3Dilution is a process in which the concentration of a solution is lowered by adding more solvent. Some of the reasons dilutions are performed are to minimize measurement errors whenpreparing a series of solutions at different concentrations, to save time and laboratory space, and to make more accurate measurements on an analytical balance when the targetconcentration is very low. The new concentration of a diluted solution can be determined from the following equation, sometimes called the dilution equation,YM₁ V₁ = M₂V₂where M₁ is the concentration of the initial solution, V₁ is the volume of the initial solution, M₂ is the concentration of the final solution, and V₂ is the volume of the final solution. Theterms M₁ and M₂2 are often expressed in molarity, M, which is equal to mol/L. Another way to write the equation is to replace subscripts of 1 and 2 with abbreviated notationsrepresenting concentrated and diluted, as follows:EMeone Veone Mail VailBased on the inverse nature of this equivalency, small volumes of concentrated solutions can be used to prepare large volumes of diluted solutions with a known concentration, which isadvantageous in areas involving solution chemistry.Part AIn order to understand how this equation is derived and why it holds true, the product of each side of the equation should be examined. What value and unit do you get when youmultiply a concentration of 0.735 M by a volume of 0.400 L ? This question can be expressed asExpress your answer to three significant figures with the appropriate units.► View Available Hint(s)M₁ V₁ = ValueSubmitPart B80$4RμAF4%5openvellum.ecollege.com→UnitsFST6?MacBook AirF6Y0.400 L X 0.735 mol =1&7P PearsonF7U8DIIFB(9DDF90Review I Constants I Periodic Table)0F10ⒸP-CF11++=F1288 BM₁V₁= ValueSubmitPart BWhen you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solution has a significantly higher concentration of the givensolute (typically 10¹ to 10 times higher than those of the diluted solutions). The high concentration allows many diluted solutions to be prepared using minimal amounts of the stocksolution.What volume of a 6.73 M stock solution do you need to prepare 100. mL of a 0.2100 M solution of KCI?Express the volume to three significant figures with the appropriate units.► View Available Hint(s)V₁ = ValueSubmit80$HÅ4ER1F4UnitsDilution as a procedure in analysisWe sometimes purposely dilute solutions if the analyte concentrations are too high to permit an accurate measurement. For example, a spectrophotometer can measure the amount oflight that passes through a sample solution for wavelengths in the visible (and sometimes ultraviolet) range. Based on the dissolved solute's properties, we can quantify that solute'sconcentration. If the solute's concentration is too high, a relatively small amount of incident light having the wavelength that corresponds to the solute's absorption behavior will reach thedetector, which results in an inaccurate measurement (consider the result if you blocked the light with an opaque disc). If the solution is diluted, less light is absorbed, and the detector inthe spectrophotometer can generate more reliable data.When talking about small quantities, it is important to understand the relationship between common prefixes. Here are some relationships that may help when working through Part C.1L 1000 mL== 1,000,000 μLOR1 mL =AnP Pearson5UnitsFSopenvellum.ecollege.comDWGw ?6F60.001 LMacBook AirTY=&744F7U8DIIF81000 μL(9F90)0AF10-PⒸ +F11+=F128

Answered: Image /qna-images/answer/cacae2b4-99d6-493c-8671-1e5bd69f6879.jpg

In order to understand how this equation is derived and why it holds true, the product of each side of the equation should be examined. What value and unit do you get when you multiply a concentration of 0.735 M by a volume of 0.400 L? This question can be expressed as 0.400 L x 1 Express your answer to three significant figures with the appropriate units. View Available Hint(s) 0.735 mol L

In order to understand how this equation is derived and why it holds true, the product of each side of the equation should be examined. What value and unit do you get when you multiply a concentration of 0.735 M by a volume of 0.400 L? This question can be expressed as 0.400 L x 1 Express your answer to three significant figures with the appropriate units. View Available Hint(s) 0.735 mol L

