The final concentration of the diluted standard is 0.2 mg/dL.
<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>
Using the dilution formula:
where
- C1 is initial concentration
- V1 initial volume
- C2 is final concentration
- V2 is final volume.
Assuming a final volume of 100 mL, and since a 1/5 dilution is made:
C1 = 1.00 mg/dL
V1 = 20
C2 = ?
V2 = 100 mL
C2 = C1V1/V2
C2 = 20 × 1/100
C2 = 0.2 mg/dL
Therefore, the final concentration of the diluted standard is 0.2 mg/dL.
Learn more about dilution at: brainly.com/question/24881505
beryllium & calcium are the two in the same column or family
Answer:
The reaction rate becomes quadruple.
Explanation:
According to the law of mass action:-
The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.
Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.
The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.
Thus,
Given that:- The rate law is:-
![r=k[A_2][B_2]](https://tex.z-dn.net/?f=r%3Dk%5BA_2%5D%5BB_2%5D)
Now,
and ![[B'_2]=2[B_2]](https://tex.z-dn.net/?f=%5BB%27_2%5D%3D2%5BB_2%5D)
So, ![r'=k[A'_2][B'_2]=k\times 2[A_2]\times 2[B_2]=4\times k[A_2][B_2]=4r](https://tex.z-dn.net/?f=r%27%3Dk%5BA%27_2%5D%5BB%27_2%5D%3Dk%5Ctimes%202%5BA_2%5D%5Ctimes%202%5BB_2%5D%3D4%5Ctimes%20k%5BA_2%5D%5BB_2%5D%3D4r)
<u>The reaction rate becomes quadruple.</u>