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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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What is the H+ concentration when pH is 2, 6, 7, 8, and 12?
Mariulka [41]

Answer:

See below.

Explanation:

pH = - log (H+)

So the pH of 1 x 10^-7 M solution is 7.

I'm sorry but I'm not sure about what the other units mean, so Im not sure of the answer to those.

If you convert the other units to the form  1 x 10^-n  then the pH will be n.

6 0
3 years ago
Read 2 more answers
Help!!!
aleksley [76]

Answer:

Q1) Distillation relies on evaporation to purify water

Explanation:

contaminated water is heated to form steam. inorganic compounds and large non-volatile organic molecules do not evaporate with the water and are left behind. The steam then cools and condenses to form purified water

3 0
3 years ago
For each of the following pairs of complexes, identify which one you would predict to have the larger Δo value, and explain why.
mash [69]

Answer:

a) [Fe(H2O)6]3+

b) [Fe(CN)6]3−

c) [Ru(CN)6]3-

Explanation:

. [Mn(H2O)6]2+ or [Fe(H2O)6]3+

The both complexes are d5 complexes with the same ligand , water. Water is a weak ligand and note that Mn^2+ often have a crystal field stabilization energy of zero hence

[Fe(H2O)6]3+ will possess a greater ∆o value.

The splitting of d orbitals according to the crystal field theory depends on the;

i)geometry of the complex

ii) nature of the metal ion,

iii)charge on the metal ion,

iv) ligands that surround the metal ion.

When the geometry and the ligands are held constant, the order of crystal field splitting is as follows;

Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+

[Fe(H2O)6]3+ or [Fe(CN)6]3−

[Fe(CN)6]3− will have a greater ∆o because the cyanide ion is a strong field ligand compared to water. A strong field ligand causes a greater splitting of the octahedral crystal field compared to a weak field ligand.

. [Fe(CN)6]3− or [Ru(CN)6]3-

[Ru(CN)6]3- will exhibit a greater crystal field splitting. Crystal field splitting increases with the second and third row transition elements when compared to the crystal field splitting of the first row transition elements. Note that, there is an increase of approximately 30%–50% in Δo on going from a first-row transition metal to a second-row metal and another 30%–50% increase on going from a second-row to a third-row metal when they have the same geometry and oxidation state.

4 0
3 years ago
Write a balanced net ionic equation to show why the solubility of Cu(OH)2(s) increases in the presence of a strong acid and calc
Helga [31]

Answer:

2.2×10^8

Explanation:

Cu(OH)2(s)<---------> Cu^2+(aq) + 2OH^-(aq) Ksp=2.2 x 10 ^-20

2H3O^+(aq) + 2OH^-(aq) <-------> 4H2O(l). Kw= 1×10^14

Cu^2+(aq) + 4H2O(l) <--------> [Cu(H2O)4]^2+(aq)

Overall ionic reaction:

Cu(OH)2(s) +2H3O^+(aq) <---------> [Cu(H20)4]^2+(aq)

Equilibrium constant for the reaction: Ksp×Kw= 2.2 x 10 ^-20 × (1/(1×10^-14))^2

Keq= 2.2×10^8

Kw= ion dissociation constant of water

3 0
2 years ago
A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,
Ad libitum [116K]
The answer is 3.68 L
8 0
2 years ago
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