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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
Answer:
480.6 g
Explanation:
Given data:
Number of molecules of methanol = 9.01 ×10²⁴
Mass in gram = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
9.01 ×10²⁴molecules ×1 mol /6.022 × 10²³ molecules
1.5 ×10¹ mol
15 mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 15 mol × 32.04 g/mol
Mass = 480.6 g
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Seconds look at the pictures how to build it.
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