Answer:
Oxygen is a colorless, odorless, tasteless gas. It changes from a gas to a liquid at a temperature of -182.96°C (-297.33°F). The liquid formed has a slightly bluish color to it. Liquid oxygen can then be solidified or frozen at a temperature of -218.4°C (-361.2°F).
a) before addition of any KOH :
when we use the Ka equation & Ka = 4 x 10^-8 :
Ka = [H+]^2 / [ HCIO]
by substitution:
4 x 10^-8 = [H+]^2 / 0.21
[H+]^2 = (4 x 10^-8) * 0.21
= 8.4 x 10^-9
[H+] = √(8.4 x 10^-9)
= 9.2 x 10^-5 M
when PH = -㏒[H+]
PH = -㏒(9.2 x 10^-5)
= 4
b)After addition of 25 mL of KOH: this produces a buffer solution
So, we will use Henderson-Hasselbalch equation to get PH:
PH = Pka +㏒[Salt]/[acid]
first, we have to get moles of HCIO= molarity * volume
=0.21M * 0.05L
= 0.0105 moles
then, moles of KOH = molarity * volume
= 0.21 * 0.025
=0.00525 moles
∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525
and when the total volume is = 0.05 L + 0.025 L = 0.075 L
So the molarity of HCIO = moles HCIO remaining / total volume
= 0.00525 / 0.075
=0.07 M
and molarity of KCIO = moles KCIO / total volume
= 0.00525 / 0.075
= 0.07 M
and when Ka = 4 x 10^-8
∴Pka =-㏒Ka
= -㏒(4 x 10^-8)
= 7.4
by substitution in H-H equation:
PH = 7.4 + ㏒(0.07/0.07)
∴PH = 7.4
c) after addition of 35 mL of KOH:
we will use the H-H equation again as we have a buffer solution:
PH = Pka + ㏒[salt/acid]
first, we have to get moles HCIO = molarity * volume
= 0.21 M * 0.05L
= 0.0105 moles
then moles KOH = molarity * volume
= 0.22 M* 0.035 L
=0.0077 moles
∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5
when the total volume = 0.05L + 0.035L = 0.085 L
∴ the molarity of HCIO = moles HCIO remaining / total volume
= 8 x 10^-5 / 0.085
= 9.4 x 10^-4 M
and the molarity of KCIO = moles KCIO / total volume
= 0.0077M / 0.085L
= 0.09 M
by substitution:
PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)
∴PH = 8.38
D)After addition of 50 mL:
from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.
the molarity of KCIO = moles KCIO / total volume
= 0.0105mol / 0.1 L
= 0.105 M
when Ka = KW / Kb
∴Kb = 1 x 10^-14 / 4 x 10^-8
= 2.5 x 10^-7
by using Kb expression:
Kb = [CIO-] [OH-] / [KCIO]
when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]
Kb = [OH-]^2 / [KCIO]
2.5 x 10^-7 = [OH-]^2 /0.105
∴[OH-] = 0.00016 M
POH = -㏒[OH-]
∴POH = -㏒0.00016
= 3.8
∴PH = 14- POH
=14 - 3.8
PH = 10.2
e) after addition 60 mL of KOH:
when KOH neutralized all the HCIO so, to get the molarity of KOH solution
M1*V1= M2*V2
when M1 is the molarity of KOH solution
V1 is the total volume = 0.05 + 0.06 = 0.11 L
M2 = 0.21 M
V2 is the excess volume added of KOH = 0.01L
so by substitution:
M1 * 0.11L = 0.21*0.01L
∴M1 =0.02 M
∴[KOH] = [OH-] = 0.02 M
∴POH = -㏒[OH-]
= -㏒0.02
= 1.7
∴PH = 14- POH
= 14- 1.7
= 12.3
Answer:
1) Maximun ammount of nitrogen gas: 
2) Limiting reagent: 
3) Ammount of excess reagent: 
Explanation:
<u>The reaction </u>

Moles of nitrogen monoxide
Molecular weight: 


Moles of hydrogen
Molecular weight: 


Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess
1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted


2) <u>Limiting reagent</u>:
3) <u>Ammount of excess reagent</u>:


The question is incomplete, here is the complete question:
A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.
<u>Answer:</u> The amount of liquid bromine produced is 4.79 moles.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:

We are given:
Given mass of liquid bromine = 766. g
Molar mass of liquid bromine,
= 159.8 g/mol
Putting values in above equation, we get:

Hence, the amount of liquid bromine produced is 4.79 moles.
The powder sugar because has more contact area