Answer:
B. halocline
Explanation:
it is a zone in the oceanic water that changes depending on the depth
Hope This Helped Sorry If Wrong
From the equation of te reaction, we know that the mass percent of NaOH in the mixture is 1.4%.
<h3>What is neutralization?</h3>
Neutralization is a reaction that occurs between an acid and a base to yield salt and water only.
In tis case, the reaction of the NaOH and HCl occurs as follows; NaOH + HCl ----> NaCl + H2O
Number of moles of HCl reacted = 15/1000 * 0.5 moldm-³ = 0.0075 moles
Since the reaction is 1:1, 0.0075 moles of NaOH reacted.
Mass of NaOH = 0.0075 moles of NaOH * 40 g/mol = 0.3 g
Percent of NaOH = 0.3 g/21.10g * 100/1 = 1.4%
Learn more about percent concentration: brainly.com/question/202460
Answer:
The cathode reaction is NiO2+H2O+2e−→Ni(OH)2+2OH−.
The question is incomplete, here is the complete question:
Consider the following chemical reaction: H₂ (g) + I₂ (g) ⇔ 2HI (g) At equilibrium in a particular experiment, the concentrations of H₂, I₂, and HI were 0.15 M, 0.033 M and 0.55 M respectively. The value of Keq for this reaction is
<u>Answer:</u> The value of
for the given reaction is 61.11
<u>Explanation:</u>
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For a general chemical reaction:

The expression for
is written as:
![K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
For the given chemical equation:

The expression of
for above equation follows:
![K_{eq}=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
We are given:
![[HI]_{eq}=0.55M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D0.55M)
![[H_2]_{eq}=0.15M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D0.15M)
![[I_2]_{eq}=0.033M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.033M)
Putting values in above expression, we get:

Hence, the value of
for the given reaction is 61.11
The answer to this is C. III & IV