Chem 12 Prov Exam PLO Review
... • reactions are the result of collisions between reactant particles • not all collisions are successful • sufficient kinetic energy (KE) and favourable geometry are required • to increase the rate of a reaction one must increase the frequency of successful collisions • energy changes are involved in ...
... • reactions are the result of collisions between reactant particles • not all collisions are successful • sufficient kinetic energy (KE) and favourable geometry are required • to increase the rate of a reaction one must increase the frequency of successful collisions • energy changes are involved in ...
Practice Exam I FR Answers and Explanations
... No two students will write the same procedure. The most essential steps that will earn points would include: (1pt) React the alloy with nitric acid. (1pt) Precipitate the silver with sodium chloride. (1pt) Dry the sample of silver chloride. ...
... No two students will write the same procedure. The most essential steps that will earn points would include: (1pt) React the alloy with nitric acid. (1pt) Precipitate the silver with sodium chloride. (1pt) Dry the sample of silver chloride. ...
NATIONAL HIGH SCHOOL CHEMISTRY EXAMINATION (1995
... 14. Two containers are filled with gas at the same temperature, one with H2 and the other with CO2 . Which one of the following statements is consistent with the kinetic theory of gases? a) The molecules of the two gases have the same average speed. b) The molecules of the two gases have the same a ...
... 14. Two containers are filled with gas at the same temperature, one with H2 and the other with CO2 . Which one of the following statements is consistent with the kinetic theory of gases? a) The molecules of the two gases have the same average speed. b) The molecules of the two gases have the same a ...
35 IChO Problems 1-13
... We assume that the rate of formation has remained constant for thousands of years and is equal to the rate of decay, hence the amount of 14C in the atmosphere has reached steady state. As a result 14C accompanies the stable isotopes 12C and 13C in the atmosphere and participates indistinguishably in ...
... We assume that the rate of formation has remained constant for thousands of years and is equal to the rate of decay, hence the amount of 14C in the atmosphere has reached steady state. As a result 14C accompanies the stable isotopes 12C and 13C in the atmosphere and participates indistinguishably in ...
DRAFT AP® CHEMISTRY 2005 SCORING GUIDELINES
... partially burned wood" in pure hydrogen! Those good students shouldn't think the experiment is done in an atmosphere of pure hydrogen or they are not that good! No mention of a balloon in my text, so where did that notion come from? "Burns" may be enough for the point without an explosion pop, thoug ...
... partially burned wood" in pure hydrogen! Those good students shouldn't think the experiment is done in an atmosphere of pure hydrogen or they are not that good! No mention of a balloon in my text, so where did that notion come from? "Burns" may be enough for the point without an explosion pop, thoug ...
Multiple-choice questions : 1. The following graph shows the volume
... Each question below consists of two separate statements. Decide whether each of the two statements is true or false; if both are true, then decide whether or not the second statement is a correct explanation of the first statement. Then select one option from A to D according to the following table: ...
... Each question below consists of two separate statements. Decide whether each of the two statements is true or false; if both are true, then decide whether or not the second statement is a correct explanation of the first statement. Then select one option from A to D according to the following table: ...
(a) From , 2012 General Chemistry I
... Suppose that 1.00 mol of ideal gas molecules maintained at 292 K and 3.00 atm expands from 8.00 L to 20.00 L and a final pressure of 1.20 atm by two different paths. (a) Path A is an isothermal, reversible expansion. (b) Path B has two parts. In step 1, the gas is cooled at constant volume until its ...
... Suppose that 1.00 mol of ideal gas molecules maintained at 292 K and 3.00 atm expands from 8.00 L to 20.00 L and a final pressure of 1.20 atm by two different paths. (a) Path A is an isothermal, reversible expansion. (b) Path B has two parts. In step 1, the gas is cooled at constant volume until its ...
Chemistry Tests Questions
... 4. When silver nitrate solution is added to iron(II) nitrate solution a grey-black precipitate is formed and after a few minutes there is no further change. The mixture is now in dynamic equilibrium. Explain what this term means. ...
... 4. When silver nitrate solution is added to iron(II) nitrate solution a grey-black precipitate is formed and after a few minutes there is no further change. The mixture is now in dynamic equilibrium. Explain what this term means. ...
hong kong diploma of secondary education examination
... N2H4(l) reacts with oxygen to give H2O(g) and N2(g) while N2H2(CH3)2(l) reacts with oxygen to give CO2(g), H2O(g) and N2(g). Given the following standard enthalpy changes of formation, decide whether the reaction of hydrazine or 1,1-dimethylhydrazine with oxygen provides more heat per gram of fuel. ...
