{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

E1_S08 - Exam#1 Chemistry 112 Spring 2008 Name_Student ID...

Info icon This preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Exam #1 Chemistry 112 Spring 2008 Name: ___________________________Student ID:____________ Show your work!!!!!! Henry’s Law: C = k P Raoult’s Law: P solvent = P° solvent χ solvent Boiling Point Elevation: Δ T B = i•m•k B Freezing Point Depression: Δ T F =i•m•k F Osmotic Pressure: Π = i•M•R•T R = 0.0821 L•atm/mol•K 1 atm = 760 torr aA + bB cC + dD K eq = [ C ] c [ D ] d [ A ] a [ B ] b K p = K c (RT) Δ n ay 2 + by + c = 0 y = b ± b 2 4ac 2a
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
1 (20 points). A 10.0 % solution of sugar (C 12 H 22 O 12 ) in water (H 2 O) has a density of 1.04 g/mL. Express this concentration as a) molality (m) b) molarity (M)
Image of page 2
2 (20 points). The vapor pressure of trichloroethane (C 2 H 3 Cl 3 ) is 100 torr at 20 °C. What is a vapor pressure of trichloroethane at 20 °C above a solution containing 2.00 g of ferrocene (Fe(C 5 H 5 ) 2 ) in 25 g of trichloroethane? Ferrocene is a non-electrolyte.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
3 (15 points). K c at 970 K is 0.38 M for: CaCO 3 (s) CO 2 (g) +CaO(s) a. Write expression for K c b. Calculate K c at 970 K for CO 2 (g) +CaO(s) CaCO 3 (s) c. Calculate K p at 970 K for CO 2 (g) +CaO(s) CaCO 3 (s)
Image of page 4
4 (15 points). The equilibrium constant, K c , for the following reaction is 9.52x10
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

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