In the year 1909, Robert A. Millikan and Harvey Fletcher performed the oil drop experiment.
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:
volume of gas = 9.1436cm³
Explanation:
We will only temperature from °C to K since the conversion is done by the addition of 273 to the Celsius value.
Its not necessary to convert pressure and volume as their conversions are done by multiplication and upon division using the combined gas equation, the factors used in their conversions will cancel out.
V1 =10.1cm³ , P1 =746mmHg, T1=23°C =23+273=296k
V2 =? , P2 =760mmmHg , T2=0°C = 0+273 =273K
Using the combined gas equation to calculate for V2;


V2=9.1436cm³
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