The answer is
<span>Fe3+ + e- Fe2+ = 0.77 V</span>
Answer:
The correct approach is Option B (Peer Review).
Explanation:
- Rather made reference to someone as a scientific peer-review, it encourages the specialist who has not been essential to the study team to analyze the study objectively and pointed out everyone's mistakes. It serves as major self-regulation for scholars and aims to make the publishing process somewhat credible. Hence, the solution to this issue is Peer Examination.
- Funding organizations rarely have the capabilities to recognize out mistakes, whereas definitive analysis is a method of study that helps to make a definitive statement. The gathering of data is simply a process of scientific study.
Other approaches do not apply to the example mentioned. Although the one mentioned is right.
Answer:
1.09 x 10⁻⁴ M
Explanation:
The equation of the reaction in given by
Ni²⁺ (aq) + 6NH₃ (aq) ⇔ Ni(NH₃)₆
At the beginning o the reaction, we have 0.18M concentration of Ni and 1.2M concentration of aqueous NH₃ and zero concentration of the product
As the reaction proceeds towards equilibrium, the concentration of the reactants decrease as the concentration of the product starts to increase
From the equation,
1 mole of Ni²⁺ reacts with 6 moles of aqueous NH₃ to give 1mole of Ni(NH3)
therefore
0.18 M of Ni would react with 1.08M (6 x 0.18M) aqueous NH₃ to give 0.18M of Ni(NH₃)6
At equilibrium,
1.08M of NH3 would have reacted to form the product leaving
(1.2 - 1.08)M = 0.12M of aqueous NH₃ left as reactant.
Therefore, formation constant K which is the ratio of the concentration of the product to that of the reactant is given by
K = [Ni(NH₃)₆} / [Ni²⁺] 6[NH₃]
5.5 x 10⁸ = 0.18 M / [Ni²⁺] [0.12]⁶
[Ni²⁺]= 0.18 M / (5.5 x 10⁸) (2.986 x 10⁻⁶)
=0.18 M / 0.00001642
= 1.09 x 10⁻⁴ M
[Ni]²⁺ = 1.09 x 10⁻⁴ M
Hence the concentration of Ni²⁺ is 1.09 x 10⁻⁴ M
Answer:
The 3 types of muscle tissue are cardiac, smooth, and skeletal.
Explanation:
