GCE Chemistry Question Paper Unit 04 - Kinetics, Equilibria
... Use the data from Experiment 1 to calculate a value for the rate constant (k) at this temperature. Deduce the units of k. Calculation ......................................................................................................................... ...
... Use the data from Experiment 1 to calculate a value for the rate constant (k) at this temperature. Deduce the units of k. Calculation ......................................................................................................................... ...
UNITS OF CONCENTRATION
... A 50.00 mL sample of water was treated as above and the I2 liberated was titrated with 8.11 mL of 0.01136 N Na2S2O3 to reach the end point. Determine the concentration of the dissolved O2 in equiv./L, moles/L and mg/L. ...
... A 50.00 mL sample of water was treated as above and the I2 liberated was titrated with 8.11 mL of 0.01136 N Na2S2O3 to reach the end point. Determine the concentration of the dissolved O2 in equiv./L, moles/L and mg/L. ...
thermdyn - chemmybear.com
... lojoules per mole, yet the standard heat of formation of (d) Gf = Hf - TSf = 533.8 - (298)(-0.5239) kJ sodium chloride from its elements in their standard state is -411 kilojoules per mole. = -377.7 kJ (a) Name the factors that determine the magnitude of the standard heat of formation of solid ...
... lojoules per mole, yet the standard heat of formation of (d) Gf = Hf - TSf = 533.8 - (298)(-0.5239) kJ sodium chloride from its elements in their standard state is -411 kilojoules per mole. = -377.7 kJ (a) Name the factors that determine the magnitude of the standard heat of formation of solid ...
materials required/recommended for this paper
... Answer the questions according to the following instructions. Section One: Answer all questions on the separate Multiple-choice Answer Sheet provided. For each questions shade the box to indicate your answer. Use only a blue or black pen to shade the boxes. If you make a mistake, place a cross throu ...
... Answer the questions according to the following instructions. Section One: Answer all questions on the separate Multiple-choice Answer Sheet provided. For each questions shade the box to indicate your answer. Use only a blue or black pen to shade the boxes. If you make a mistake, place a cross throu ...
H 2 (g)
... Specific heat capacity, c of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius (Jg 1C1). Heat capacity, C Heat capacity,C is the amount of heat required to raise the temperature of a given quantity of the substance by one de ...
... Specific heat capacity, c of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius (Jg 1C1). Heat capacity, C Heat capacity,C is the amount of heat required to raise the temperature of a given quantity of the substance by one de ...
Chemistry-Maths-Student-Guide
... It is also perfectly acceptable to work in quantities measured in grams. The expected amount formed is found from the initial amount in moles of the reactant (the one that is not in excess) and the chemical equation for the reaction. Note that questions will be set both on calculating yields and cal ...
... It is also perfectly acceptable to work in quantities measured in grams. The expected amount formed is found from the initial amount in moles of the reactant (the one that is not in excess) and the chemical equation for the reaction. Note that questions will be set both on calculating yields and cal ...
Chapter 9 Stoichiometry
... Another example If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? How many grams of solid? How much excess reagent remains? ...
... Another example If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? How many grams of solid? How much excess reagent remains? ...
Section 3_Energetics
... Direct determination of lattice energy is very difficult because it is very difficult to get isolated sodium and chloride ions. Therefore the values are usually calculated from other experimentally determined data by applying the Hess Law. The Born-Haber Cycle is a technique of applying Hess‘s Law t ...
... Direct determination of lattice energy is very difficult because it is very difficult to get isolated sodium and chloride ions. Therefore the values are usually calculated from other experimentally determined data by applying the Hess Law. The Born-Haber Cycle is a technique of applying Hess‘s Law t ...
Chapter 4: Solution Chemistry: The Hydrosphere
... 4.1 SOLUTIONS ON EARTH AND OTHER PLACES aqueous solution: a solution where water is the dissolving medium (the solvent) – For example, when table salt (NaCl) is dissolved in water, it results in an aqueous solution of sodium chloride, NaCl(aq), with Na+ and Cl- ions dissolved in water. – Note: The p ...
