MISSED THIS? Read Section14.5 (Pages 594-601); Watch KCV 14.5,IWE 14.4.An aqueous NaCl solution is made using 139 g ofNaCl diluted to a total solution volume of 1.20 L.Calculate the molarity of the solution.Express your answer in moles per liter to three significant figures.▸ View Available Hint(s)M =SubmitΕΠΙ ΑΣΦ?mol/L

Answered: Image /qna-images/answer/8254be81-4338-47fc-a5e0-ca4199b2d679.jpg

Answered: MISSED THIS? Read Section 14.5 (Pages…

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...

See similar textbooks

Similar questions

What mass of NaNO3 can be dissolved in 100 mL H2O at 10oC?
1
Data of NaOH Trial 1: mass=0.537g Volume =23.6ml Trial 2: mass=0.566g Volume = 26.8ml Trial 3: mass= 0.611g Volume = 23.4ml 1) calculate the molarity of NaOH for each trial and find the average molarity.
V,M, = V,M, %3D Commercial grade sulfuric acid has a molarity of 19 M. If 127 mL of H2SO4 is added to water to bring the total volume of the solution to 516 mL, what is the resulting concentration? Report your answer with 4 decimal places. • Previous Next Chp 口85
QUESTION 9 (4b-301-196-3.69) A 196 gram sample of glucose (C6H1206) was used to make a 3.69 M solution. What is the volume of that solution? Show all work and give your answer with three sig figs. Note M(C6H1206) = 180.16 g/mol. QUESTION 10 (4b-401-1.53-1) An intern working in a research lab knocks over a beaker containing an important solution. The beaker originally contained 1.50 L of a 1.53 M solution, but after it spilled it only contained 1 L of solution. The intern decides to hide his mistake by refilling the beaker to contain 1.50 L again by adding pure water (not realizing that this will dilute the solution). What is the new concentration of the solution? Show all work and give your answer with three sig figs. Click Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answers ADR 1600 4101 1O Aa MacBook Air 80 000 O00 DII DD F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 %23 2.
Please don't provide handwritten solution .....
water. Part B Calculate the mass percent of a NaCl solution in which 185 g of NaCl is dissolved in 1.89 L of water at 25 °C (density 0.997 g/mL). %3D Express your answer numerically. > View Available Hint(s) ΑΣφ %(m/m) Submit P Pearson
Part A The forensic technician at a crime scene has just prepared a luminol stock solution by adding 13.0 g of luminol into a total volume of 75.0 mL of H₂O. What is the molarity of the stock solution of luminol? Express your answer with the appropriate units. ► View Available Hint(s) Submit molarity of luminol solution = Value Part B 10 moles of luminol = μA Submit Before investigating the scene, the technician must dilute the luminol solution to a concentration of 6.00 × 10-2 M. The diluted solution is then placed in a spray bottle for application on the desired surfaces. How many moles of luminol are present in 2.00 L of the diluted spray? Express your answer with the appropriate units. ► View Available Hint(s) Value Units Units ? P
Calculate the concentration of sodium hydroxide solution in g/l and in mol/L
Please answer #2 & #3! Thank you!
A student determined the average molarity of their sample of acetic acid to be 0.536 M. Calculate the (w/w)% of the solution. Assume the density of this solution is that of pure water, dH2O = 1000. g/L. Enter a number without the '%' symbol. (I'm struggling to find the Volume from the attached equation(s) that are "meant" to be used for this problem)
Homework Example Sample: NaC1+NaBr + inert excess AGNO, U = 1.000 g Ag: 107.87, Cl: 35.45 AgCl(s) + AgBr(s) = 0.5260 g Na: 22.99 Cl, treatment I AgCl(s) + AgCl(s) Br: 79.90 = 0.4260 g Nacl = ? % : NaBr = ? % Solution if NaCl=x g. NaBr =y g yx187.77 102.89 xx143.32 g of AgCl(s) = : g of AgBr(s) = 58.44 xx143.32 yx187.77 + xx100 = 0.5260 g % NaCl = 58.44 102.89 1.000 xx143.32 ух143.32 = 0.4260 g yx100 % NaBr =: 58.44 102.89 1.000

SEE MORE QUESTIONS

Recommended textbooks for you

Chemistry

ISBN:

9781305957404

Author:

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

Publisher:

Cengage Learning

Chemistry

ISBN:

9781259911156

Author:

Raymond Chang Dr., Jason Overby Professor

Publisher:

McGraw-Hill Education

Principles of Instrumental Analysis

Chemistry

ISBN:

9781305577213

Author:

Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:

Cengage Learning

Chemistry

ISBN:

9781305957404

Author:

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

Publisher:

Cengage Learning

Chemistry

ISBN:

9781259911156

Author:

Raymond Chang Dr., Jason Overby Professor

Publisher:

McGraw-Hill Education

Principles of Instrumental Analysis

Chemistry

ISBN:

9781305577213

Author:

Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:

Cengage Learning

Organic Chemistry

Chemistry

ISBN:

9780078021558

Author:

Janice Gorzynski Smith Dr.

Publisher:

McGraw-Hill Education

Chemistry: Principles and Reactions

Chemistry

ISBN:

9781305079373

Author:

William L. Masterton, Cecile N. Hurley

Publisher:

Cengage Learning

Elementary Principles of Chemical Processes, Bind…

Chemistry

ISBN:

9781118431221

Author:

Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard

Publisher:

WILEY

SEE MORE TEXTBOOKS