Answer:
Smithers thinks that a special juice will increase the productivity of workers. He creates two groups of 50 workers each and assigns each group the same task (in this case, they're ... Group A is given the special juice to drink while they work. Group ... 1. Control Group. Group B. 2. Independent Variable. volume of special juice.
Explanation:
Answer: If more than one variable is changed in an experiment, scientist cannot attribute the changes or differences in the results to one cause. By looking at and changing one variable at a time, the results can be directly attributed to the independent variable.
Answer:
Kb
Explanation:
The correct question is
What is Kb for N2H4 if the pH of a 0.158M solution of N2H5Cl is 4.5?
Solution-
N2H4Cl hydrolyses on addition of water
The reaction equation is as follows -
N2H4+ + H2O ----> N2H4 + H3O+

pH
![[H+] = 10^{-pH} = 10^{-4.5}](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%2010%5E%7B-pH%7D%20%3D%2010%5E%7B-4.5%7D)

Kb
Answer:
![[Ag^+]=2.82x10^{-4}M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%3D2.82x10%5E%7B-4%7DM)
Explanation:
Hello there!
In this case, for the ionization of silver iodide we have:
![AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)\\\\Ksp=[Ag^+][I^-]](https://tex.z-dn.net/?f=AgI%28s%29%5Crightleftharpoons%20Ag%5E%2B%28aq%29%2BI%5E-%28aq%29%5C%5C%5C%5CKsp%3D%5BAg%5E%2B%5D%5BI%5E-%5D)
Now, since we have the effect of iodide ions from the HI, it is possible to compute that concentration as that of the hydrogen ions equals that of the iodide ones:
![[I^-]=[H^+]=10^{-3.55}=2.82x10^{-4}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D%5BH%5E%2B%5D%3D10%5E%7B-3.55%7D%3D2.82x10%5E%7B-4%7DM)
Now, we can set up the equilibrium expression as shown below:

Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:
![x=[Ag^+]=2.82x10^{-4}M](https://tex.z-dn.net/?f=x%3D%5BAg%5E%2B%5D%3D2.82x10%5E%7B-4%7DM)
Best regards!
Answer:
1.3×10⁻³ M
Explanation:
Hello,
In this case, given the dissociation reaction of acetic acid:

We can write the law of mass action for it:
![Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BCH_3CO_2%5E-%5D%7D%7B%5BCH_3CO_2H%5D%7D)
Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change
due to the dissociation extent, we are able to rewrite it as shown below:

Thus, via the quadratic equation or solve, we obtain the following solutions:

Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.
Regards.