... 4.1 SOLUTIONS ON EARTH AND OTHER PLACES aqueous solution: a solution where water is the dissolving medium (the solvent) – For example, when table salt (NaCl) is dissolved in water, it results in an aqueous solution of sodium chloride, NaCl(aq), with Na+ and Cl- ions dissolved in water. – Note: The p ...
File
... The concentration of a solution is the amount of solute, in moles, dissolved in each dm3 (1000cm3) of solution. The number of moles in a solution can be calculated using this formula: Number of moles = Concentration x Volume in dm3 In order to obtain the volume in dm3 from cm3, divide the volume in ...
... The concentration of a solution is the amount of solute, in moles, dissolved in each dm3 (1000cm3) of solution. The number of moles in a solution can be calculated using this formula: Number of moles = Concentration x Volume in dm3 In order to obtain the volume in dm3 from cm3, divide the volume in ...
handout 4
... Lecture Examples: In each of the following examples, determine has been oxidized and what has been reduced and which is the oxidizing agent and which is the reducing agent. a. Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g) ...
... Lecture Examples: In each of the following examples, determine has been oxidized and what has been reduced and which is the oxidizing agent and which is the reducing agent. a. Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g) ...
Acids - Beck-Shop
... 1 Add a measured volume of one solution to a conical flask using a pipette. 2 Add the other solution to a burette, and record the initial burette reading to the nearest 0.05 cm3. 3 Add a few drops of an indicator to the solution in the conical flask. 4 Run the solution in the burette into the so ...
... 1 Add a measured volume of one solution to a conical flask using a pipette. 2 Add the other solution to a burette, and record the initial burette reading to the nearest 0.05 cm3. 3 Add a few drops of an indicator to the solution in the conical flask. 4 Run the solution in the burette into the so ...
1. (a) Propan-1ol, C2H5CH2OH can be oxidised to propanoic acid
... X reacts with sodium hydrogencarbonate solution to give a gas which turns lime water milky. It also reacts with a solution of sodium nitrite and hydrochloric acid between 0 ºC and 5 ºC to produce a substance which reacts with phenol to give an orange ...
... X reacts with sodium hydrogencarbonate solution to give a gas which turns lime water milky. It also reacts with a solution of sodium nitrite and hydrochloric acid between 0 ºC and 5 ºC to produce a substance which reacts with phenol to give an orange ...
Document
... In the process of cheese making a bacterial culture is added to milk, which causes the milk to separate into the curds (solid cheese) and whey (remaining liquid). During this process the bacteria convert the lactose present in milk, into lactic acid, CH3CHOHCOOH. Lactic acid is a weak, monoprotic, o ...
... In the process of cheese making a bacterial culture is added to milk, which causes the milk to separate into the curds (solid cheese) and whey (remaining liquid). During this process the bacteria convert the lactose present in milk, into lactic acid, CH3CHOHCOOH. Lactic acid is a weak, monoprotic, o ...
Document
... Changes in the states of matter are often shown on phase diagrams, and you will probably see at least one of two different types of phase diagrams. Let’s start with the phase diagram for water. The phase diagram for water is a graph of pressure versus temperature. Each of the lines on the graph repr ...
... Changes in the states of matter are often shown on phase diagrams, and you will probably see at least one of two different types of phase diagrams. Let’s start with the phase diagram for water. The phase diagram for water is a graph of pressure versus temperature. Each of the lines on the graph repr ...
0.08206 L atm/K mol - Arizona State University
... A. In the gaseous state, the CO2 molecules have a lower mass. B. In the gaseous state, the CO2 molecules have a greater mass. C. In the gaseous state, there is more space between the CO2 molecules. D. In the gaseous state, there is less space between the CO2 molecules. E. In the gaseous state, the C ...
... A. In the gaseous state, the CO2 molecules have a lower mass. B. In the gaseous state, the CO2 molecules have a greater mass. C. In the gaseous state, there is more space between the CO2 molecules. D. In the gaseous state, there is less space between the CO2 molecules. E. In the gaseous state, the C ...
