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3 years ago

What happens in a Lewis acid-based reaction

Chemistry

2 answers:

AleksAgata [21]3 years ago

7 0

Answer:

When an acid accepts an electron pair from a base

Explanation:

Just did a test with this exact same question :)

Mkey [24]3 years ago

6 0

A Lewis acid-base reaction occurs when a base donates a pair of electrons to an acid. ... A Brønsted-Lowry acid such as HCl is an acid-base adduct according to the Lewis concept, and proton transfer occurs because a more stable acid-base adduct is formed.

Example: Water (H2O) has two nonbonding pairs of electrons, so it donates one pair to the carbocation. As a result, one bond is formed and shared between carbon and oxygen. Water acts as a Lewis base because it is the electron pair donor in this reaction.

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The equilibrium constant, Kc, for the following reaction is 56.0 at 278 K. 2CH2Cl2(g) CH4(g) + CCl4(g) When a sufficiently large

BabaBlast [244]

Answer:

21.5 M

Explanation:

1) Equilibrium equation (given):

2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)

2) Write the concentration changes when some concentration, A, of CH₂Cl₂ (g) sample is introduced into an evacuated (empty) vessel:

2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)

A - x x x

3) Replace x with the known (found) equilibrium concentraion of CCl₄ (g) of 0.348 M

2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)

A - 0.3485 0.348 0.348

4) Write the equilibrium constant equation, replace the known values and solve for the unknown (A):

Kc = [ CH₄ (g) ] [ CCl₄ (g) ] / [ CH₂Cl₂ (g) ]²

56.0 = 0.348² / A²

A² = 56.0 / 0.348² = 462.

A = 21.5 M ← answer

7 0

3 years ago

At 25 °c, what is the hydroxide ion concentration, [oh–], in an aqueous solution with a hydrogen ion concentration of [h ] = 3.0

chubhunter [2.5K]

As,

Water has a pkw=14

so it can be represented as,

[H+] [OH-] = 1*10^-14

If [H+] = 3*10^-5M

[OH-] = (1*10^-14) / ( 3*10^-5)

[OH-] = 3.3*10^-9 M

5 0

3 years ago

Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota

Ghella [55]

Answer:

(a) Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = -2.881 J/K; total change in entropy = 0

(b)Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = 0 ; total change in entropy = 2.881 J/K

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = $V_{1}$