Ch 17 practice assessment w
... package and sealing. Some perishable items can be sensitive to changes in temperature and humidity. If they are to stay fresh for the longest possible time, they need to be kept in a controlled environment. But, how can this be accomplished if they are traveling in a truck through different weather ...
... package and sealing. Some perishable items can be sensitive to changes in temperature and humidity. If they are to stay fresh for the longest possible time, they need to be kept in a controlled environment. But, how can this be accomplished if they are traveling in a truck through different weather ...
Final Exam Review Sheets
... b. high specific heat capacity (absorbs a lot of heat as it warms up): amount of heat required to change the temperature of an object or body by a given amount c. cohesive forces: tendency of similar or identical particles/surfaces to cling to one another; hydrogen bonds increase cohesiveness of wat ...
... b. high specific heat capacity (absorbs a lot of heat as it warms up): amount of heat required to change the temperature of an object or body by a given amount c. cohesive forces: tendency of similar or identical particles/surfaces to cling to one another; hydrogen bonds increase cohesiveness of wat ...
Chapter 4 Notes
... Example: The atoms in He and N2, for example, have oxidation numbers of 0. 3. The oxidation number of a monatomic ion equals the charge of the ion. Example: oxidation number of Na+ is +1; the oxidation number of N3- is -3. 4. The oxidation number of oxygen in compounds is usually ...
... Example: The atoms in He and N2, for example, have oxidation numbers of 0. 3. The oxidation number of a monatomic ion equals the charge of the ion. Example: oxidation number of Na+ is +1; the oxidation number of N3- is -3. 4. The oxidation number of oxygen in compounds is usually ...
Semester Exam Review
... formation, Hf, of propane given that Hf of H2O(l) = -285.3 kJ/mol and Hf of CO2(g) = -393.5 kJ/mol. (d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00 kilograms of water (specific heat = 4.18 J/g.K), calculate the increase in temperature of water. ...
... formation, Hf, of propane given that Hf of H2O(l) = -285.3 kJ/mol and Hf of CO2(g) = -393.5 kJ/mol. (d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00 kilograms of water (specific heat = 4.18 J/g.K), calculate the increase in temperature of water. ...
Belarus, National Final, 2001 (PDF 149K).
... creased by acidifying the solution (using, for example, nitric acid). a) Calculate the molar concentration of silver acetate in a solution saturated at 20o C, if the density of the solution is 1.01 g/cm3. b) Calculate the solubility product constant for silver acetate. c) What is the pH of a solutio ...
... creased by acidifying the solution (using, for example, nitric acid). a) Calculate the molar concentration of silver acetate in a solution saturated at 20o C, if the density of the solution is 1.01 g/cm3. b) Calculate the solubility product constant for silver acetate. c) What is the pH of a solutio ...
Chemistry Final Exam Review
... PV = nRT → (0.967 atm)V = (0.0612mol )(0.0821 mol • K ( 295 K ) → V = 1.53L ...
... PV = nRT → (0.967 atm)V = (0.0612mol )(0.0821 mol • K ( 295 K ) → V = 1.53L ...
SOLLIQSOL questions
... The normal boiling and freezing points of argon are 87.3 K and 84.0 K, respectively. The triple point is at 82.7 K and 0.68 atmosphere. (a) Use the data above to draw a phase diagram for argon. Label the axes and label the regions in which the solid, liquid and gas phases are stable. On the phase di ...
... The normal boiling and freezing points of argon are 87.3 K and 84.0 K, respectively. The triple point is at 82.7 K and 0.68 atmosphere. (a) Use the data above to draw a phase diagram for argon. Label the axes and label the regions in which the solid, liquid and gas phases are stable. On the phase di ...
precipitate - UniMAP Portal
... most determination the precipitate is of such low solubility that losses from dissolution are negligible. An additional factor is the common ion effect, this further decrease the solubility of the precipitate. E.g. When Ag+ is precipitated out by addition of ClAg+ + Cl- = AgCl The low solubility of ...
... most determination the precipitate is of such low solubility that losses from dissolution are negligible. An additional factor is the common ion effect, this further decrease the solubility of the precipitate. E.g. When Ag+ is precipitated out by addition of ClAg+ + Cl- = AgCl The low solubility of ...
Chemical Equilibrium – Le Chatelier`s Principle
... constant temperature, then the equilibrium is “shifted to the right”, which means that the new equilibrium concentrations are obtained by a net increase of the forward reaction until the new equilibrium is established. The equilibrium constant remains unchanged. If, however, the temperature is chang ...
... constant temperature, then the equilibrium is “shifted to the right”, which means that the new equilibrium concentrations are obtained by a net increase of the forward reaction until the new equilibrium is established. The equilibrium constant remains unchanged. If, however, the temperature is chang ...
