A concept map for four types of intermolecular forces and a certain type of bond is shown. Compare the relative strength of the two forces B and C.explain how you determine this comparison but identifying the forces.
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- A concept map for four types of intermolecular forces and a certain type of bond is shown. Compare the relative strength of the two forces B and C.explain how you determine this comparison but identifying the forces.
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- In a river with a constant flow rate, the factory releases wastewater containing NaCl at a concentration of 200 mmol / L to the river at 20.0 L / s. The Na+ and Cl- concentrations at the downstream observation point were 1.00 mmol / L and 0.800 mmol / L, respectively. What is the river flow Q and Cl- concentration before the inflow of drainage? The river originally contains 0.500 mmol / L of Na+.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); ____https://cxp.cengage.com/contentservice/assets/owms01h/references/chemtables/genChemIndex.html
- https://cxp.cengage.com/contentservice/assets/owms01h/references/chemtables/gen_chem/eo.html ^ table image: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…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.
- Add Stoichmetric factor: ______CH3CH2OH + _______O2 yields ________CO2 + _______H20Please answer g please In the table below, use your most important scientific instrument (your calculator) to fill in the little boxesthat are highlighted:2. Data TableIdentity of Dehydrated salt (for afterwards) ________________________________ Mass of empty dish 70.874g Mass of dish + hydrated salt 71.346g Mass of hydrated salt Mass of dish + dehydrated salt XXXXXXX a) first heating 71.194g b) second heating 71.174g Mass of dehydrated salt Mass of water lost Wt. % of water in hydrated salt Moles of dehydrated salt Moles of water driven off Empirical formula of hydrated salt a. Show the calculation for obtaining the mass of your hydrated salt used.b. Show the calculation for obtaining the mass of your dehydrated salt after the second heating.c. Show the calculation for obtaining the mass of water that was driven off (based on secondheating).d. Show the calculation for obtaining the weight percent of water that was in your hydrated…In biological systems, enzymes are used to accelerate the rates of certain biological reactions. Glucoamylase is an enzyme that aids in the conversion of starch to glucose (a sugar that cells use for energy). Experiments show that 1 μg mol of glucoamylase in a 4% starch solution results in a production rate of glucose of 0.6 μg mol/(mL)(min). Determine the production rate of glucose for this system in units of lb mol/(ft3)(day). A. 0.0015 lb mol/(ft3)(day) B. 0.0315 lb mol/(ft3)(day) C. 0.0539 lb mol/(ft3)(day) D. 0.3569 lb mol/(ft3)(day)
- When concentrated nitric is sold, the label contains no mention of the molarity of the acid. Instead, the label normally lists the concentration of nitric acid as a wt/wt percent and gives the specific gravity of the solution. If the solution is 68.0 (wt/wt %) and nitric acid and has a specific gravity of 1.41 kg/L, calculate the molarity of 68.0 % concentrated nitric acid. FW HNO3=63.01g/mol203.77miles/sec is converted to cm/nsec?Produce Water mass “X” is known to be a mixture of water masses “1”, “2”, and “3”. Using the data plotted below, calculate the composition of “X” (i.e. determine the fraction from each of the 3 water masses that makes up water mass X). Watermass#1– SF6 =25 S=33.5 Watermass#2– SF6 =75 S=32 Watermass#3– SF6 =50 S=36.5 Water mass X – SF6 =60 S= 34