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.
★ « <em><u>what is oxidation number of S in H2SO5??</u></em><em><u> </u></em><em><u>»</u></em><em><u> </u></em><em><u>★</u></em>
- <em><u>it's </u></em><em><u> </u></em><em><u>6</u></em><em><u>!</u></em><em><u>!</u></em>
Explanation:
- <em>Oxidation number of S in H2SO5 is 6 .</em>
The scale of most metal characteristics goes from the bottom left-hand corner.
The least metallic is the top right-hand.
So then that means that
Calcium-YES, second column
Germanium-No, to far, in the middle
Arsenic-Non-metal,
Bromine, same for this
Calcium
Answer:
(CH3)2CHCH2CHO
Explanation:
The reaction sequence begins with the reaction of isopropanol with phosphorus tribromide to yield isopropyl bromide. This is followed by reaction with Magnesium in ether solution giving a grignard reagent, isopropyl magnesium bromide. This is now attacked by oxirane and the epoxide ring opens, hydrolysis of the product, followed by oxidation using pyridinium chlorochromate (PCC) yields the final product- (CH3)2CHCH2CHO
The detailed reaction mechanism is attached to this answer.
Answer is: a. Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Alkaline metals (group 1), in this example rubidium, have lowest ionizations energy and easy remove valence electrons (one electron), they are most reactive metals.
Earth alkaline metals (group 2), in this example strontium, have higher ionization energy than alkaline metals, because they have two valence electrons, they are less reactive.
Rubidium electron configuration: ₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹; one valence electron is 5s¹ orbital.
Strontium electron configuration: ₃₈Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²; two valence electrons is 5s² orbital.