In dilutions the amount of solute is held constant, but the volume is changed. Since (concentration)(volume) is equal to the amount of solute, we can derive an expression: M1V1=M2V2 where M represents molarity in mol/L and V is volume of the solution in Liters. THIS ONLY WORKS FOR DILUTIONS. Show you work and use the Molarity Simulation to check your answers. If you wanted 0.80L of a 0.25 M CuSO4 solution, what volume of 1.0 M CuSO4 solution would you dilute?

In dilutions the amount of solute is held constant, but the volume is changed. Since (concentration)(volume) is equal to the amount of solute, we can derive an expression: M1V1=M2V2 where M represents molarity in mol/L and V is volume of the solution in Liters. THIS ONLY WORKS FOR DILUTIONS. Show you work and use the Molarity Simulation to check your answers. If you wanted 0.80L of a 0.25 M CuSO4 solution, what volume of 1.0 M CuSO4 solution would you dilute?

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter3: Mass Relations In Chemistry; Stoichiometry

Section: Chapter Questions

Problem 23QAP

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3.68 Magnesium is lighter than other structural metals, so it is increasingly important in the design of more efficient vehicles. Mg2+ ions are present in seawater, and the metal is often prepared by "harvesting" these ions and converting them to Mg metal. The average magnesium content of the oceans is about 1270 g Mg2+ per ton of seawater, and the density of seawater is about 1.03 g/mL. What is the molarity of Mg2+ ions in seawater? The design for a concept car calls for 103 kg of magnesium per vehicle. How many gallons of seawater would be required to supply enough magnesium to build one of these ears?
In dilutions the amount of solute is held constant, but the volume is changed. Since (concentration)(volume) is equal to the amount of solute, we can derive an expression: M1V1=M2V2 where M represents molarity in mol/L and V is volume of the solution in Liters. THIS ONLY WORKS FOR DILUTIONS. Show you work and use the Molarity Simulation to check your answers. If 0.50 L of 1.0 M CoCl2 is diluted to 1.0L what is the new concentration (in Molarity)?
In dilutions the amount of solute is held constant, but the volume is changed. Since (concentration)(volume) is equal to the amount of solute, we can derive an expression: M1V1=M2V2 where M represents molarity in mol/L and V is volume of the solution in Liters. THIS ONLY WORKS FOR DILUTIONS. Show you work and use the Molarity Simulation to check your answers. If 0.20 L of 2.5 M K2CrO4 is diluted to 0.769L what is the new concentration?
Describe any visual differences between the hydrated sample and the dried, anhydrous form. How would the following errors affect the empirical formula for the compound? a. The student ran out of time and did not do the second heating. Explain how this error will affect the calculation for the number of moles of water in the hydrate? Will the final answer be artificially high or low? How do you know? b. The student recorded the mass of the cup + sample incorrectly and started with 2.2 g of hydrated compound but used 2.0 g in the calculations. Explain how this error will affect the calculation for the number of moles of water in the hydrate? Will the final answer be artificially high or low? How do you know?
In chemistry laboratories, it is faster and more convenient to measure out liquids by volume with pipets than by mass with an electronic balance. Therefore, when prepare solutions, volume ratio, instead of mass ratio is used sometimes. We can define a new term, volume % of a component, similar to mass % of a component. Let’s use a solution that is made of A and B as an example to explain this concept. If the volume % of A is 70%, the solution is made of 70 mL A with 30 mL B. A solution is prepared by mixing acetonitrile, CH3CN, and methanol, CH3OH, together. The volume percentage of acetonitrile is 35.84%. The density of acetonitrile is 0.786 g/mL and the density of methanol is 0.791 g/mL. Assume that the volume is additive. What is the mass percentage of acetonitrile in the solution? ____________% What is the molarity of the solution? __________ M What is the mole fraction of acetonitrile in the solution? ____________ What is the molality of the solution? __________ m
How would the following errors affect the empirical formula for the compound? The student ran out of time and did not do the second heating. Explain how this error will affect the calculation for the number of moles of water in the hydrate? Will the final answer be artificially high or low? How do you know? The student recorded the mass of the cup + sample incorrectly and started with 2.2 g of hydrated compound but used 2.0 g in the calculations. Explain how this error will affect the calculation for the number of moles of water in the hydrate? Will the final answer be artificially high or low? How do you know?
A chemist prepares a solution of iron III chloride FeCl3 by measuring out 2.70g of FeCl3 into a 300.mL volumetric flask and filling to the mark with distilled water. Calculate the molarity of Cl− anions in the chemist's solution. Be sure your answer is rounded to the correct number of significant digits.
I am trying to ascertain the density of 0.9-M HCl. The book provides a density of 1.0-M HCl as 1.016g/mL. I know the Molecular Weight aka Molar Mass is 36.46g/mol I also know that the equation to calculate the density in this instand is: Density=(Molarity x Molecular Weight)/WeightI have found examples online, and one answer provided here that shows the Density of 0.9-M HCl is 0.9115g/mL. That answer was arrived at as follows:Density = (0.9M x 36.46g/mol)/36 = 32.814/36 = 0.9115g/mL My problem is that I don't understand where the 36 came from for "weight", and I need to be able to show all unit measurements for each step. So my issue has 2 parts: 1) How was the Weight value determined to be 36 (I need to see the formula)2) Where in the step-by-step process is mL in there so that after all cancellations/conversions are completed, the only remaining units are g/mL to show how 0.9115g/mL was arrived at.
Commercial bleaches such as Clorox are said to contain 5.25% NaOCl by weight. Is this correct? Use your results and a density of 1.0 g/mL for the bleach in your calculations. These calculations should reflect the precision of the assumed density.
(When answering this problem, report the answer with the appropriate number of significant figures. When entering units, use proper abbreviated units with proper capitalization.)A beverage drink mix sample is prepared by mixing a packet of beverage powder in a gallon (3.78 L) of water. A sample for analysis is prepared by diluting 10.00 mL of this solution in a 50.0 mL volumetric flask. A student calculates that the concentration of dye in the analyzed beverage mix solution is (2.59x10^-5) mol/L. Determine the mass (in grams) of food dye in the beverage mix packet. Hint: The answer is assumed to be reduced to the highest power possible.
What is the concentration, in units of grams of dye per liter of solution (g/L), of the dilute dye solution that is produced as follows? Transfer 16.51 mL of the stock solution (6.12 g/L) to a 200.0 mL volumetric flask and dilute to the 200.0 mL line with deionized water. Your answer should be reported to the correct number of significant figures.
The density of toluene 1C7H82 is 0.867 g>mL, and the density of thiophene 1C4H4S2 is 1.065 g>mL. A solution is made by dissolving 8.10 g of thiophene in 250.0 mL of toluene. Assuming that the volumes of the solute and solvent are additive, what is the molarity of thiophene in the solution?