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Exam 1 with Answers - General Chemistry for Scientists and Engineers I | CHEM 201, Exams of Chemistry

University of Nevada - Reno (UNR)Chemistry

Material Type: Exam; Class: Gen Chem Scien Engr I; Subject: Chemistry; University: University of Nevada-Reno; Term: Fall 2015;

Typology: Exams

2014/2015

Uploaded on 12/14/2015

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CHEM 201, Fall 2015
Practice Exam 1 (100 points)
Name:____________KEY____________________________
(PRINT, last name first)
Please do not start working on the exam until you are told to begin. Use the last page with
periodic table as a scratch paper.
pf3
pf4
pf5

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Download Exam 1 with Answers - General Chemistry for Scientists and Engineers I | CHEM 201 and more Exams Chemistry in PDF only on Docsity!

CHEM 201 , Fall 2015 Practice Exam 1 (100 points) Name: ____________KEY____________________________ (PRINT, last name first) Please do not start working on the exam until you are told to begin. Use the last page with periodic table as a scratch paper.

A. Multiple choice. Circle the most correct answer AND write your letter choice on the line. (3 points each) ___A___1. Which of the following contains the largest number of atoms? (A) 1g H 2 (B) 1g F 2 (C) 1g Ne (D) 1g He ___A___2. An example of an alkaline earth metal is: (A) magnesium (B) fluorine (C) zinc (D) lithium ___B____3. Which of the following is a non-­‐metal? (A) Ca (B) S (C) Cd (D) In ___B___4. Which molecular formula is also an empirical formula? (A) P 4 O 10 (B) C 3 H 8 (C) C 5 H 10 (D) N 2 O 4 ___C___5. The hypochlorite ion is: (A) ClO 3 –^ (B) ClO 2 –^ (C) ClO–^ (D) Cl– ___D___6. An atom with 23 protons and 28 neutrons is an isotope of which element? (A) S (B) Se (C) Sn (D) V ___B___7. Which of the following statements about carbon is TRUE? (A) Every carbon atom has a mass of 12.01 amu. (B) The atomic mass, 12.01 amu, is an average value based on isotopic composition. (C) The mass of one mole of 12 C atoms is 12.01 grams. (D) 12 C has 7 neutrons. ___C___8. In the following redox reaction occurring in acidic solution, what is the coefficient for Cl 2 in the balanced equation? Cr 2 O 72 –^ ( aq ) + Cl−^ ( aq ) -­‐-­‐-­‐> Cl 2 ( g ) + Cr3+^ ( aq ) (A) 1 (B) 2 (C) 3 (D) 4 ___B___9. In the reaction in #8 above, which species is oxidized? (A) Cr 2 O 7 2−^ (B) Cl–^ (C) Cl 2 (D) Cr3+ ___B___10. Which compound is soluble in water? (A) AgCl (B) Na 2 SO 4 (C) MgS (D) CaCO 3

C. Answer the following questions in the space provided. To receive full credit, report your answers with units to the proper number of significant figures.

  1. Mixing solutions of aqueous copper(II) nitrate and aqueous sodium hydroxide forms a solid precipitate. (10 points total) (a) Write the balanced molecular equation for the reaction. Include state symbols. ( points) Cu(NO 3 ) 2 (aq) + 2NaOH(aq) → Cu(OH) 2 (s) + 2NaNO 3 (aq) (b) Write the net ionic equation for the reaction. Include state symbols. (3 points) Cu^2 +(aq) + 2OH-­‐(aq) → Cu(OH) 2 (s) (c) While performing this reaction, you started with 75.0 mL of a 0.250 M copper(II) nitrate solution. If the volume of the sodium hydroxide solution used to complete the above (part a) precipitation reaction was 84.2 mL (to the equivalence point), what is the concentration of the sodium hydroxide solution? ( 7 points) mol of Cu(NO 3 ) 2 = 0.250 mol/L * 0.075 L = 0.01875 mol mol of NaOH = 0.01875 mol * 2 = 0.0375 0 mol Molarity of NaOH = 0.0375 0 mol / 0.0842 L = 0.445 M
  2. Aluminum readily oxides in air to form aluminum oxide. In a certain experiment, 25.0 g aluminum was exposed to 25.0 g oxygen gas, forming the oxide. (10 points total) (a) Write a balanced equation for the reaction. (3 points) 4Al + 3O 2 → 2Al 2 O 3 (b) What is the limiting reactant? (3 points) mol of Al = 25.0 g / 26.98 g/mol = 0.9266 mol mol of O 2 = 25.0 g / 32.00 g/mol = 0.7813 mol 0.9266 mol Al / 0.7813 mol O 2 = 1.186 < 4/3 Al – limiting reactant (c) What is the theoretical yield of aluminum oxide in grams? (7 points) mol of Al 2 O 3 = 0.9266 mol / 2 = 0.4633 mol g of Al 2 O 3 =0.4633 mol * 101.96 g/mol = 47.2 g
  1. Combustion analysis reveals that an unknown compound is composed of 93.71% C and 6.29 0 % H. The mass spectrum shows that its molar mass is 128.17 g/mol. Determine the empirical formula (7 pts.) and the molecular formula (7 pts). (14 points total) Assuming 100.0 g sample of unknown compound: 93.71 g C / 100.0 g sample * 128.17 g/mol = 120.11 g C/ mol sample 6.290 g H / 100.0 g sample * 128.17 g/mol = 8.0619 g H / mol sample 120.11 g C / mol sample * 1 mol C / 12.01 g C = 10.00 mol C / mol sample 8.0619 g H / mol sample * 1 mol H / 1.008 g H = 7.998 mol H / mol sample Empirical Formula Molecular Formula C 5 H 4 C 10 H 8