Solution
... A 0.5662-g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3. If 1.0882 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound? ...
... A 0.5662-g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3. If 1.0882 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound? ...
GCE Chemistry Question Paper Unit 05 - Energetics, Redox
... The diagram shows a non-rechargeable cell that can be used to power electronic devices. The relevant half-equations for this cell are equations 2 and 4 in the table above. ...
... The diagram shows a non-rechargeable cell that can be used to power electronic devices. The relevant half-equations for this cell are equations 2 and 4 in the table above. ...
SAMPLE AP CHEMISTRY EXAM QUESTIONS
... decreases, what behavior of water in the beaker will be observed? Explain why this occurs. ...
... decreases, what behavior of water in the beaker will be observed? Explain why this occurs. ...
Chemistry II Exams and Keys Corrected 2016 Season
... 11. Europium has two stable isotopes. A sample of elemental Eu is found to have 2.83034×1023 atoms of Eu-151. If elemental Europium is found to have a mass of 151.96 amu on earth, what is the natural abundance of Eu-153? A. 48.0 % B. 50.0 % C. 52.0 % D. 54.0% 12. The following figure depicts the two ...
... 11. Europium has two stable isotopes. A sample of elemental Eu is found to have 2.83034×1023 atoms of Eu-151. If elemental Europium is found to have a mass of 151.96 amu on earth, what is the natural abundance of Eu-153? A. 48.0 % B. 50.0 % C. 52.0 % D. 54.0% 12. The following figure depicts the two ...
AP Chemistry
... a. when energy required to break bonds > energy released to form new bonds, +H (endothermic) 1. products at a higher energy state than reactants (weaker bonds) 2. surroundings lose energy (cool down) b. when energy required to break bonds < energy released to form new bonds, –H (exothermic) 1. pro ...
... a. when energy required to break bonds > energy released to form new bonds, +H (endothermic) 1. products at a higher energy state than reactants (weaker bonds) 2. surroundings lose energy (cool down) b. when energy required to break bonds < energy released to form new bonds, –H (exothermic) 1. pro ...
Chapters 12 – 20 Practice Problems
... 30. Calculate the entropy change of the surroundings at 25C for the reaction below. C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) ∆Hrxn = −2044 kJ A) 1.30 kJ/K B) 15.5 kJ/K C) 6.86 kJ/K D) 10.4 kJ/K E) 20.5 kJ/K 31. The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has ∆H = +95.7 kJ and ∆S = +142.2 J/K at ...
... 30. Calculate the entropy change of the surroundings at 25C for the reaction below. C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) ∆Hrxn = −2044 kJ A) 1.30 kJ/K B) 15.5 kJ/K C) 6.86 kJ/K D) 10.4 kJ/K E) 20.5 kJ/K 31. The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has ∆H = +95.7 kJ and ∆S = +142.2 J/K at ...
Solutes
... 3. A student measured the conductivity in water, of unlabeled liquids, after each added drop. The following graph was produced... a. Identify the line that represents: aluminum chloride AlCl3, 4 ions water H2O, no change in H2O magnesium chloride MgCl2, 3 ions sugar C6H12O6, 1 particle sodium chlor ...
... 3. A student measured the conductivity in water, of unlabeled liquids, after each added drop. The following graph was produced... a. Identify the line that represents: aluminum chloride AlCl3, 4 ions water H2O, no change in H2O magnesium chloride MgCl2, 3 ions sugar C6H12O6, 1 particle sodium chlor ...
Mathematical Operations
... power to which 10 must be raised to equal the number. For example, the common logarithm of 1000 (written log 1000) is 3 because raising 10 to the third power gives 1000. 103 = 1000, therefore, log 1000 = 3 Further examples are log 105 ...
... power to which 10 must be raised to equal the number. For example, the common logarithm of 1000 (written log 1000) is 3 because raising 10 to the third power gives 1000. 103 = 1000, therefore, log 1000 = 3 Further examples are log 105 ...
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.