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:
For finding frequency, we need to first find the period of the graph.
The period of a sinusoidal graph is the time interval in which it repeats its pattern.
In the graph, we can see, after 
time, it repeats its pattern.
Hence the period of the graph is 
.
Now we need to find its frequency 
The formula for frequency is 
This is the answer
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C. 28 KJ
AMU of H2O2 = 2(1) + 2(16) = 34 g/mol
10 g / 34 g/mol = 0.294 mol H2O2
0.294 mol / H = 2 mol / 190 KJ
H = 28.9 KJ
