Exam 3 - Version 090 Exam 3 Lyon (52370) 1 This print-out...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 090 Exam 3 Lyon (52370) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. PLEASE BE SURE YOU BUBBLE YOUR NAME, EID AND EXAM VERSION NUM- BER CORRECTLY ON YOUR ANSWER SHEET. 001 10.0 points 50 g of Fe reacts with 43 g of S to produce Fe 2 S 3 , the only product, with no reactant left over. How much Fe 2 S 3 would be produced from 78 g of Fe and excess S? 1. 90.7 g 2. 128 g 3. 169 g 4. 145 g correct 5. 93 g 6. 121 g 7. 0.0184 g Explanation: m ini , Fe = 50 g m ini , S = 43 g m Fe = 78 g By the law of conservation of mass, 50 g of Fe and 43 g of S would produce 93 g of Fe 2 S 3 , if no Fe and S are remaining at the end of the reaction. By the law of definite proportions then, the mass ratio of product Fe 2 S 3 to reactant Fe is unchanging. Thus by setting up a ratio, we can determine the amount of product formed, if all of the Fe is consumed: 93 g Fe 2 S 3 50 g Fe = x g Fe 2 S 3 78 g Fe x = 145 g Fe 2 S 3 . 002 10.0 points Which of the following is NOT a strong acid? 1. chloric acid 2. nitric acid 3. hydroiodic acid 4. hydrofluoric acid correct 5. hydrochloric acid 6. hydrobromic acid 7. perchloric acid 8. sulfuric acid Explanation: Hydrofluoric acid is the only acid that is NOT on the list of 7 strong acids. It is a weak acid. 003 10.0 points The following experiment was carried out us- ing a newly synthesized chlorofluorocarbon. Exactly 50 mL of the gas effused through a porous barrier in 157 s. The same volume of argon effused in 76 s under the same condi- tions. Which compound is the chlorofluoro- carbon? 1. C 2 Cl 3 F 3 2. C 2 Cl 4 F 2 3. C 2 Cl 2 F 4 correct 4. C 2 Cl 5 F 5. C 2 ClF 5 Explanation: t CFC = 157 s V = 15 mL t Ar = 76 s MM Ar = 40 g/mol MMs for each gas (in g/mol): C 2 ClF 5 : 154 . 466 C 2 Cl 2 F 4 : 170 . 921 C 2 Cl 3 F 3 : 187 . 376 C 2 Cl 4 F 2 : 203 . 83 C 2 Cl 5 F : 220 . 285 The rate r = V t 1 T for constant volume, so Version 090 Exam 3 Lyon (52370) 2 r Ar r CFC = t CFC t Ar By Grahams Law, r Ar r CFC = radicalBigg MM CFC MM Ar = t CFC t Ar MM CFC = parenleftbigg t CFC t Ar parenrightbigg 2 MM Ar = parenleftbigg 157 s 76 s parenrightbigg 2 (40 g / mol) = 170 . 699 g / mol , which closely matches C 2 Cl 2 F 4 in molecular mass. 004 10.0 points At constant temperature, the volume occu- pied by a definite mass of a gas is inversely proportional to the applied pressure. This is a statement of which law? 1. Avogadros 2. Charles 3. Grahams 4. Boyles correct 5. Daltons Explanation: Boyles Law relates the volume and pres- sure of a fixed (definite) mass of gas at con- stant temperature....
View Full Document

Page1 / 8

Exam 3 - Version 090 Exam 3 Lyon (52370) 1 This print-out...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online