Review questions and answers
Review questions and answers Chem 110
Loyola Marymount University
Popular in General Chemistry I
verified elite notetaker
Popular in Science
This 2 page Study Guide was uploaded by Alyssa Weisblatt on Sunday October 9, 2016. The Study Guide belongs to Chem 110 at Loyola Marymount University taught by Dr. Jennifer Casey in Fall 2016. Since its upload, it has received 42 views. For similar materials see General Chemistry I in Science at Loyola Marymount University.
Reviews for Review questions and answers
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/09/16
Midterm Review – Questions and Problems 1. Formula for Molarity? 2. How to classify a precipitate reaction? 3. Learn solubility rules!!!!! 4. What is important in a net ionic equation? 5. Acid vs. Base in reactions? 6. Learn acid/base rules!!!! 7. How to achieve neutralization? 8. Oxidized vs. Reduced? 9. Reducing agent vs. oxidizing agent? 10.Formula for Boyle’s Law? 11.Formula for Charle’s Law? 12.Formula for Avogadro’s Law? 13.Formula for ideal gas law? 14.What is STP? 15.Formula for gas mixtures? 16.Formula for Dalton’s Law? 17.Formula for rms speed? 18.Formula for effusion? 19.Formula for real gases? 1. 255 g KBr in water to make 1.75 L of solution, what is the molarity? 2. Have 10.0 M of CaCl₂ solution and want 3.00 L of .500 M, how much do you need to transfer? 3. AgNO₃ (aq) + KCl (aq) → ? is this a precipitate reaction? 4. Pb(NO₃)₂ (aq) + 2KI (aq) → ? is this a precipitate reaction? If so, what’s the net ionic equation? 5. HCl (aq) + H₂O (l) → ? which is the acid? Which is the base? 6. 20 mL of H₂SO₄ with unknown M, requires 22.87 mL of .158 M KOH to reach equivalence point, what is the value of the unknown M? 7. 2Na (s) + 2H₂O (l) → 2NaOH (aq) + H₂ (g) what is the oxidizing agent? 8. Temperature= 125° C, Pressure= 755 mmHg, what is the density of N₂? 9. CO (g) + 2H₂ (g) → CH₃OH (g) Have 35.7 g CH₃OH, what is the volume of H₂ at a temperature of 355 K and a pressure of 738 mmHg? 10.1.00 L of He, Ar, and Ne with total pressure equaling 662 mmHg, temperature of 298 K, pressure of He being 341 mmHg, and pressure of Ne being 112 mmHg, how many grams of Ar are there? 11.How fast is O₂ going at 25° C? 12.An unknown gas effuses at a rate that is 0.462 times that of N₂ at the same temperature, what is the molar mass of the unknown gas? 13.At 27° C, 130 atm, 1.50 L, and 10.0 mol, is this gas ideal? Midterm Review – Answers 1. Amount of solute (mol) / amount of solution (L) 2. A solid will be formed 3. Learn solubility rules!!! 4. The reactants that will form the solid 5. Acids are proton donors and bases are proton acceptors 6. Learn acid/base rules!!! 7. Moles of acid = moles of base to achieve neutralization 8. If a species loses electrons it is oxidized; if it gains electrons, it is reduced 9. The reducing agent is the chemical that was oxidized; the oxidizing agent was reduced 10.P₁V₁= P₂V₂ 11.V₁ / T₁ = V₂ / T₂ 12.V₁ / n₁ = V₂ / n₂ 13.PV = nRT 14.Standard temperature and pressure, with temperature being 0° C and pressure being 1 atm 15.n(total) (RT/V) 16.Pₐ = χₐP₊ 17.√3RT/MM 18.√MM(B) / MM(A) 19.[P + a(n/V)²][V – nb] = nRT 1. .122 M 2. 0.150 L 3. AgCl (s) + KNO₃ (aq) yes 4. PbI₂ (s) + 2KNO₃ (aq) yes Pb²⁺ (aq) + 2I⁻ (aq) → PbI₂ (s) 5. HCl is acid, H₂O is base 6. H₂SO₄ + 2KOH → 2H₂O + 2K⁺ + SO₄²⁻ molarity of H⁺ = .181 .181 M H⁺ ( 1 M H₂SO₄ / 2 M H⁺ ) = .0903 M 7. Oxidizing agent: H₂O (l) 8. (n/V) = .993 atm/ (.08206 L atm/ mol K)(398 K) = .0304 mol/L .0304 mol N₂/L (28.014 g N₂ / 1 mol N₂) = .852 g/L 9. 2.23 mol H₂ 2.23(.08206 L atm/ mol K)(355 K) / .971 atm = 66.9 L 10.P(Ar) = .275 atm .275 atm(1.00 L) / (.08206 L atm / mol K)(298 K) = . 0112 mol Ar .0112 mol Ar (39.95 g Ar / 1 mol Ar) = 0.449 g Ar 11.482 m/s 12.131 g/mol 13.P = 164 atm not ideal because pressures are different
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'