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 )
Well it breaks down into small parts
I would think it is a heterogeneous mixture since it can't be an element since there are more than one type of atom, it can't be a compound since the leaves are not bonded together, and it can not be a homogeneous mixture since the leaves don't all blended together (the pile is not uniform) and you can distinguish all the different parts of the mixture. It can be considered a heterogeneous mixture since the leaves are mixed together (along with other things like dirt) in a non-uniform way so that you can point out the parts of the mixture and it does not look like one thing.
I hope this helps. Let me know in the comments if anything is unclear.
Answer:
Explanation:
see answer below in the attached file.
Answer:
Prompt Neutrons
Explanation:
Principle. Using uranium-235 as an example, this nucleus absorbs thermal neutrons, and the immediate mass products of a fission event are two large fission fragments, which are remnants of the formed uranium-236 nucleus. These fragments emit two or three free neutrons (2.5 on average), called prompt neutrons.
