Acid-Base Reactions Worksheet #2 - Mro
... When methane, CH4, reacts with molecular chlorine, hydrogen chloride and carbon tetrachloride are produced. If 49.62 mL of methane reacts at a temperature of 465.° C and at a pressure of 1400. torr, with 75.00 mL of molecular chlorine, how many grams of carbon tetrachloride will be produced? ...
... When methane, CH4, reacts with molecular chlorine, hydrogen chloride and carbon tetrachloride are produced. If 49.62 mL of methane reacts at a temperature of 465.° C and at a pressure of 1400. torr, with 75.00 mL of molecular chlorine, how many grams of carbon tetrachloride will be produced? ...
Thermodynamics
... Thermodynamics states: In a spontaneous process, there is a net increase of entropy (taking into account system and surroundings). ...
... Thermodynamics states: In a spontaneous process, there is a net increase of entropy (taking into account system and surroundings). ...
AP Chemistry
... A(g) + B(g) C(g) + D(g) For the gas-phase reaction represented above, the following experimental data were obtained. Exp. Initial [A] Initial [B] ...
... A(g) + B(g) C(g) + D(g) For the gas-phase reaction represented above, the following experimental data were obtained. Exp. Initial [A] Initial [B] ...
matter
... • Chemical reactions occur to produce a more stable product than the existing reactants – Ex: 2Na(s) + Cl2(g) 2NaCl(s) *The sodium is highly unstable and the chlorine gas is somewhat unstable. The resulting Sodium Chloride is VERY stable. **It is important to understand that the products have tota ...
... • Chemical reactions occur to produce a more stable product than the existing reactants – Ex: 2Na(s) + Cl2(g) 2NaCl(s) *The sodium is highly unstable and the chlorine gas is somewhat unstable. The resulting Sodium Chloride is VERY stable. **It is important to understand that the products have tota ...
Water is nature`s most wonderful, abundant and useful compound
... and then dilute with distilled water to 1 Litre. 5. Standardization of EDTA solution: Rinse and fill the burette with EDTA solution. Pipette out 50 mL of standard hard water in a conical flask. Add 10-15 mL of buffer solution and 4 to 5 drops indicator. Titrate with EDTA solution till wine-red colou ...
... and then dilute with distilled water to 1 Litre. 5. Standardization of EDTA solution: Rinse and fill the burette with EDTA solution. Pipette out 50 mL of standard hard water in a conical flask. Add 10-15 mL of buffer solution and 4 to 5 drops indicator. Titrate with EDTA solution till wine-red colou ...
CHEMICAL REACTIONS
... ________ 21. Double-replacement reactions are generally driven by the formation of: a. a precipitate. c. water. b. a gaseous product. d. all of the above ________ 22. If butane (C4H10) undergoes complete combustion: a. 8 CO2 is one product. c. 9 O2 is one reactant. b. 8 CO is one product. d. 10 O2 i ...
... ________ 21. Double-replacement reactions are generally driven by the formation of: a. a precipitate. c. water. b. a gaseous product. d. all of the above ________ 22. If butane (C4H10) undergoes complete combustion: a. 8 CO2 is one product. c. 9 O2 is one reactant. b. 8 CO is one product. d. 10 O2 i ...
Unit F335/01
... (c) At 500 K, the equilibrium constant for equation 1.1 is 7.76 × 10–3. In an equilibrium mixture at 500 K, the concentrations of hydrogen and carbon dioxide are: [H2] = 1.00 × 10–5 mol dm–3 [CO2] = 3.46 × 10–5 mol dm–3 Calculate the equilibrium concentrations of H2O and CO at 500 K. Assume the H2O ...
... (c) At 500 K, the equilibrium constant for equation 1.1 is 7.76 × 10–3. In an equilibrium mixture at 500 K, the concentrations of hydrogen and carbon dioxide are: [H2] = 1.00 × 10–5 mol dm–3 [CO2] = 3.46 × 10–5 mol dm–3 Calculate the equilibrium concentrations of H2O and CO at 500 K. Assume the H2O ...
Final Review Answers
... 2) Differentiate between ionic, covalent, and metallic bonding in terms of electron location and types of atoms combined. Ionic - M/NM, e- donated; Covalent - NM, e- shared; Metallic - M, valence e- move freely 3) How many valence electrons do each of the following atoms have? a) sodium 1 Na b) argo ...