Chemistry

10th Edition

ISBN:9781305957404

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

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

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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What is the weight in grams of 3.98 × 10^13 CF, molecules? Hint: must use Avogadro's number Put vour answer in box in scientific notation with one past the decimal in this format xx*10^-X As always, this may seem like the wrong sig figs but this is how the formula questions work--I must choose one amount for everybody despite the amount of sig figs Canvas gives you
Suppose the following chemical reaction can take place in this mixture
Converting Concentrations to Different Units Introduction: Depending on how the concentration unit is defined, concentration can serve as a conversion between the amount of a solute and the amount of a solution or solvent. Concentrations can therefore be helpful in a number of stoichiometry issues. The original definition of the concentration unit is frequently preferred because it contains the conversion factor. Today, for chemists, it is suitable to make a solution in the lab using a specific concentration unit like molarity, and then convert the concentration to another unit. Today in the lab, I will prepare a sodium bicarbonate (baking soda) solution with a specific molarity. I will then translate the concentration into molality, mass percent, and mole fraction. Procedure The first step of this lab was to make sure I had everything I needed and that my surroundings were safe. I then looked at an empty beaker weighing paper and a balance. I first took a piece of weighing…
1. A bright yellow precipitate is produced in a laboratory reaction, when lead(II) nitrate solution is reacted with potassium iodide solution. How many moles of a precipitate are produced when 256 mL of a 1.5 mol/L potassium iodide solution is reacted with excess lead(II) nitrate solution? Hint: Write a balanced chemical reaction equation first.(Record only your final answer with the correct number of significant digits and the proper units.)
How many grams of silver chloride will form when 20 mL of 0.10 M sodium chloride reacts with excess silver nitrate? Follow the steps: How many moles of silver chloride is produced from 1 mole of sodium chloride Number of moles of silver chloride formed in the reaction Molar mass of silver chloride Mass of silver chloride produced Unit 13 24 0.5 0.0025 0.002 0.0036 243.32 223.36 || 143.32 || 141.32 0.31 || 2.31 || 0.26 || 0.29 Kg g/mol mol
Read the information below, then answer the question. A laboratory assistant was experimenting with chemical reactions when she combined a small amount of zinc (Zn) with hydrochloric acid (HCI). She discovered that zinc chloride (ZnCl2) and hydrogen (H2) were produced. Equations Which of the following substances are the products? O ZnCl2 + H2 O ZnCl2 + HCI O Zn + H2 O Zn + HCI 6. 7 8. 10 11 12 13 14 Help Center | Schoology Blog | PRIVACY POLICY I Terms W pe here to search 近

When solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates out of solution according to the equation AGNO3 (aq) + NaCI(aq)→AgCl(s) + NaNO3(aq) Part A What mass of silver chloride can be produced from 1.69 L of a 0.214 solution of silver nitrate? Express your answer with the appropriate units. • View Available Hint(s) ? mass of AgCl = Value Units Submit Part B The reaction described in Part A required 3.48 L of sodium chloride. What is the concentration of this sodium chloride solution? Express your answer with the appropriate units. • View Available Hint(s) HA ? Value Units Submit
< For the following reaction, 62.5 grams of silver nitrate are allowed to react with 27.9 grams of copper(II) chloride. silver nitrate (aq) + copper(II) chloride (s) → silver chloride (s) + copper(II) nitrate (aq) What is the maximum amount of silver chloride that can be formed? [Review Topics] [References] Use the References to access important values if needed for this question. What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? Submit Answer Retry Entire Group Show Hint 9 more group attempts remaining grams grams Previous Next Sve and Exit Save and Exit
Write out the balanced equations between 1) calcium carbonate and hydrochloric acid (HCl), and 2) sodium carbonate and hydrochloric acid. You then need to determine how much (volume in mL) of HCl is required to react with 2 g of calcium carbonate, and 2 g of sodium carbonate. Think about calculating moles and finding volumes.
She tells you that instead of titrating metal carbonates, it should be possible to react them with an excess of acid and collect the carbon dioxide gas that is generated by the reaction in a eudiometer. From the volume of carbon dioxide, you can calculate the moles of metal carbonate that reacted and use this information to determine the molar mass of the metal carbonate. She gives you a sample of another unknown metal carbonate to test this approach. You weigh out 0.1836 g of the unknown sample and react it with an excess of hydrochloric acid inside a eudiometer at 18.4 °C and 0.956 atm. Once the reaction is complete, you find that 17.20 mL of carbon dioxide gas were produced. a) How many moles of metal carbonate reacted? mol b) What is the molar mass of the metal carbonate? g/mol c) What is the chemical symbol of the metal cation?