I.E-E.A)

The electron affinity and ionization energy values of the given elements is as follows.

(In attachment)

First we have to find the value of constant by using the fluorine atom to whom the electronengativity taken as "4".

<u>Fluorine:</u>

4=constant[1678-(-327.8)]

Constant=0.0019942168

By using this constant values we can find electronegatvity values of remaining elements.

<u>Chlorine:</u>

Electronegativity=0.0019942168[1255+348.7]=3.1980\sim 3

Therefore, electronegativity of chlorine is 3.

<u>Bromine:</u>

Electronegativity=0.0019942168[1138+324.5]=2.91\sim 2.9

Therefore, electronegativity of bromine is 2.9.

<u>Iodine:</u>

Electronegativity=0.0019942168[1007+295.7]=2.59\sim 2.5

Therefore, electronegativity of iodine is 2.5.

8 0
3 years ago
Which of the following equations demonstrates the law of conservation of mass?
VLD [36.1K]

Answer:

Mg + O2 ---> MgO

Explanation:  V2O5 + CaS → CaO + V2S5. 5V2O5 + 5CaS → 10CaO + 5V2S5. 3V2O5 + 3CaS → 3CaO + 3V2S5.

7 0
3 years ago
Calculate the reaction rate given the following:
klasskru [66]

Answer:

0.0123 M/s

Explanation:

4 0
3 years ago
Consider the following reaction:2NOBr(g) 2NO(g) + Br2(g)If 0.412 moles of NOBr(g), 0.678 moles of NO, and 0.224 moles of Br2 are
Vaselesa [24]

Answer: K_c=0.0587M

Explanation: The given chemical reaction is:

2NOBr(g)\rightleftharpoons 2NO(g)+Br_2(g)

Equilibrium constant (Kc) in general is written as:

K_c=\frac{[products]}{[reactants]}

Note:- Coefficients are written as their powers

So, the Kc expression for the above reaction will be:

K_c=\frac{[NO]^2[Br_2]}{[NOBr]^2}

Equilibrium moles are given for all of them. Let's divide the moles by given liters to get the concentrations.

[NOBr]=\frac{0.412mol}{10.3L}   = 0.040 M

[NO]=\frac{0.678mol}{10.3L}   = 0.0658 M

[Br_2]=\frac{0.224mol}{10.3L}   = 0.0217 M

Plug in the values in the equilibrium expression to calculate Kc.

K_c=\frac{(0.0658M)^2(0.0217M)}{(0.040M)^2}

K_c=0.0587M

4 0
3 years ago
How are electricity and magnetism related?
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Answer:

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5 0
3 years ago
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