Answer:
The order will be:
CCH > CHCH₂ > CH₂CH₃> CH₃
Explanation:
According to Cahn-Ingold-Prelog system we rank the groups based on the atomic number of directly attached atom with the chiral carbon.
For example: between C and H, we rank Carbon first.
If the same atoms are attached for different groups then we prioritized based on the second element with highest atomic number.
For example:
Among CH₃ and C₂H₅, the priority will be given to C₂H₅.
If an atom is double or triple bonded to the directly attached atom then each pi bond is considered to be a new atom.
Hence CH=CH₂ means, that there are two carbons attached to CH carbon.
So the order based on above selection rules will be:
CCH > CHCH₂ > CH₂CH₃> CH₃
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
Because potassium is more reactive than hydrogen
Hello!
Find the Energy of the Photon by Planck's Equation, given:
E (photon energy) =? (in Joule)
h (Planck's constant) = 
f (radiation frequency) = 
Therefore, we have:
I Hope this helps, greetings ... DexteR! =)