... N2H4(l) reacts with oxygen to give H2O(g) and N2(g) while N2H2(CH3)2(l) reacts with oxygen to give CO2(g), H2O(g) and N2(g). Given the following standard enthalpy changes of formation, decide whether the reaction of hydrazine or 1,1-dimethylhydrazine with oxygen provides more heat per gram of fuel. ...
SAMPLE EXERCISE 4.5 Comparing Acid Strengths
... Solve: (a) When bonded to a nonmetal, hydrogen has an oxidation number of +1 (rule 3b). Because the H 2S molecule is neutral, the sum of the oxidation numbers must equal zero (rule 4). Letting x equal the oxidation number of S, we have 2(+1) + x = 0. Thus, S has an oxidation number of –2. (b) Becaus ...
... Solve: (a) When bonded to a nonmetal, hydrogen has an oxidation number of +1 (rule 3b). Because the H 2S molecule is neutral, the sum of the oxidation numbers must equal zero (rule 4). Letting x equal the oxidation number of S, we have 2(+1) + x = 0. Thus, S has an oxidation number of –2. (b) Becaus ...
Document
... Solve: (a) When bonded to a nonmetal, hydrogen has an oxidation number of +1 (rule 3b). Because the H 2S molecule is neutral, the sum of the oxidation numbers must equal zero (rule 4). Letting x equal the oxidation number of S, we have 2(+1) + x = 0. Thus, S has an oxidation number of –2. (b) Becaus ...
... Solve: (a) When bonded to a nonmetal, hydrogen has an oxidation number of +1 (rule 3b). Because the H 2S molecule is neutral, the sum of the oxidation numbers must equal zero (rule 4). Letting x equal the oxidation number of S, we have 2(+1) + x = 0. Thus, S has an oxidation number of –2. (b) Becaus ...
b - PianetaChimica
... Mass spectrometry is an important tool in the determination of the structures of organic compounds. The process begins with the ionisation of the sample to form a positively charged ion, the molecular ion. At this stage, the molecular ion commonly fragments to form additional cations. These cations ...
... Mass spectrometry is an important tool in the determination of the structures of organic compounds. The process begins with the ionisation of the sample to form a positively charged ion, the molecular ion. At this stage, the molecular ion commonly fragments to form additional cations. These cations ...
Name: ______KEY__________________ Date: ______ CHM 130
... 14. (5 pts) An oven cleaning solution is 40.0 % by mass of NaOH. If one jar of this product contains 454 grams of solution, how many grams of NaOH does it contain? Show all work. Mass solute / mass solution x 100 = (x g / 454 g) x 100 = 40.0% ...
... 14. (5 pts) An oven cleaning solution is 40.0 % by mass of NaOH. If one jar of this product contains 454 grams of solution, how many grams of NaOH does it contain? Show all work. Mass solute / mass solution x 100 = (x g / 454 g) x 100 = 40.0% ...
Answer Key, Problem Set 6 – complete, with explanations
... The ions are freely moving about in the solution; I try to show this by showing them randomly distributed throughout. A solution is a homogeneous mixture. Even though there are two different types of ions (as well as water molecules, which are not shown), if you had a macroscopic sample of the solut ...
... The ions are freely moving about in the solution; I try to show this by showing them randomly distributed throughout. A solution is a homogeneous mixture. Even though there are two different types of ions (as well as water molecules, which are not shown), if you had a macroscopic sample of the solut ...
Chapter 8 and 9 – Energy Balances
... solidification) refers to the amount of heat (change in enthalpy) that must be added to a unit amount of material to cause it to melt. Contributions to latent heats include changes in molecular level interactions as well as changes in specific volume that accompany the phase change. Latent heats are ...
... solidification) refers to the amount of heat (change in enthalpy) that must be added to a unit amount of material to cause it to melt. Contributions to latent heats include changes in molecular level interactions as well as changes in specific volume that accompany the phase change. Latent heats are ...
standard enthalpy change of reaction
... using experimental data on temperature changes, quantities of reactants and mass of water. 5.2.4 Evaluate the results of experiments to determine enthalpy changes. Students should be aware of the assumptions made and errors due to ...
... using experimental data on temperature changes, quantities of reactants and mass of water. 5.2.4 Evaluate the results of experiments to determine enthalpy changes. Students should be aware of the assumptions made and errors due to ...
Insertion of SO2 into the Metal−Carbon Bonds of Rhodium and
... proposed that still holds in the majority of cases.1a While current interest in this reaction has decreased, SO2 remains the subject of numerous studies2 because of its diverse coordination properties and its role in acid rain production. In this paper, we present two new systems that are able to cl ...
... proposed that still holds in the majority of cases.1a While current interest in this reaction has decreased, SO2 remains the subject of numerous studies2 because of its diverse coordination properties and its role in acid rain production. In this paper, we present two new systems that are able to cl ...
