The molecule is stable and can exist even though the number of valence electrons around central atom in the molecule are less than 8.
<h3>Is BF3 molecule stable or not?</h3>
BF3 molecule is a stable molecule because all the electrons present in the outermost shell of boron are covalently bonded with fluorine. Boron in BF3, three bonds is the maximum possible because boron only has 3 electrons to share.
So we can conclude that the molecule is stable and can exist even though the number of valence electrons around central atom in the molecule are less than 8.
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Answer:
Velocity
Explanation:
The change of velocity rate it's the definition of acceleration.
While increasing or decreasing velocity, it means that an acceleration it's happening.
Using some math:

This can be read as "the acceleration it's equal to te variation of velocity with respect to time"
Answer:
Few important points related to
reaction:
1.
is a one-step reaction that follows second order kinetics.
2. In
reaction, a transition state is formed in situ.
3. Strong nucleophiles like
are used in case of bi-molecular nucleophilic substitution reaction.
Ethyl acetate can be prepared by a second-order nucleophilic substitution reaction between acetic acid and ethyl bromide.
The reaction between acetic acid and ethyl bromide is drawn below:
Answer:
Here's what I get
Explanation:
1. Write the chemical equation
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻; Kₐ = 2 × 10⁻⁵
Let's rewrite the equation as
A⁻ + H₂O ⇌ HA + OH⁻
2. Calculate Kb

3. Set up an ICE table
A⁻ + H₂O ⇌ HA + OH⁻
I/mol·L⁻¹: 0.35 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.35-x x x
4. Solve for x
![\dfrac{\text{[HA ][OH$^{-}$]}}{\text{[A$^{-}$]}} = \dfrac{x^{2}}{0.35-x} = 5 \times 10^{-10}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7B%5BHA%20%5D%5BOH%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%20%3D%20%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.35-x%7D%20%3D%205%20%5Ctimes%2010%5E%7B-10%7D)
Check for negligibility,
![\dfrac{\text{[HA]}}{K_{\text{b}}} = \dfrac{0.35}{5 \times 10^{-10}} = 7 \times 10^{8}> 400\\\\\therefore x \ll 0.35\\\\\dfrac{x^{2}}{0.35} = 5 \times 10^{-10}\\\\x^{2} = 0.35 \times 5 \times 10^{-10} = 1.8\times 10^{-10}\\\\x = \sqrt{1.8\times 10^{-10}} = \mathbf{1 \times 10^{-5}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7B%5BHA%5D%7D%7D%7BK_%7B%5Ctext%7Bb%7D%7D%7D%20%3D%20%5Cdfrac%7B0.35%7D%7B5%20%5Ctimes%2010%5E%7B-10%7D%7D%20%3D%207%20%5Ctimes%2010%5E%7B8%7D%3E%20400%5C%5C%5C%5C%5Ctherefore%20x%20%5Cll%200.35%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.35%7D%20%3D%205%20%5Ctimes%2010%5E%7B-10%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.35%20%5Ctimes%205%20%5Ctimes%2010%5E%7B-10%7D%20%3D%201.8%5Ctimes%2010%5E%7B-10%7D%5C%5C%5C%5Cx%20%3D%20%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-10%7D%7D%20%3D%20%5Cmathbf%7B1%20%5Ctimes%2010%5E%7B-5%7D%7D)
5. Calculate the pOH
[OH⁻] = 1 × 10⁻⁵ mol·L⁻¹
pOH = -log[OH⁻] = -log(1 × 10⁻⁵) = 4.88
6. Calculate the pH.
pH + pOH = 14.00
pH + 4.88 = 14.00
pH = 9.12
Note: The answer differs from that given by Silberberg because you used only one significant figure for the Kₐ of acetic acid.