The answer for this two us wrong can u fix it please the answer is for 16 is 113 ml for 15 is 800km show me how u can get that 15)500 miles ta kilameterImile=L6a34 Km500 Miles l.Ga934Xgou> B04.672km16.)7.90table spoen to mLItable spoon=d 19.7868 mL79 table spo0=14.7868 7.9-l10.901 mLDotokoLlee Dhete

Answered: 15) S00 miles ta kilameter Imile=L6a34…

15) S00 miles ta kilameter Imile=L6a34 Km 500Miles I.Ga934Xgou B04.672km 16.)7.90table spoek Itable spoonz 19.7868 mL 79 table spo0214.7868+7.9 -ll0,901 mL to mL

World of Chemistry, 3rd edition

3rd Edition

ISBN:9781133109655

Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste

Chapter1: Chemistry: An Introduction

Section: Chapter Questions

Problem 7A

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Convert 45 mi/h to m/s, showing how the units cancel appropriately.
Given the following data for Mass of test tube and stearic acid = 14.17 gMass of test tube = 11.40 gFreezing point of strearic acid = 69.59o CMass of weighing paper + naphthalene =1.230 gMass of weighing paper = 0.920 gFreezing point solution = 64.00o CKf = 4.5o C/m Determine the following1. mass of stearic acid in g (2 decimal places); 2.77g2. mass of naphthalene in g (2 decimal places); 0.31g3. freezing point depression (2 decimal places); 3.93oC4. molality of solution (3 significant figures); _____5. moles of naphthalene (3 significant figures); _____6. molar mass of naphthalene, experimentally (3 significant figures); _____7. % error if theoretical molar mass of naphthalene is 128.17 g/ mole, USE ABSOLUTE VALUE (3 significant figure); _____
Stock iron(II) solution (200Ug mL-1 Fe) ferrous ammonium sulfate hexahydrate mass= 0.1437g, transfer it to a 100 ml beaker. add 15 ml approx of water and 15m1 'approx of dilute sulphuric acid (2M H2SO.). then transfer FeII to 100 ml flask makeup to the mark with water. calculate the moles of ferrous ammonium sulfate hexahydrate solution in unit ug/mL.
You were tasked to prepare 590 ml of 3.00 M HNO3. If you were provided a stock solution of 70.5% (w/w) HNO3 (MW = 63.0 g) with a specific gravity of 1.42, what volume of this solution (in ml) will you be using?
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.
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 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…
-If ethanol (S.G. = 0.790) were used rather than water in measuring the density of the irregular objects, how would that have affected the result? Explain. -When you were measuring the volume of the irregular solid you were told to stir to eliminate bubbles. If the bubbles were not eliminated, how would that affect the density? Explain. -Suppose the density of 65.0 mL of some liquid was found to be 0.748 g/mL. What would be the specific gravity of 45.0 mL of this liquid?
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+.
Qs/ A 0.2121-g sample of pure Na,C,0, (134.00 g/mol) was t ed with 43.31 mL of KMnO;. What is the normality of the KMnO, solution? The chemical reac- tion is 2MnOj; + 5C,0i + 16H* — 2Mn?* + 10CO, + 8H,O
A volumetric calcium analysis on triplicate samples of the blood serum of a patient believed to be suffering from a hyperparathyroid condition produced the following data: mmol Ca/L = 3.55, 3.65, 3.14. What is the 95% confidence interval for the mean of the data, assuming(a) No prior information about the precision of the analysis?(b) s → σ = 0.056 mmol Ca/L?
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/mol