Week 4 Text Exercises-Part III - Week 4 Text Exercises Part...

This preview shows page 1 out of 1 page.

Week 4 Text Exercises, Part III Prepare written answers to the following assignments from Introductory Chemistry: Exercise 50 in Ch. 13 A saline solution contains 1.1% NaCl by mass. How much NaCl is present in 87.2 g of this solution? Answer: 1.1 g / 100 g = 0.011 0.011 * 87.2 g =0.9592 =0.96 g Exercise 56 in Ch. 14 A 5.00-mL sample of an H 3 PO 4 solution of unknown concentration is titrated with a 0.1003 M NaOH solution. A volume of 6.55 mL of the NaOH solution was required to reach the endpoint. What is the concentration of the unknown H 3 PO 4 solution? Answer: 3NaOH(aq) + H3PO4(aq) --> Na3PO4(aq) + 3H2O(l) 3 moles NaOH are required to neutralise 1 mole of H3PO4 moles NaOH used = molarity x litres = 0.1003 M x 0.00655 L = 0.000656965 moles NaOH 3 moles NaOH neutralise 1 mole H3PO4, then
Image of page 1

Unformatted text preview: moles H3PO4 = 1/3 x moles NaOH = 0.000218988 moles This is how many moles of H3PO4 were in the 5.00 ml sample titrated Molarity H3PO4 = moles / litres = 0.000218988 mol / 0.00500 L = 0.0438 M Exercise 62 in Ch. 15 An equilibrium mixture of the following reaction was found to have [SO 3 ] = 0.391 M and [O 2 ] = 0.125 M at 600 °C. What is the concentration of SO 2 ? 2 SO 2 ( g ) + O 2 ( g ) ⇌ 2 SO 3 ( g ) K eq = 4.34 at 600 °C Answer: As Keq = [SO3]^2 / [O2] / [SO2]^2 4.34=[(0.391^2)/0.125]*[SO2]^2 4.34*0.125=(0.391^2)*[SO2]^2 (0.391^2) / [4.34*0.125]=[SO2]^2 0.152881 / 0.5425 =SO2^2 0.28180 = SO2^2 SO2=0.531 M Exercise 52 in Ch. 16 What is the oxidation state of Cr in each of the following compounds? A. CrO = +2 B. CrO 3 = +6 C. Cr 2 O 3 = +3...
View Full Document

  • Winter '12
  • EarlGurley
  • Chemistry, Mole, Sodium hydroxide, Molar concentration

{[ snackBarMessage ]}

Get FREE access by uploading your study materials

Upload your study materials now and get free access to over 25 million documents.

Upload now for FREE access Or pay now for instant access
Christopher Reinemann
"Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

Ask a question for free

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern