Answer:
the partial pressure of the O2 is 0.167 atm
Explanation:
The computation of the partial pressure of the O2 is shown below:
As we know that
P = P_N2 + P_O2 + P_CO2
P_O2 = P - P_N2 - P_CO2
= (1.007 - 0.79 - 0.05)
= 0.167 atm
Hence, the partial pressure of the O2 is 0.167 atm
we simply applied the above formula
Preparing 15 mg/gl working standard solution from a 20 mg/dl stock solution will require the application of the dilution principle.
Recalling the principle:
initial volume x initial molarity = final volume x final molarity
Since we were not given any volume to work with, we can as well just take an arbitrary volume to be prepared. Let's assume that the stock solution is 10 mL and we want to prepare 15 mg/gl from it:
Applying the dilution principle:
10 x 20 = final volume x 15
final volume = 200/15
= 13.33 mL
This means that in order to prepare 13.33 mL, 15 mg/l working standard solution from 10 ml, 20 mg/dl stock solution, 3.33 mL of the diluent must be added to the stock solution.
More on dilution principle can be found here: brainly.com/question/11493179
The Arrhenius definition of acid-base reactions, which was devised by Svante Arrhenius, is a development of the hydrogen theory of acids. ... An Arrhenius base is a substance that dissociates in water to form hydroxide (OH -) ions. In other words, a base increases the concentration of OH - ions in an aqueous solution.
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