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 )
0.424209104545485 is the answer my friend lol or at least what I got
Answer:
Zn =⇒ Zn+2(0.10) + 2e- (anode)
Zn+2(?M) + 2e- === Zn(s) (cathode)
Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn
E = E^o -0.0592 log Q; in this case E^o is zero.
E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2
23 mV x 1 volt/1000mv = 0.023 Volts
0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
0.023 = -0.0296 { log 0.10 – log [Zn+2] }
0.023 = -0.0296{ -1 - log[Zn+2] }
0.023 = +0.0296 + 0.0296log[Zn+2]
-0.0066 = 0.0296log[Zn+2]
-0.22= log[Zn+2]
[Zn+2] = 10^-0.22 = 0.603 Molar
Answer:
2Al + 3CuSO4 → Al2(SO4)3 + 3Cu
Explanation:
I THINK
1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
also remember that 1 kg= 2.20 lbs
2. 0.94 g/mL----> 0.94 grams= 1 mL
1 Liters= 1000 mL
1kg= 1000 grams
