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- using exactly 5.00 mL of 0.0400 M stock CuSO4 solution. Add 100 mL of water. Data for Part IMass of empty dish: 32.470 g empty dishVolume of 0.0400 M solution: 5.00 mL CuSO4 solution.Mass of dish and 0.0400 M solution: 37.497 g dish and solutionMass of dish and CuSO4 solid: 32.503 g dish and CuSO4 solidCalculations for Part I1. Calculate the mass of solution? 5.027 g 2. Calculate the mass of solid CuSO4 dissolved in the solution? 0.033 g 3. Calculate the number of moles of solid CuSO4 dissolved in the solution? 2.07 x 10-4 mol4. Calculate the mass of water evaporated from the solution? 0.0414 M5. Calculate the density of solution, (g solution/mL solution)?6. Calculate the % by mass, CuSO4 in solution (100 x g CuSO4/g solution)?7. Calculate the molality of solution (moles CuSO4/kg solvent)?8. Calculate the molarity of solution (moles CuSO4/L solution)?9. Given that the true molarity is 0.0400 M, calculate the percent error…Given the following data forMass of test tube, beaker and cyclohexane = 100.17 gMass of test tube and beaker = 84.07 gFreezing point of cyclohexane = 6.59 oCMass of weighing paper + naphthalene =1.080 gMass of weighing paper = 0.928 gFreezing point solution = 5.11oCKf = 20.8oC/mDetermine the followinga. mass of cyclohexane in g (2 decimal places); _____b. mass of naphthalene in g (4 decimal places); _____c. freezing point depression (2 decimal places); _____d. molality of solution (3 significant figures); _____e. moles of naphthalene (3 significant figures); _____f. molar mass of naphthalene, experimentally (3 significant figures); _____g. % error if theoretical molar mass of naphthalene is 128.17 g/ mole, USE ABSOLUTE VALUE (3 significant figure); ____using exactly 5.00 mL of 0.0400 M stock CuSO4 solution. Add 100 mL of water. Data for Part IMass of empty dish: 32.470 g empty dishVolume of 0.0400 M solution: 5.00 mL CuSO4 solution.Mass of dish and 0.0400 M solution: 37.497 g dish and solutionMass of dish and CuSO4 solid: 32.503 g dish and CuSO4 solidCalculations for Part I1. Calculate the mass of solution2. Calculate the mass of solid CuSO4 dissolved in the solution.3. Calculate the number of moles of solid CuSO4 dissolved in the solution.4. Calculate the mass of water evaporated from the solution.5. Calculate the density of solution, (g solution/mL solution).6. Calculate the % by mass, CuSO4 in solution (100 x g CuSO4/g solution).7. Calculate the molality of solution (moles CuSO4/kg solvent).8. Calculate the molarity of solution (moles CuSO4/L solution).9. Given that the true molarity is 0.0400 M, calculate the percent error of your result.
- 1. Calculate the experimental density of a salt solution and the percent error (same as relative error percent) using some or all the data given below. solubility of NaCl salt in water: 0.357 g/mLmass of empty graduated cylinder: 25.19g mass of graduated cylinder + salt solution: 30.47g total volume of salt solution: 4.98 mLtrue density of salt solution: 1.07 g/mLA tank with a capacity of 74 liters, initially contains 20 kilograms of salt dissolved in 50 liters of water. Another solution, containing 3 kilograms of dissolved salt per liter, runs into the tank at 4 liters every minute and that the mixture (kept uniform by stirring) runs out of the tank at the rate of 1 liter per minute. What is the concentration of the solution in the tank at point of overflow? Intermediate values should be rounded only to two decimal places. Round off the final answer to two decimal places.A. B. and C. already solved (https://www.bartleby.com/questions-and-answers/chemistry-question/a89691d6-2162-4677-b0cc-dc84fcceda08). Please answer E, F and GCircle in pencil in the image on th eirght cide of the diagram is a "2"( Pb(NO3)2 ).
- What is an intensive property that can be measured wInstructions for this problem say, "Solve each problem and report your answer to correct significant figure and units. Remember that units multiply, divide, add, subtract, and simplify just like variables." (23.0 lb/gal)(2 ft) = Do I need to convert any of the units? Or can I simply multiply, leaving me with a unit of "ft lb/gal"? None of the other problems in this section have space or required a lot of work to be shown. Image of my attempt attached.3. A student weighed an unknown mass directly on an electronic balance of intermediate sensitivity, and got a result of 28.774 g. The student then realized a mistake had been made, and weighed the same unknown by difference on the same balance with the following results:Mass of beaker plus unknown = 59.121 g.Mass of empty beaker = 30.346 g.When the sensitivity of the balance is taken into account, the two values obtained for the unknown mass are considered to be:a. Differentb. The SameExplain your answer.
- so for the standard division for data B, i got 0.248573 how do i round it would it be 0.249 or 0.248When answering, please not only provide the calculation work, but the reasoning and basic concepts behind the answer. This will significantly help me out. Thanks!Round to correct sig fig, please be accurate. Units in kJ