Answer:this case, the mass is 2.0g, the specific heat capacity of water is 4.18J/g/K, and the change in temperature is 5.0°C=5K , therefore the energy needed to raise it is: 5×2×4.18=41.8J
Explanation:
To determine the absolute pressure of this gas, all you need to do is to add the value of atmospheric pressure and the value of gage pressure.
Atmospheric pressure is equivalent to 100 kPa.
Gage pressure is 276 kPa.
Then, we add both values.
N = 100 kPa + 276 kPa
N = 376 kPa
The absolute pressure of this gas is 376 kPa.
Hope this helps :)
Explanation:
Given the mass of HCl is ---- 0.50 g
The volume of solution is --- 4.0 L
To determine the pH of the resulting solution, follow the below-shown procedure:
1. Calculate the number of moles of HCl given by using the formula:
2. Calculate the molarity of HCl.
3. Calculate pH of the solution using the formula:
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Since HCl is a strong acid, it undergoes complete ionization when dissolved in water.
Thus, ![[HCl]=[H^+]](https://tex.z-dn.net/?f=%5BHCl%5D%3D%5BH%5E%2B%5D)
Calculation:
1. Number of moles of HCl given:
2. Concentration of HCl:
3. pH of the solution:
![pH=-log[H^+]\\=-log(0.003425)\\=2.47](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D%5C%5C%3D-log%280.003425%29%5C%5C%3D2.47)
Hence, pH of the given solution is 2.47.
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:
i) 0,7 molH20/s
ii)11,2 g O/s
iii)1,4 g H/s
Explanation:
i) To find the molar flow rate of water, we just convert the mass of water to moles of water using its molecular weight(g/mol) and changing to the proper units (lb to grames and hours to seconds):
ii) Now we just consider the oxygen in the water stream (for 1 mole of water there is 1 mole of oxygen):
iii)Just considering the hydrogen in the stream (for 1 mole of water there is 2 moles of hydrogen):
