Answer:
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
Explanation:
A buffer system is formed in 1 of 2 ways:
- A weak acid and its conjugate base.
- A weak base and its conjugate acid.
Determine whether mixing each pair of the following results in a buffer.
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
YES. NH₃ is a weak base and NH₄⁺ (from NH₄Cl ) is its conjugate base.
b. 50.0 mL of 0.10 M HCl with 35.0 mL of 0.150 M NaOH.
NO. HCl is a strong acid and NaOH is a strong base.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
YES. HF is a weak acid and it reacts with NaOH to form NaF, which contains F⁻ (its conjugate base).
d. 175.0 mL of 0.10 M NH₃ with 150.0 mL of 0.12 M NaOH.
NO. Both are bases.
- The mass percent of
Pentane in solution is 16.49%
- The mass percent of
Hexane in solution is 83.51%
<u>Explanation</u>:
- Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.
-
Convert these values to mol% using their molecular weights:
Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol
Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol
Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%
Hexane mol%: yh = 100 - 39.68 = 60.32%
Pp-vap = 425 torr = 0.555atm
Ph-vap = 151 torr = 0.199atm
-
From Raoult's law we know:
Pp = xp
Pp - vap = yp
Pt (1)
Ph = xh
Ph - vap = yh
Pt (2)
-
Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:
(1 - xp)
Ph - vap = (1 - yp)
Pt (3)
-
Substituting (1) into (3) we get:
(1-xp)
Ph - vap = (1 - yp)
xp
Pp - vap / yp (4)
xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap) (5)
-
Subbing in the values we find:
Pentane mol% in solution: xp = 19.08%
Hexane mol% in solution: xh = 80.92%
-
Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:
mp = 0.1908 mol
72.15 g/mol
= 13.766 g
mh = 0.8092 mol
86.18 g/mol
= 69.737 g
-
Mass% of Pentane solution = 13.766/(13.766+69.737)
= 16.49%
-
Mass% of Hexane solution = 83.51%
Gravity forces earth to move around the sun so A
Answer:
(a) Rate at which
is formed is 0.050 M/s
(b) Rate at which
is consumed is 0.0250 M/s.
Explanation:
The given reaction is:-

The expression for rate can be written as:-
![-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
Given that:-
(Negative sign shows consumption)
![-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![-(-0.050\ M/s)=\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=-%28-0.050%5C%20M%2Fs%29%3D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\frac{d[NO_2]}{dt}=0.050\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D0.050%5C%20M%2Fs)
(a) Rate at which
is formed is 0.050 M/s
![-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-0.050%5C%20M%2Fs%3D-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
![\frac{d[O_2]}{dt}=0.0250\ M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D0.0250%5C%20M%2Fs)
(b) Rate at which
is consumed is 0.0250 M/s.
Answer:
The specific heat of aluminium is 0.8792 J/g °C or 0.21 Cal/g °C
Explanation:
Step 1 : Write formule of specific heat
Q=mcΔT
with Q = heat transfer (J)
with m = mass of the substance
with c = specific heat ⇒ depends on material and phase ( J/g °C)
with ΔT = Change in temperature
For this case :
Q = 1680 Calories = 7033.824 J ( 1 calorie = 4.1868 J)
m = 100.0g
c= has to be determined
ΔT = 100 - 20 = 80°C
<u>Step 2: Calculating specific heat</u>
⇒ via the formule Q=mcΔT
7033.824 J = 100g * c * 80
7033.824 = 8000 *c
c = 7033.824 /8000
c = 0,879228 J/g °C
or 0.21 Cal / g°C
The specific heat of aluminium is 0.8792 J/g °C or 0.21 Cal/g °C