Answer:
The arm that was not sprayed with anything
Explanation:
The control group would be <u>the arm that was not sprayed with anything</u>.
<em>The control group during an experiment is a group that forms the baseline for comparison in other to determine the effects of a treatment. The control group does not include the variable that is being tested and as such, it provides the benchmark to measure the effects of the tested variable on the other group - the experimental group. In this case, the experimental group would be the arm that was sprayed with the repellent.</em>
Answer:
The correct answer is: Ka= 5.0 x 10⁻⁶
Explanation:
The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:
HA ⇄ H⁺ + A⁻
t= 0 0.200 M 0 0
t -x x x
t= eq 0.200M -x x x
At equilibrium, we have the following ionization constant expression (Ka):
Ka= ![\frac{ [H^{+}] [A^{-} ]}{ [HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BH%5E%7B%2B%7D%5D%20%20%5BA%5E%7B-%7D%20%5D%7D%7B%20%5BHA%5D%7D)
Ka= 
Ka= 
From the definition of pH, we know that:
pH= - log [H⁺]
In this case, [H⁺]= x, so:
pH= -log x
3.0= -log x
⇒x = 10⁻³
We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:
Ka= 
= 
= 5.025 x 10⁻⁶= 5.0 x 10⁻⁶
Answer:
6.0 L
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 0.075 M
V1 = 200 L
M2 = 2.5 M
V2 = ?
Solve for V2 --> V2 = M1V1/M2
V2 = (0.075 M)(200 L) / (2.5 M) = 6.0 L
<span>Well if you're looking for grams, all you need to do is cancel out units.
(ml)(g/ml)=g because the ml cancels out.
Thus, multiply: (85.32ml)(1.03g/ml)=...I'll let you solve this. :)
Good luck! Hope that helped. When in doubt, look at the units.</span>
