Answer:
Explanation:
To calculate pH you need to use Henderson-Hasselbalch formula:
pH = pka + log₁₀ ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Where HA is the acid concentration and A⁻ is the conjugate base concentration.
The equilibrium of acetic acid is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75
Where <em>CH₃COOH </em>is the acid and <em>CH₃COO⁻ </em>is the conjugate base.
Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:
pH = 4,75 + log₁₀ ![\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
a) The pH is:
pH = 4,75 + log₁₀ ![\frac{[2 mol]}{[2 mol]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2%20mol%5D%7D%7B%5B2%20mol%5D%7D)
<em>pH = 4,75</em>
<em></em>
b) The pH is:
pH = 4,75 + log₁₀ ![\frac{[2 mol]}{[1mol]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2%20mol%5D%7D%7B%5B1mol%5D%7D)
<em>pH = 5,05</em>
<em></em>
I hope it helps!
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent Bond
Between 0.4 and 1.7 then it is Polar Covalent Bond
Greater than 1.7 then it is Ionic
For Br₂;
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar Covalent Bond)
For MgS;
E.N of Sulfur = 2.58
E.N of Magnesium = 1.31
________
E.N Difference 1.27 (Ionic Bond)
For SO₂;
E.N of Oxygen = 3.44
E.N of Sulfur = 2.58
________
E.N Difference 0.86 (Polar Covalent Bond)
For KF;
E.N of Fluorine = 3.98
E.N of Potassium = 0.82
________
E.N Difference 3.16 (Ionic Bond)
Result: The Bonds in Br₂ and SO₂ are Covalent in Nature.
What you have to do is find a periodic table and add the mass of each atom that the compound is made of.
Ca= 40.1
O= 16.0
H= 1.01
keep in mind that you have to also account for how many atoms of each there are in the molecule. for example, in Ca(OH)2, there are one Ca, two O and two H
so the molar mass of Ca(OH)2= 40.1 + (2 x 16.0) + (2 x 1.01)= 74.12 g/mol
Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
- CH3Cl + H2O → CH3OH + HCl
at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
- K(T) = A e∧(-Ea/RT)
- Ln K = Ln(A) - [(Ea/R)(1/T)]
∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
Answer:
The print is so little i cant read it. :( Sorry!
Explanation:
