<h2>Question:</h2>
- A solution that is able to dissolve additional solute is best described as _____.
<h2>Answer:</h2>
<h2>D.) Unsaturated </h2>
- <u>Unsaturated</u> means the substance is poured into the solvent that can be dissolved. It is a solution (with less solute than the saturated solution) that completely dissolves, leaving with no remaining substances.
<h2>For example:</h2>
- If you were to pour olive oil into a glass of water, it will dissolve, so it is <u>unsaturated</u>.
_________
#LetsStudy
A) according to this reaction:
by using ICE table:
NH2OH(aq) + H2O(l) → HONH3+(aq) + OH-
initial 0.4 M 0 0
change -X +X +X
Equ (0.4-X) X X
when Kb = [OH-][HONH3+]/[NH2OH]
when we have Kb = 1.1x10^-8 so,
by substitution:
1.1x10^-8 = X^2/(0.4-X) by solving this equation for X
∴X = 6.6x10^-5 M
∴[OH] = 6.6x10^-5 M
when POH = - ㏒[OH]
∴POH = -㏒(6.6x10^-5)= 4.18
∴PH = 14 - POH = 14 - 4.18
= 9.82
when PH = -㏒[H+]
∴[H+] = 10^9.82 = 1.5x10^-10 M+0.02molHcl
= 0.02
∴ the new value of PH = -㏒(0.02)
∴PH = 1.7
B) according to this reaction:
by using ICE table:
HONH3+(aq) → H+(aq) + HONH2(aq)
intial 0.4 0 0
change -X +X +X
Equ (0.4-X) X X
when Ka HONH3Cl = 9.09x10^-7
and Ka = [H+][HONH2] / [HONH3+]
So by substitution and we can assume [HONH3+] = 0.4 as the value of Ka is so small so,
9.09x10^-7 = X^2 / 0.4 by solving for X
∴ X = 6 x 10 ^-4
∴[H+] = 6x10^-4
PH = -㏒[H+]
= -㏒ (6x10^-4) = 3.22
when [H+] = 6x10^-4 + 0.02 m HCl
∴new value of PH = -㏒(6x10^-4+0.02)
= 1.69
C) when we have pure H2O and PH of water = 7
So we can get [H+] when PH = -㏒[H+]
∴[H+] = 10^-7 + 0.02MHCl
= 0.02
∴new value of PH = -㏒0.02
PH = 1.7
d) when HONH2 & HONH3Cl have the same concentration and Hcl added to them so we can assume that PH=Pka
and when we have Ka for HONH3Cl = 9.09x10^-7
So we can get the Pka:
Pka = -㏒Ka
= -㏒9.09x10^-7
= 6.04
∴PH = 6.04
and because of the concentration of the buffer components, HONH2 & HONH3Cl have 0.4 M and the adding of HCl = 0.02 M So PH will remain very near to 6
It would use rapid combustion
Answer: 322.56 Kelvin
Explanation:
Use the Ideal Gas Law
R is the gas constant
T is the temperature in Kelvins
P is the pressure in atmospheres
V is the volume in liters
n is the number of moles of gas
First, the mm of mercury need to be converted to atmospheres using the conversion factor 1atm = 760 torr.
Now plug everything in
An allylic carbocation is a reactive intermediate in the reaction of 1,3-diene with her, resulting in 1,4-addition.
<h3>What is
carbocation?</h3>
- A molecule called a carbocation has three bonds and a positively charged carbon atom.
- They are essentially carbon cations, to put it simply.
- It was once referred to as carbonium ion.
- Any even-electron cation with a sizable positive charge on the carbon atom is now referred to as a carbocation.
<h3>Why are carbohydrate molecules crucial?</h3>
- Because charge can be exchanged between many atoms when the vacant p orbital of a carbocation overlaps with the p orbitals of another carbon-carbon double or triple bond, carbocations next to other carbon-carbon double or triple bonds are very stable.
Learn more about carbocation here:
brainly.com/question/13164680
#SPJ4
