Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5
Answer:
The answer is C. An electrochemical cell.
Explanation:
The aluminum ion react with the sulfide to form aluminum sulfide.
Complete Question:
Suppose a cobalt atom in the +3 oxidation state formed a complex with two bromide (Br-) anions and four ammonia (NH3) molecules. write the chemical formula of this complex.
Answer:
[Co(NH₃)₄]⁺Br₂
Explanation:
The cobalt atom with +3 oxidation is represented as Co⁺³, and if it's bonded to two bromide ions, and four ammonia molecules. The molecules that are bonded to the metal atom (Co) are called complexing agents.
In the representation, we first put the molecules that surround the metal atom, forming an anion with the oxidation of the metal:
[Co(NH₃)₄]⁺³
Then, the ions are put in the formula. Because there are two bromides ion, each one with 1 minus charge, only 2 plus charged will be neutralized, and the complex will be:
[Co(NH₃)₄]⁺Br₂
Answer:
II
Explanation:
We must have a good idea of the fact that there are two mechanisms that come into play when we are discussing about the addition of hydrogen halides to alkenes. The first is the ionic mechanism and the second is the radical mechanism.
The ionic mechanism is accounted for by the Markovnikov rule while the radical mechanism occurs in the presence of peroxides and is generally referred to as anti Markovnikov addition.
The intermediate in anti Markovnikov addition involves the most stable radical, in this case, it is a tertiary radical as shown in the images attached. The most stable radical is II hence it leads to the major product shown in the other image.
