If this lab’s procedure were carried out using 38.77 mL each of 0.16 M K2CrO4 and 0.248 M Pb(NO3)2. How many mmol of Pb(NO3)2 were used in this reaction? Report your answer with 3 significant figures
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A: Mass of K2Cr2O7 = 30 g Mass of water = 50 g Mass precipitated out at 20°C = ?
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A: Given data, 3.00 M CH3OH solution contains 0.270 mol CH3OH
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Q: A student, performing the same experiment as you, recorded the following information. The original…
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A: We’ll answer the first question since the exact one wasn’t specified. Please submit a new question…
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Q: PRE-LAB QUESTIONS 1. Calculate the molarity of a 6.15% (w/v%) sodium hypochlorite (NAOCI) solution.…
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Q: What is the concentration of AlF63- in 1.00 L of aqueous solution made with 39.1 mmole of Al(NO3)3…
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Q: Describe how you would prepare 350.0 mL of 0.100 M C12H22O11 starting with 3.00 L of 1.50 МС12Н2О1-
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Q: Consider the reaction of 75.0 mL of 0.350 M C₅H₅N (Kb = 1.7 x 10⁻⁹) with 100.0 mL of 0.405 M HCl.…
A: Given: The volume of C5H5N is 75 mL. The molarity of C5H5N is 0.350 M. The volume of HCl is 100 mL.…
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Q: Consider the reaction of 75.0 mL of 0.350 M C₅H₅N (Kb = 1.7 x 10⁻⁹) with 100.0 mL of 0.411 M HCl.…
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Q: Which of the following will result to a dimly lit light bulb when subjected to the conductivity…
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Q: 0.350 M
A: Volume of acid ( HCl ) = 100.0mL Volume of base ( C5H5N ) = 75.0mL
Q: If this lab's procedure were carried out using 20.51 ml each of 0.249 M K2CrO4 and 0.159 M Pb(NO3)2.…
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Q: How would you use a volumetric flask to prepare 250.0 mL of 0.150 0 M K2SO4?
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- You have to make 50 mL of 0.05M HCl from a 10M HCl stock solution, but the only measuring devices available are 100mL graduated cylinders and 10mL serological pipets. How could you accurately make the dilute HCl solution (clearly explain)?The first goal is to make the oxalic acid standard solution. You measure 1.5232 g of oxalic acid on an analytical balance, add it to a 250-mL volumetric flask and add deionized H2O to a final volume of 250.0 mL. Molar mass of H2C2O4•2H2O = 126.07 g/mol Mass of H2C2O4•2H2O = 1.5232g what is the Number of moles of H2C2O4•2H2O?Consider the reaction of 75.0 mL of 0.350 M C₅H₅N (Kb = 1.7 x 10⁻⁹) with 100.0 mL of 0.437 M HCl. What would the total volume of the solution be after the reaction, in mL?
- A student, performing the same experiment as you, recorded the following information. The original solution was diluted by a factor of 5.Calculate the volume of NaOH used to reach the equivalence point. Volume of diluted vinegar sample (mL) 25.00 Molarity of NaOH 0.0900 Initial buret reading (mL) 16.50 Final buret reading (mL) 38.00 Volume of NaOH3. Describe the characteristics that a crystal obtained through recrystallization must have, the final crystal, not the solvent.8. Find the final concentration of ethanol as a percent (v/v) when 30 mL of ethanol 50 % (v/v), 60 mL of ethanol 70 % (v/v), and 40 mL of ethanol 95 % (v/v) are mixed together.
- Which of the following will result to a dimly lit light bulb when subjected to the conductivity test? a. 1.0 M citric acid b. Glacial acetic acid c. Mixture of 10.0 mL 1.0 M Mg(OH)2 and 10.0 mL 1.0 M HNO3 d. 70% (v/v) ethanol7. a) What weight of anhydrous copper(II) sulfate would be required to make 1.5 liters of solution containing 10.0 grams of Cu2+ ions per liter? 7 b) What weight of copper(II) sulfate pentahydrate would be required to make the solution in 7a) above?A student prepared benzoic acid using the method described in the procedure using 4.05 g KMnO4, 2.00 mL benzyl alcohol and 150 mL 3.0M H2SO4. How many molar equivalences of hydrogen ions were added to the reaction?
- Chemistry 1. What is the concentration of AlF63- in 1.00 L of aqueous solution made with 39.1 mmole of Al(NO3)3 and 0.250 moles of NaF? Report the concentration in mM to 1 decimal place. Kf = 6.3 x 1019 for AlF63-How to prepare these solutions? 250.0 mL 0.125 M stock Na2S2O3 solution from Na2S2O3·5H2O crystals 250 mL 0.10 M NaOHNOTE: Use the 1.0 M NaOH prepared250.0 mL standard 2500 ppm Cu(II) stock solutiona. Weigh and dissolve appropriate amount of Cu(NO3)2·5H2O crystals in enough distilled water.Step 1: In tube 1, you add 100 ul of plasma to 900 ul of distilled water. Step 2: In tube 2, you then take 100 ul from tube 1 and add 900 ul of distilled water. Step 3: In tube 3, you then take 100 ul from tube 2 and add 900 ul of distilled water. What is the final/total dilution factor in tube 3 at the end of step 3?