Answer: 4 N-2 N=2 Nto the left. Change in motion: The box will move to the left. 7.)
Explanation: Please mark branliest.
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This
electronic transition would result in the emission of a photon with the highest
energy:
4p
– 2s
<span>This
can be the same with the emission of 4f to 2s which would emit energy in the
visible region. The energy in the visible region would emit more energy than in
the infrared region which makes this emission to have the highest energy.</span>
Answer:
Carbon and Oxygen, Argon and Helium.
Explanation:
noble gases have full outer shells of electrons, and so cannot share other atoms' electrons to form bonds. sodium and chlorine form an ionic bond.
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
