▼ ▼ Part A In order understand how this equation derived and why it holds true, the product f each side of the equation should be examined. What value and unit do you get when you multiply a concentration of 0.687 M by a volume of 0.500 L ? This question can be expressed as 0.500 L 0.687 mol X = T Express your answer to three significant figures with the appropriate units. ▸ View Available Hint(s) μA 1 C ? M₁V₁= Value Units Submit Part B When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solution has a significantly higher concentration of the given solute (typically 10¹ to 10¹ times higher than those of the diluted solutions). The high concentration allows many diluted solutions to be prepared using minimal amounts of the stock solution. What volume of a 6.61 M stock solution do you need to prepare 500. mL of a 0.0579 M solution of MgSO4? Express the volume to three significant figures with the appropriate units. ► View Available Hint(s) Di V₁ = Units Submit Value

▼ ▼ Part A In order understand how this equation derived and why it holds true, the product f each side of the equation should be examined. What value and unit do you get when you multiply a concentration of 0.687 M by a volume of 0.500 L ? This question can be expressed as 0.500 L 0.687 mol X = T Express your answer to three significant figures with the appropriate units. ▸ View Available Hint(s) μA 1 C ? M₁V₁= Value Units Submit Part B When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solution has a significantly higher concentration of the given solute (typically 10¹ to 10¹ times higher than those of the diluted solutions). The high concentration allows many diluted solutions to be prepared using minimal amounts of the stock solution. What volume of a 6.61 M stock solution do you need to prepare 500. mL of a 0.0579 M solution of MgSO4? Express the volume to three significant figures with the appropriate units. ► View Available Hint(s) Di V₁ = Units Submit Value

Chemical Principles in the Laboratory

11th Edition

ISBN:9781305264434

Author:Emil Slowinski, Wayne C. Wolsey, Robert Rossi

Publisher:Emil Slowinski, Wayne C. Wolsey, Robert Rossi

Chapter29: Synthesis And Analysis Of A Coordination Compound

Section: Chapter Questions

Problem 1ASA

See similar textbooks

Similar questions

Gallium chloride is formed by the reaction of 2.6 L of a 1.44 M solution of HCl according to the following equation: 2Ga+6HCl2GaCl3+3H2. (a) Outline the steps necessary to determine the number of moles and mass of gallium chloride. (b) Perform the calculations outlined.
On Easter Sunday, April 3, 1983, nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: 2HNO3(aq)+Na2CO3(aq)2NaNO3(aq)+H2O(l)+CO2(g) a. Calculate H for this reaction. Approximately 2.0 104 gal nitric acid was spilled. Assume that the acid was an aqueous solution containing 70.0% HNO3 by mass with a density of 1.42 glcm3. What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? (Hf for NaNO3(aq) = 467 kJ/mol) b. According to The Denver Post for April 4, 1983, authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of H, what was their major concern?
Compute the needed amount of H2SO4 (stock concentration: 18.1 M or 98%; Molar Mass = 98.07848 g/mol) to prepare the above concentrations at 200 mL (% weight).
Using this balanced equation: NaHCO3 + CH3COOH H2O + CO2 + NaC2H3O2 In an experiment , the following mass measured 3.0 grams, 5.5 grams and 7.0 grams of Baking soda is mixed with 0.5 mol of acetic acid (vinegar) to form carbon dioxide as a product formed Name of Bottle NaHCO3 (Baking Soda) Acetic Acid Calculation of Amount of Product Formed Mass Mole Volume Mole Limiting Excess A 3.0 g 50 mL 0.5 mol B 5.5 g 50 mL 0.5 mol C 7.0 g 50 mL 0.5 mol
You are asked to weigh 2 g of lithium and add it to 1500 ml of distilled water +2drops of indicator, knowing that the density of water is 1 at room temperaturewhile conducting this experiment. After calculating the amount of LiOH, you need to dilute to prepare 0.25 M of LiOH in a 3000 ml solution. Show your calculation and units.Knowing that the initial volume that you used was 1500 ml. Use this equation to calculate ( CLiOH x VLiOH) before dilution = (CLiOH x VLiOH) after dilution
What volume of 0.220 M HBr solution is required to produce 0.0190 moles of HBr? Answer should be in mL
What is the correct volume of 2.5 M PhMgBr in THF is needed to completely react with 1.5 mL of methyl benzoate? You do not need to consider an excess and should calculate (and choose) the minimum amount as required by the reaction stoichiometry). Enter your answer to one decimal place WITH the proper unit of measurement.
Concentration is measured by molarity (M), or moles per liter. Brackets are also used to symbolize molarity. For example, if 0.6 moles of HNO3 are dissolved in a liter of water, you would say [HNO3] = 0.6 M. Because HNO3 is a strong acid, it dissociates almost completely in water. That means the concentration of H+ is very nearly equal to that of HNO3. What is [H+] if [HNO3] is 0.01 M? The pH of a solution is equal to the negative log of H+ concentration: pH = –log[H+] What is the pH of this solution? What is the pH of a 0.6 M HNO3 solution?
How many millilitres of 0.450 mol L−1 H2SO4(aq) are required to completely neutralize 46.7 mL of 0.930 mol L−1 KOH(aq) ? Hint: Write a balanced chemical equation for the reaction. Give your answer in mL, accurate to three significant figures. Do not include units as part of your answer.
The information is in the pic attached and the question is What pressure of Cl2 (g) is required to maintain this concentration?
During the dilution process, the amount of solvent remains the same. Question options: True False
Assume a starting amount of 20 mL of 6M AgNO3 and 40 mL of 5M Na2CO3. Caluclate the limiting reactant and theoretical yield for this reaction.