Answer:
Equilibrium constant Kc for the reaction will be 1.722
Explanation:
O2(g)+NO(g)→CO(g)+ NO2(g)
0.88 3.9 --- ---
0.88x 3.9-x x x
GIVEN:
0.88X-X= 0.11
⇒ X=0.77
CO2(g)+NO(g) → CO(g) + NO2(g)
0.88 3.9 --- ---
0.88-x 3.9-x x x
= 3.13 0.77 0.77
=0.11
Kc = ![\frac{[CO] *[NO2]} {[CO2]*[NO]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO%5D%20%2A%5BNO2%5D%7D%20%7B%5BCO2%5D%2A%5BNO%5D%7D%20)
=
= 1.722
Answer:
Kₐ = 6.7 x 10⁻⁴
Explanation:
First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:
HF + H₂O ⇄ H₃O⁺ + F⁻
Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]
Since we are given the pH we can calculate the [ H₃O⁺ ] ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1 relation , we will also have [F⁻ ]. The [ HF ] is given in the question so we have all the information that is needed to compute Kₐ.
pH = -log [ H₃O⁺ ]
1.68 = - log [ H₃O⁺ ]
Taking antilog to both sides of this equation:
10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻² M= [ H₃O⁺ ]
[ F⁻ ] = 2.1 X 10⁻² M
Solving for Kₐ :
Kₐ = ( 2.1 X 10⁻² ) x ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴
(Rounded to two significant figures, the powers of 10 have infinite precision )
Answer:
The conversion factor is 14.79 mL/Tbsp.
Explanation:
To do an unity conversiton, we can make a factor by a ratio transformation:
So, the conversion factor is 14.79 mL/Tbsp and 3 Tbsp has 44.37 mL.
