Explanation:
The equation for the average is: (sum of all data values/ (total number of data values)
Answer:
A. (CH3)3C-I reacts by SN1 mechanism whose rate is independent of nucleophile reactivity.
Explanation:
We must recall that (CH3)3C-I is a tertiary alkyl halide. Tertiary alkyl halides preferentially undergo substitution reaction via SN1 mechanism.
In SN1 mechanism, the rate of reaction depends solely on the concentration of the alkyl halide (unimolecular mechanism) and is independent of the concentration of the nucleophile. As a result of this, both Br^- and Cl^- react at the same rate.
The answer is: the distance between two nuclei is 2.35×10⁻¹⁰ m.
r(Na⁺) = 1.16×10⁻¹⁰ m; radius of sodium cation.
r(F⁻) = 1.9×10⁻¹⁰ m; radius of fluoride anion.
d(NaF) = r(Na⁺) + r(F⁻).
d(NaF) = 1.16×10⁻¹⁰ m + 1.9×10⁻¹⁰ m.
d(NaF) = 2.35×10⁻¹⁰ m; distance between two nuclei.
The sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.
Hello!
I believe the correct answer to this question is H+ and H2O.
I hope you found this helpful! :)
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
[OH-] = 10^-pOH = 10^-3 = x
Considering the sodium salt NaA in water, we have the equation
NaA → Na+ + A-
hence, [A-] = 0.0100 M
Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
A- + H2O ⇌ HA + OH-
Initial 0.0100 0 0
Change -x +x +x
Equilibrium 0.0100-x x x
We can now calculate the Kb for A-:
Kb = [HA][OH-] / [A-]
= x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4
We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
Ka = Kw / Kb
= 1.00x10^-14 / 1.00x10^-4
= 1.00x10^-10