Stoichiometry - HCC Learning Web
... Step 1: Write the balanced chemical equation for the reaction. Step 2: Calculate the moles of "given" substance. If more than one reactant amount is given, calculate the moles of each to determine which is the limiting reactant. Step 3: Calculate the moles of "desired" substance from your answer in ...
... Step 1: Write the balanced chemical equation for the reaction. Step 2: Calculate the moles of "given" substance. If more than one reactant amount is given, calculate the moles of each to determine which is the limiting reactant. Step 3: Calculate the moles of "desired" substance from your answer in ...
Ch 10 - Enrico Fermi High School
... 1. What effect (increase, decrease, no change) will a decrease in temperature have on K? 2. What effect (inc, dec, none) will removing H2 have on the equilibrium constant, K? 3. In which direction will the reaction shift if gaseous H2 is removed from the system? 4. Adding a catalyst (a gold surface) ...
... 1. What effect (increase, decrease, no change) will a decrease in temperature have on K? 2. What effect (inc, dec, none) will removing H2 have on the equilibrium constant, K? 3. In which direction will the reaction shift if gaseous H2 is removed from the system? 4. Adding a catalyst (a gold surface) ...
AP `94 Multiple Choice
... 16. Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, 20. … Ag+ + … AsH3(g) + … OH– … Ag(s) HC2H3O2. For 20.0 milliliters of the vinegar, 26.7 + … H3AsO3(aq) + … H2O milliliters of 0.600–molar NaOH solution was When the equation above is balanced with lowe ...
... 16. Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, 20. … Ag+ + … AsH3(g) + … OH– … Ag(s) HC2H3O2. For 20.0 milliliters of the vinegar, 26.7 + … H3AsO3(aq) + … H2O milliliters of 0.600–molar NaOH solution was When the equation above is balanced with lowe ...
Common Student Misconceptions
... Compounds whose aqueous solutions conduct electricity poorly are called weak electrolytes • These substances exist as a mixture of ions and un-ionized molecules in solution. • The predominant form of the solute is the un-ionized molecule. • Example: acetic acid, HC2H3O2. HC2H3O2(aq) ⇋ H+(aq) + C2H3O ...
... Compounds whose aqueous solutions conduct electricity poorly are called weak electrolytes • These substances exist as a mixture of ions and un-ionized molecules in solution. • The predominant form of the solute is the un-ionized molecule. • Example: acetic acid, HC2H3O2. HC2H3O2(aq) ⇋ H+(aq) + C2H3O ...
Document
... Another way to state Hess’s law is: If two or more equations with known enthalpy changes can be added together to form a new “target” equation, then their enthalpy changes may be similarly added together to yield the enthalpy change of the target equation. Hess’s law can also be written as an equati ...
... Another way to state Hess’s law is: If two or more equations with known enthalpy changes can be added together to form a new “target” equation, then their enthalpy changes may be similarly added together to yield the enthalpy change of the target equation. Hess’s law can also be written as an equati ...
CHAPTER 12 Study Guide
... 55. a. Initially, the amount of NaCl formed increases as the amount of Na used increases. For this part of the curve, sodium is the limiting reagent. Beyond a mass of about 2.5 g of Na, the amount of product formed remains constant because chlorine is now the limiting reagent. b. Chlorine becomes th ...
... 55. a. Initially, the amount of NaCl formed increases as the amount of Na used increases. For this part of the curve, sodium is the limiting reagent. Beyond a mass of about 2.5 g of Na, the amount of product formed remains constant because chlorine is now the limiting reagent. b. Chlorine becomes th ...
Practice Test Material - Directorate of Education
... When an ideal gas expands in vacuum there is neither absorption nor evolution of heat but when a real gas expands cooling is observed. ...
... When an ideal gas expands in vacuum there is neither absorption nor evolution of heat but when a real gas expands cooling is observed. ...
Thermometric titration
A thermometric titration is one of a number of instrumental titration techniques where endpoints can be located accurately and precisely without a subjective interpretation on the part of the analyst as to their location. Enthalpy change is arguably the most fundamental and universal property of chemical reactions, so the observation of temperature change is a natural choice in monitoring their progress. It is not a new technique, with possibly the first recognizable thermometric titration method reported early in the 20th century (Bell and Cowell, 1913). In spite of its attractive features, and in spite of the considerable research that has been conducted in the field and a large body of applications that have been developed; it has been until now an under-utilized technique in the critical area of industrial process and quality control. Automated potentiometric titration systems have pre-dominated in this area since the 1970s. With the advent of cheap computers able to handle the powerful thermometric titration software, development has now reached the stage where easy to use automated thermometric titration systems can in many cases offer a superior alternative to potentiometric titrimetry.The applications of thermometric titrimetry discussed on this page are by no means exhaustive. The reader is referred to the bibliography for further reading on the subject.