S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
<u>Explanation:</u>
S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
To measure the rate of this reaction we must measure the rate of concentration change of one of the reactants or products. To do this, we will include (to the reacting S₂O₈
²⁻ and I⁻
i) a small amount of sodium thiosulfate, Na₂S₂O₃,
ii) some starch indicator.
The added Na₂S₂O₃ does not interfere with the rate of above reaction, but it does consume the I₂ as soon as it is formed.
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
This reaction is much faster than the previous, so the conversion of I2 back to I⁻ is essentially instantaneous.
![rate = \frac{dI2}{dt} = \frac{1/2 [S2O3^2^-]}{t}](https://tex.z-dn.net/?f=rate%20%3D%20%5Cfrac%7BdI2%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%2F2%20%5BS2O3%5E2%5E-%5D%7D%7Bt%7D)
Answer:
Element A = Oxygen
Element H =
Element B = Aluminum
Element J = Magnesium
Element C = Selenium
Element L = Carbon
Element D = Sodium
Element Q = Francium
Element F = Antimony
Element R = Calcium
Element G = Chlorine
Element S = Tellurium
Explanation:
Element A is Oxygen because: oxygen 6 valence electrons
; is a gas at room temperature
; and is transported in blood to cells.
Element H is Neon because: Neon is a noble gas
; qppears as red light when charged with electricity (Neon light signs) and it has the second highest Ionization energy of the elements
Element B is Aluminum because: Aluminum is a metal and its ion has charge of +3. It is also located on the borders of the Metalloid staircase
.
Element J is Magnesium because its ion has charge of 2+ and is isoelectronic with Neon because it loses two electrons to now have 10 electrons.
Element C is Selenium because its ion that has a charge of -2 is formed by gaining two electrons in order to have 36 electrons which is isoelectronic with Kr
ypton
Element L is Carbon because carbon has the smallest atomic radius of any member in the Carbon family because it is the first member of the family and atomic radius increases on going down the group.
Element D is Sodium because its ion has charge of +1 and it has 2 inner core levels
, the 1 and 2 energy levels.
Element Q is Francium because it has the largest radius and lowest ionization energy of any element
Element F is Antimony. It is a member of Nitrogen family and has the second highest ionization energy level in family
.
Element R is calcium because its on has charge of +2 which is isoelectronic with Argon
. Calcium also has atomic radius is larger than Ar
gon.
Element G is Chlorine. It has the second to the smallest radius of elements in the 3rd period as the second to the last element in the period because atomic radius decreases across a period from left to right.
Element S is Tellurium. It has atomic mass larger than Iodine just to the right of it and is found in the 5th period
A i believe is the answer
Answer: The given pairs of electrons most likely reside in 
type of orbital.
Explanation:
As it is given that two lone pair of electrons are present on the oxygen atom of ketone (such as cyclohexanone).
Also, there will be one bond pair between carbon and oxygen atom.
Hence, total electrons present in the domain are as follows.
2 lone pairs + 1 bon pair of electron = 3 electron domains
This means that there will be type of orbital present.
Thus, we can conclude that given pairs of electrons most likely reside in type of orbital.
