An A horizon is a mineral horizon, It forms at the surface as topsoil. O horizon turns eventually into topsoil
Answer: pH = 6.77
Explanation:
1) <u>Chemical equilibrium</u>
- 2 H₂O (l) ⇄ H₃O⁺ (aq) + OH⁻ (aq)
2) <u>Equilibrium constant, Kw</u>
- By stoichiometry [H₃O⁺] = [OH⁻]. Call it x
- x = √ (2.92 × 10⁻¹⁴) = 1.709 × 10⁻⁷ M = [H₃O⁺]
3)<u> pH</u>
- pH = - log [H₃O⁺] = - log (1.709 × 10⁻⁷) = 6.77
LiOH, LiNO₂, LiHCO₃, Li₂SO₃, Li₂HPO₄
PH = pKa + log
![\frac{[base]}{[Acid]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Bbase%5D%7D%7B%5BAcid%5D%7D%20)
Acid is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
so pH = 4.74 + log (0.2/0.2) = 4.74
This is called maximum buffer capacity (when acid conc. and base conc. are equal) the pH = pKa in this case
Answer:
The answer to your question is 2.32 atm
Explanation:
Data
P = ?
n = 0.214
V = 2.53 L
T = 61°C
R = 0.082 atm L/mol°K
Formula
PV = nTR
solve for P
P = nRT/V
Process
1.- Calculate the temperature in K
°K = °C + 273
°K = 61 + 273
= 334
2.- Substitution
P = (0.214 x 0.082 x 334) / 2.53
3.- Simplification
P = 5.86/2.53
4.- Result
P = 2.32 atm