Answer is: pH of aniline is 9.13.<span>
Chemical reaction: C</span>₆H₅NH₂(aq)+
H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).
pKb(C₆H₅NH₂) = 9.40.
Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.
c₀(C₆H₅NH₂) = 0.45 M.
c(C₆H₅NH₃⁺) = c(OH⁻) = x.
c(C₆H₅NH₂) = 0.45 M - x.
Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).
4·10⁻¹⁰ = x² / (0.45 M - x).
Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.
pOH = -log(0.0000134 M.) = 4.87.
pH = 14 - 4.87 = 9.13.
You will need the equation PV = nRT
P = Pressure in kPa
V = Volume in L
n = moles
R = 8.314 (constant)
T = Temperature in Kelvin
First convert 2.5 atm into kPa:
2.5 X 101.3 = 253.25 kPa
Convert 125 Celsius into Kelvin:
125 + 273 = 398 K
Convert Gallons to Litres:
1.25 X 3.79 = 4.74 L
Plug your values into the equation to solve for n:
(253.25)(4.74) = n(8.314)(398)
n = (253.25)(4.74)/(8.314)(398)
n = 0.362 moles
Now use M = m/n to solve for the mass of O2
M = Molar Mass
M = mass
n= moles
32 = m/(0.362)
m = (32)(0.362)
m = 11.58g
The correct answer is (3)
I-131 and P-32
The explanation:
according to attached table:
- we can see that the half life of p 32 is 14.28d (more than one hour)
- and the half life of I-131 is 8.021 d
(more than one hour)
and They both have β- decay mode and with half-lives greater than hour.
Answer:
Do you need the definition or the equation to find it?
Explanation:
Equation:
Mass=volume x density
