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Helen [10]
1 year ago
13

1. Ethanol and 2,6-dimethylphenol both have hydroxyl groups that can be deprotonated by a base. Why does ethanol act as a solven

t but not as a nucleophile under the conditions used in this experiment
Chemistry
1 answer:
Volgvan1 year ago
3 0

Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive. Since soft nucleophiles are less strongly solvated than hard nucleophiles, these solvents boost the relative reactivity of soft anions.

<h3>Ethanol is either a nucleophile or a base.</h3>

The ethanol is a base  Because carbocation is an extremely reactive species, a base or nucleophile as weak as ethanol can replace or remove it. SN1 and E1 would not be conceivable without the carbocation or a strong departing group.

<h3>How do solvents impact anionic nucleophile's reactivity?</h3>

In polar aprotic solvents, nucleophilic substitution reactions of anionic nucleophiles often proceed more quickly. The normal relative reactivity order in such solvents (like DMSO)is Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive.

Learn more about nucleophiles here:-

brainly.com/question/27127109

#SPJ4

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7 0
3 years ago
All of the alkali earth metals, Group 2, have two valence electrons. Which of these would represent the oxidation number of the
chubhunter [2.5K]
Since Group 2 alkali earth metals have 2 valence electrons, they tend to lose those 2 when forming ionic bonds. And the Loss of Electrons = Oxidation (L.E.O. for short). Therefore this group, including Mg and Ca, have an oxidation of [+2].
So the correct answer is C) +2
5 0
3 years ago
Read 2 more answers
How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me
Georgia [21]
First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.


Second, calculate the amount of heat required for each step.

1) Liquid heating

Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol

=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

Third, add up the heats for every steps:

Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J

Fourth, divide the total heat by the heat rate:

Time = 22,466.3 J / (6000.0 J/min) = 3.7 min

Answer: 3.7 min


3 0
3 years ago
According to Le Chatelier’s principle, increased pressure will _______.
defon

The correct answer is D.

Increasing the pressure shifts the position of equilibrium towards the side with fewer gas molecules.

7 0
3 years ago
A sample of gas at a pressure of 121.59 kPa, a volume of 31 L, and a temperature of 360 K contains how many moles of gas? A. 0.7
natali 33 [55]

Answer:

C. 1.3 mol

Explanation:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

Given:

P = 121.59 kPa

V = 31 L

T = 360 K

R = 8.3145 L kPa / mol / K

Find: n

n = PV / (RT)

n = (121.59 kPa × 31 L) / (8.3145 L kPa / mol / K × 360 K)

n = (3769.29 L kPa) / (2993.22 L kPa / mol)

n = 1.26 mol

Round to two significant figures, there are 1.3 moles of gas.

4 0
3 years ago
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