Solution dilution calculator mg ml

    • [DOC File]QUALITY ASSURANCE MANUAL

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      Dilution 1 volume used is 150 mL BOD5 = 20 mg/L. Dilution 2 volume used is 100 mL BOD5 = 22 mg/L . Dilution 3 volume used is 75 mL BOD5 = 15 mg/L. Assuming all dilutions had an oxygen depletion of 2.0 mg/L and a final DO of greater than 1.0 mg/L, the reportable BOD5 is 19 mg/L.

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    • [DOC File]Buffers and stocks

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      Ethidium Bromide Stock solution, 10 mg/ml. Ethidium Bromide 400 mg. Water 40 ml. Store at 4 ˚C. 5x Laemmli Sample buffer. SDS 1.5g. 1M Tris pH 6.8 3.75ml. Bromophenol blue 0.015g. DTT 1.16g. Glycerol 7.5ml. Mix thoroughly with stirring. Add water until 15ml total volume. Make 1ml aliquots and store at -20˚C. 1M MgCl2. MgCl2, 6H2O 10.17g

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    • [DOC File]USING A CONDUCTIVITY PROBE

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      Complete all calculations for Dilution Procedure #1 as follows: Use the equation C1V1 = C2V2 to calculate the amount of 1000 ppm NaCl stock solution required to make 50 mL each of 200 ppm, 150 ppm, 100 ppm, 50 ppm, and 25 ppm NaCl solutions.

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    • [DOC File]DILUTIONS AND STANDARDS

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      By adding 1 mL of solution to 9 mL of water, you have prepared a 1:10 dilution. Therefore: 4 g/mL x 1/10 = 4 g/mL/10 = 0.4 g/mL. Example 2: If you had a 100 mg/dL solution of glucose and made a 1:5 dilution, what concentration of glucose is contained in the dilution? 100 mg/dL x 1/5 = 100 mg/dL/5 = 20 mg/dL . Problems: 7.

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    • [DOC File]Determination of Biochemical Oxygen Demand (BOD5)

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      5. Transfer 203 ml of sample into a white 500 ml casserole dish and titrate with 0.0250N sodium thiosulfate to a pale straw color. Add 1-2 ml of starch solution and continue to titrate to first disappearance of the blue color. (200 ml of original dilution water is equal to 203 ml of dilution water plus reagents.) Titrate two of the three samples.

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    • [DOC File]Calculations

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      Many methods express concentration as mg/mL or % (w/v) or even more confusingly g/dL. These units make it easier to weigh out and make up the solution but more difficult to work out the stoichiometry of the reaction. % (w/v) is actually g/100 mL. g/dL is also g/100 mL (dL is a deciliter or 100 mL) mg/mL is the same as g/L or g/ L

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    • [DOCX File]k12s.phast.umass.edu

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      You send a sample of rice to the lab for the determination of the total arsenic content. The analyst dissolves 725 mg of the rice in 5.0 mL of concentrated acid and then dilutes the solution to a final volume of 20.0 mL. When analyzed by ICP-MS, the solution concentration is found to be 25 µg L-1.

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    • [DOC File]Comparing Mitosis and Meiosis

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      Measurement Conversion 75.34 mg g 4.3 mL μL 440.3 mL L 3.33 g μg 0.34 g mg 34.0 g kg 0.004 L mL 80.34 μL mL 34 mg μg 0.25 g ng Part B. Making Solutions of Differing Mass/Volume Concentrations. Describe how you would prepare 25 mL of a NaCl solution at a concentration of 2.5 g/mL.

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    • [DOC File]BITC1311 Introduction to Biotechnology

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      Apr 26, 2006 · *Calculate the volumes of acid and water needed to prepare 50 mL each of a 0.1 M solution, a 0.01 M solution and a 0.001 M solution through serial dilution. Show the calculations in your notebook. Label 3 small beakers with the concentrations from step 2 and add the correct amount of water to each one, using the appropriate measuring device.

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    • [DOC File]USING A CONDUCTIVITY PROBE

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      Use graduated cylinders to prepare the 50 ppm NaCl solution by adding the appropriate amount of 100 ppm NaCl solution to enough distilled water to make up 150 mL of the solution. Label the beaker. Use the equation C1V1 = C2V2 to calculate the amount of 50 ppm NaCl solution required to make 150 mL of a 25 ppm NaCl solution.

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