E:\My Documents\sch4u\SCH4U review McKay answers.wpd
... n = )H / )Hrxn ()Hrxn = )H/n rearranged) = -201 kJ / -1487 kJ/mol = 0.135 mol 4) From your text on pages. 414-416 answer 13: a) rate doubles, b) rate drops by half, c) no change (reactants are not gases). 14: a) a series of steps, since there are 6 moles of reactants, see back of text for b) 15: a) ...
... n = )H / )Hrxn ()Hrxn = )H/n rearranged) = -201 kJ / -1487 kJ/mol = 0.135 mol 4) From your text on pages. 414-416 answer 13: a) rate doubles, b) rate drops by half, c) no change (reactants are not gases). 14: a) a series of steps, since there are 6 moles of reactants, see back of text for b) 15: a) ...
Solutions
... One of the main difficulties in the study of solutions is the wide variety of variables and units used to specify concentration (in the case of ideal mixtures, only molar fractions, and sometimes mass fractions, were common). Generically speaking, concentrations in a mixture express the quantitative ...
... One of the main difficulties in the study of solutions is the wide variety of variables and units used to specify concentration (in the case of ideal mixtures, only molar fractions, and sometimes mass fractions, were common). Generically speaking, concentrations in a mixture express the quantitative ...
Gas Laws
... 18. The rate of effusion of an unknown gas was determined to be 2.92 times faster than that of ammonia. What is the approximate molecular weight of the unknown gas? 19. In the reaction, N2 + H2 NH3, how many mL of nitrogen, measured at STP, are required to produce 400 mL of NH3, measured at STP? H ...
... 18. The rate of effusion of an unknown gas was determined to be 2.92 times faster than that of ammonia. What is the approximate molecular weight of the unknown gas? 19. In the reaction, N2 + H2 NH3, how many mL of nitrogen, measured at STP, are required to produce 400 mL of NH3, measured at STP? H ...
Gas Laws
... 15. If excess hydrochloric acid is added to 13.5 grams of Al, what volume of hydrogen gas will be produced if the gas is collected at a temperature of 80.0 oC and a pressure of 750. torr? 22.1 L 16. At a certain temperature, the velocity of chlorine molecules is 0.0410 m/s. What is the velocity of s ...
... 15. If excess hydrochloric acid is added to 13.5 grams of Al, what volume of hydrogen gas will be produced if the gas is collected at a temperature of 80.0 oC and a pressure of 750. torr? 22.1 L 16. At a certain temperature, the velocity of chlorine molecules is 0.0410 m/s. What is the velocity of s ...
File
... c) NH4Cl (s) NH3 (g) + HCl(g) 5. At 25 °C, Kc =0.0146 for the following reaction: PCl5 PCl3 + Cl2 If, at equilibrium, the molar concentrations for PCl5 and PCl3 are 0.500 M and 0.200 M respectfully, calculate the concentration of chlorine gas. (0.0365M) 6. Consider the reaction: CO + 2H2 CH ...
... c) NH4Cl (s) NH3 (g) + HCl(g) 5. At 25 °C, Kc =0.0146 for the following reaction: PCl5 PCl3 + Cl2 If, at equilibrium, the molar concentrations for PCl5 and PCl3 are 0.500 M and 0.200 M respectfully, calculate the concentration of chlorine gas. (0.0365M) 6. Consider the reaction: CO + 2H2 CH ...
Chapter 7 Review
... a) Write the chemical reaction for the Haber process and write a K equation to describe it. (2) b) Why was this reaction so important when it was developed back in 1909? (2) c) How did Haber manage to keep this reaction moving forward to produce ammonia? (4) ...
... a) Write the chemical reaction for the Haber process and write a K equation to describe it. (2) b) Why was this reaction so important when it was developed back in 1909? (2) c) How did Haber manage to keep this reaction moving forward to produce ammonia? (4) ...
COMPCHEM5_2011
... • an energy cutoff in the size of the solute-solvent interaction energy. • those solvent molecules that make contact with the exposed van der Waals surface area of the solute • those components of the free energy that correlate statistically with the solvent accessible surface area or the van der Wa ...
... • an energy cutoff in the size of the solute-solvent interaction energy. • those solvent molecules that make contact with the exposed van der Waals surface area of the solute • those components of the free energy that correlate statistically with the solvent accessible surface area or the van der Wa ...
Chapter 4. Aqueous Reactions and Solution Stoichiometry
... • Use the stoichiometric coefficients to move between reactants and products. • This step requires the balanced chemical equation. • Convert the laboratory units back into the required units. • Convert moles to grams using molar mass. • Convert moles to molarity or volume using M = mol/L. ...
