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
The answer is .0004704. i hope that helped.
Answer:
25,050 calories.
Explanation:
A calorie is the amount of energy needed to raise the temperature of one gram of water 1 degree centigrade. If we are raising 835 grams of water 30 degrees then we multiply 835*30 to get 25,050 calories.
Density is weight by volume.
First. If you divide the weight by density you can find the volume
Second you must convert the ML in to Liters.
1L=1000ml
0.38214 Liters.
Answer: the answer is d
Explanation: hurry and answer
