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Sladkaya [172]
3 years ago
9

For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugati

on, withdraws electrons by resonance, or donates electrons by resonance. (Effects should be compared with that of a hydrogen; remember that many substituents can be characterized in more than one way.)
a. Br
b. CH2CH3
c. NHCH3
d. OCH3
Chemistry
1 answer:
AveGali [126]3 years ago
3 0

Answer:

a. withdraws electrons inductively

b. donates electrons by hyperconjugation

c. donates electrons by resonance

d.  withdraws electrons inductively

Explanation:

a.  The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:

F > Cl > Br>I

b.  The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons.  Because of the delocalization, hyperconjugation is the result effect.

c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.

d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.

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4 0
2 years ago
Which is evidence that the reaction below is a redox reaction? 2Na + Cl2 → 2NaCl
Mamont248 [21]
A redox reaction --> a reaction whereby oxidation & reduction occurs
Reduction: 
Charge of Cl2 = 0
Charge of Cl- in NaCl = -1
Hence, since charge of Cl2 decreased from 0 in Cl2 to -1 in NaCl, reduction occured. 
Oxidation:
Charge of Na = 0
Charge of Na+ in NaCl = +1
Hence, since charge of Na increased from 0 in Na to +1 in NaCl, oxidation occured.
Since both oxidation & reduction occured in the reaction, it is a redox reaction.  
3 0
3 years ago
The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?
Agata [3.3K]

<u>Answer:</u> The molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

PbI_2\rightleftharpoons Pb^{2+}+2I^-

                s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Pb^{2+}][I^-]^2

We are given:

K_{sp}=7.9\times 10^{-9}

Putting values in above equation, we get:

7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L

Hence, the molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

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3 years ago
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balu736 [363]

Answer:

Explanation:

Option B is the correct answer

8 0
2 years ago
Read 2 more answers
Attempt 2
Assoli18 [71]
Lets let our mass equal 3 on alletals and solve using d=m/v equation

Aluminum
V=3/2.70=1.11
Silver
V=3/10.5=.286
Rhenium
V=3/20.8=.144
Nickel
V=3/8.90=.337
This gives us the following list from largest to smallest Aluminum, Nickel, Silver, and Rhenium
4 0
3 years ago
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