... • Use the stoichiometric coefficients to move between reactants and products. • This step requires the balanced chemical equation. • Convert the laboratory units back into the required units. • Convert moles to grams using molar mass. • Convert moles to molarity or volume using M = mol/L. ...
Types of Reactions and Solution Chemistry
... In an acid-base titration, an indicator is used to show the change from an acidic situation (all acid), and as the base is slowly added and neutralization occurs, the color shift will be towards the basic side. When the moles of acid = moles of base neutralization is said to occur. We note this by t ...
... In an acid-base titration, an indicator is used to show the change from an acidic situation (all acid), and as the base is slowly added and neutralization occurs, the color shift will be towards the basic side. When the moles of acid = moles of base neutralization is said to occur. We note this by t ...
File
... If we have a chemical compound like NaCl, the molar mass will be equal to the molar mass of one atom of sodium plus the molar mass of one atom of chlorine. If we write this as a calculation, it looks like this: (1 atom x 23 grams/mole Na) + (1 atom x 35.5 grams/mole Cl) = 58.5 grams/mole NaCl For o ...
... If we have a chemical compound like NaCl, the molar mass will be equal to the molar mass of one atom of sodium plus the molar mass of one atom of chlorine. If we write this as a calculation, it looks like this: (1 atom x 23 grams/mole Na) + (1 atom x 35.5 grams/mole Cl) = 58.5 grams/mole NaCl For o ...
Test3_sp2012with answers
... _B__11. If H2O(s) → H2O(l) is endothermic and H2O(l) → H2O(g) is endothermic then H2O(g) → H2O(s) A) is endothermic B) is exothermic C) could be either exo or endothermic D) is equal to the sum of the first two reactions. _B__12. The temperature of a liquid is decreased. What happens to the vapor p ...
... _B__11. If H2O(s) → H2O(l) is endothermic and H2O(l) → H2O(g) is endothermic then H2O(g) → H2O(s) A) is endothermic B) is exothermic C) could be either exo or endothermic D) is equal to the sum of the first two reactions. _B__12. The temperature of a liquid is decreased. What happens to the vapor p ...
Multiple Choice Practice. A) P B) S C) Cl D) Li E) 1 F 1. Has the
... D) The solid phase melts if the pressure increases at constant temperature E) The liquid phase vaporizes if the pressure increases at constant temperature 31. Which of the following compounds is most ionic? A) SiCl4 B) BrCl C) PCl3 D) Cl2O E) CaCl2 32. The simplest formula for an oxide of nitrogen t ...
... D) The solid phase melts if the pressure increases at constant temperature E) The liquid phase vaporizes if the pressure increases at constant temperature 31. Which of the following compounds is most ionic? A) SiCl4 B) BrCl C) PCl3 D) Cl2O E) CaCl2 32. The simplest formula for an oxide of nitrogen t ...
Document
... present in a reaction mixture (i.e., solid, liquid, gas, aqueous solution). • If we are to understand reactivity, we must be aware of just what is changing during the course of a reaction. • Sometimes there is no visible change in the solution, but the reaction still occurred ...
... present in a reaction mixture (i.e., solid, liquid, gas, aqueous solution). • If we are to understand reactivity, we must be aware of just what is changing during the course of a reaction. • Sometimes there is no visible change in the solution, but the reaction still occurred ...
No Slide Title
... Determining Direction of Reaction • Q < Kc:ratio of products to reactants is too small, reaction will proceed in forward direction to reach equilibrium. • Q = Kc:the system is at equilibrium. • Q > Kc:ratio of products to reactants is too large, reaction will proceed in reverse direction to reach e ...
... Determining Direction of Reaction • Q < Kc:ratio of products to reactants is too small, reaction will proceed in forward direction to reach equilibrium. • Q = Kc:the system is at equilibrium. • Q > Kc:ratio of products to reactants is too large, reaction will proceed in reverse direction to reach e ...
Microsoft Word
... All acetates are soluble except for Be(CH3COO)2 All phosphates are insoluble except for those of Group I elements and NH4+. All carbonates are insoluble except for those of Group I elements and NH4+. All hydroxides are insoluble except for those of NH4+, Group I, Sr(OH)2, and Ba(OH)2; Ca(OH)2 is sli ...
... All acetates are soluble except for Be(CH3COO)2 All phosphates are insoluble except for those of Group I elements and NH4+. All carbonates are insoluble except for those of Group I elements and NH4+. All hydroxides are insoluble except for those of NH4+, Group I, Sr(OH)2, and Ba(OH)2; Ca(OH)2 is sli ...