How many milliliters of 0.85 M NAOH are required to react with 30.83 mL of 2.5 M HCl?The balanced equation is HCl + NAOH --> H20 + NaClYou can set this up as a stoichiometry problem: mL HCl --> mol HCI --> mol NaOH --> mL NaOHmL NAOH solutionSubmit Question

Given, Balanced chemical equation: HCl + NaOH ➝ NaCl + H2O Note: 1000 mL =1L Molarity of HCl…

How many milliliters of 0.85 M NAOH are required to react with 30.83 mL of 2.5M HCI? The balanced equation is HCI + NAOH --> H20 + NaCl You can set this up as a stoichiometry problem: mL HCl --> mol HCI --> mol NaOH --> mL NaOH mL NaOH solution Submit Question

How many milliliters of 0.85 M NAOH are required to react with 30.83 mL of 2.5M HCI? The balanced equation is HCI + NAOH --> H20 + NaCl You can set this up as a stoichiometry problem: mL HCl --> mol HCI --> mol NaOH --> mL NaOH mL NaOH solution Submit Question

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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If 0.100 grams of unknown solid are dissolved in water and titrated with 4.68 mL of the NaOH, what is the molar mass of the solid? HINTS: Molar mass is grams / moles. We have grams from the weighed mass, we can calculate moles of acid from the titration. Monoprotic means that it produces 1 H+ per molecule. This means that our moles of NaOH added and moles of acid are the same. You will need to use your concentration of NaOH from the previous question
50.0mL of 0.180 M sodium hydroxide solution is added to 50mL of .200 M lactic acid solution. The reaction goes to completion. It can be written CH3CHOH2H + OH ---> CH3CHOHCO2 + H2O What is the limiting reactant? How many moles of it were used?
Will the calculated Molarity of NaOH be too high or too low or unaffected if the following happen: When you answer the question, consider how the situation affects the calculation of molarity of NaOH and moles NaOH, which is calculated from the mass of KHP. For each answer, you must supply a clear explanation for your answer. Often times, students will restate the facts rather than explaining why they chose the answer they chose. a) You add the weighed KHP to a flask containing a 60mL of water rather than 50 mL of water. Explain. Your answer must contain a well-thought-out and clearly written explanation. b) The buret is still wet with water on the inside when you add your NaOH solution. Explain c) The KHP is wet when you weigh it. Explain. d) You titrate past the equivalence point by 0.50mL Answer a-d
Choi is fond of carbonated beverages so he crafted his own soda formulation called “Sodalicious.” Ulan, who is a fellow enthusiast, also made her own formulation which she called “Soda Pop!” to challenge his friend Choi. To know which formulation is better, they sought to determine the carbonic acid content of their soda formulations using titration. In the first part of their experiment, a 0.5 Mlabeled NaOH solution was standardized against 0.350 grams of KHP (MW = 204.22 g/mol) primary standard that is 99.6% pure. The titration required 3.16 mL of the NaOH solution to reach the phenolphthalein endpoint. In the second part of their experiment, Choi and Ulan independently prepared their samples by taking 25.0-mL of their soda sample and diluting it to 30.0 mL. Using the same titrant and indicator, Choi’s “Sodalicious” required 1.39 mL of the standardized titrant to reach the endpoint, while Ulan’s “Soda Pop!” required 0.926 mL of the same titrant to reach the endpoint. A. What is the…
Choi is fond of carbonated beverages so he crafted his own soda formulation called “Sodalicious.” Ulan, who is a fellow enthusiast, also made her own formulation which she called “Soda Pop!” to challenge his friend Choi. To know which formulation is better, they sought to determine the carbonic acid content of their soda formulations using titration. In the first part of their experiment, a 0.5 Mlabeled NaOH solution was standardized against 0.350 grams of KHP (MW = 204.22 g/mol) primary standard that is 99.6% pure. The titration required 3.16 mL of the NaOH solution to reach the phenolphthalein endpoint. In the second part of their experiment, Choi and Ulan independently prepared their samples by taking 25.0-mL of their soda sample and diluting it to 30.0 mL. Using the same titrant and indicator, Choi’s “Sodalicious” required 1.39 mL of the standardized titrant to reach the endpoint, while Ulan’s “Soda Pop!” required 0.926 mL of the same titrant to reach the endpoint. D. Calculate…
Volume of an unknown used was 30 mL, Initial Buret volume was 0 and the Final Buret volume was 8.5 mL. What is the molarity of the unknown solution if the Net volume of NaOH being used was 8.5 mL and Millimoles (mmoles) NaOH reacted was 0.791?
The molar mass of carbon is 12 g/mol. How many moles are there in 3grams of carbon?.please encode all your solution thank you! Approximately, how many grams of phosphoric acid (MW=98 g/mol) are in 175 ml of a 3.5 M phosphoric acid solution?.please encode all your solution thank you! What mass (in grams) of glucose would you need to have 0.8 mol, giventhat the molar mass of glucose is 180g/mol?.please encode all your solution thank you!
If 0.8010 g of sodium oxalate, Na2C2O4, requiresof a 22.96 mL KMnO4 solution to reach the end point, what is the molarity of the KMnO4 solution? _______ (value units)
was made up by dissolving 155 mg of KHP to a final volume of 10.00 mL. Titration of 1.00 mL of this KHP solution with KOH solution required 0.78 mL to reach the end point. What is the molarity of the KOH solution?
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ou 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. first, Determine the amounts of the product; (LiOH) After knowing 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