... 2) Differentiate between ionic, covalent, and metallic bonding in terms of electron location and types of atoms combined. Ionic - M/NM, e- donated; Covalent - NM, e- shared; Metallic - M, valence e- move freely 3) How many valence electrons do each of the following atoms have? a) sodium 1 Na b) argo ...
CHM 151LL: States of Matter: Physical and Chemical Changes
... States of Matter Substances can exist in three physical states: solid, liquid, and gas. Some of the differences between these states of matter are 1) the atoms’ or molecules’ freedom of movement and 2) the amount of space between the atoms or molecules. The physical state of a substance at a specif ...
... States of Matter Substances can exist in three physical states: solid, liquid, and gas. Some of the differences between these states of matter are 1) the atoms’ or molecules’ freedom of movement and 2) the amount of space between the atoms or molecules. The physical state of a substance at a specif ...
Chemistr.e1a.chapter.4.new2015
... 2. Monatomic ions: For monatomic ions in ionic compounds, the oxidation number equals the charge on the ion. All group I ions have an oxidation number of +1 and all group II ions have an oxidation number of +2. Examples: In NaCl both Na+ and Cl- are monatomic ions (not polyatomic ions). The oxidati ...
... 2. Monatomic ions: For monatomic ions in ionic compounds, the oxidation number equals the charge on the ion. All group I ions have an oxidation number of +1 and all group II ions have an oxidation number of +2. Examples: In NaCl both Na+ and Cl- are monatomic ions (not polyatomic ions). The oxidati ...
Chapter 5HW_Ans
... b) conversion factor: x X moles of C2H2; therefore 7 moles of O2 2molesC 2 H 2 as moles of C2H2 cancel as they are in both numerator and denominator 2molesC 2 H 2 c) conversion factor: x X moles of H2O; therefore 0.5 moles of 2molesH 2 O C2H2 as moles of H2O cancel as they are in both numerator and ...
... b) conversion factor: x X moles of C2H2; therefore 7 moles of O2 2molesC 2 H 2 as moles of C2H2 cancel as they are in both numerator and denominator 2molesC 2 H 2 c) conversion factor: x X moles of H2O; therefore 0.5 moles of 2molesH 2 O C2H2 as moles of H2O cancel as they are in both numerator and ...
Ch_5_OpenStax_Chemistry edited
... The calorimeter contains water and/or other materials of known heat capacity. The walls of the calorimeter are insulated so all of the heat flow is between the process and the calorimeter components. System = Substance(s) undergoing the chemical or physical change. Surroundings = Calorimeter ...
... The calorimeter contains water and/or other materials of known heat capacity. The walls of the calorimeter are insulated so all of the heat flow is between the process and the calorimeter components. System = Substance(s) undergoing the chemical or physical change. Surroundings = Calorimeter ...
Practice Test #1
... The average velocity of the gas particles is directly proportional to the pressure. Gas particles are very small compared with the average distance between particlesGas particles collide with the walls of their container and in doing so give rise to pressure. Gasesare made up of tiny particles in co ...
... The average velocity of the gas particles is directly proportional to the pressure. Gas particles are very small compared with the average distance between particlesGas particles collide with the walls of their container and in doing so give rise to pressure. Gasesare made up of tiny particles in co ...
AP CHEMISTRY – Source: 1999 AP Exam CHAPTER 8 PRACTICE
... i. Finding rules for computing the oxidation numbers for the atoms in O2F2. 1. The oxidation number of Fluorine is always -1. 2. Therefore, the oxidation number of F in O2F2 is -1. ii. Found rules to determine that: the oxidation number of F is -1. The oxidation state O(x) of O in O2F2 can be comput ...
... i. Finding rules for computing the oxidation numbers for the atoms in O2F2. 1. The oxidation number of Fluorine is always -1. 2. Therefore, the oxidation number of F in O2F2 is -1. ii. Found rules to determine that: the oxidation number of F is -1. The oxidation state O(x) of O in O2F2 can be comput ...
CHEM 30 REDOX
... During this reaction, the reddish-orange dichromate ion changes color to the green chromium ion when it reacts with the alcohol; the degree of the color change is directly related to the level of alcohol in the expelled air. To determine the amount of alcohol in that air, the reacted mixture is comp ...
... During this reaction, the reddish-orange dichromate ion changes color to the green chromium ion when it reacts with the alcohol; the degree of the color change is directly related to the level of alcohol in the expelled air. To determine the amount of alcohol in that air, the reacted mixture is comp ...
advanced chemistry may 2011 marking scheme
... showing products starting from zero concn and remaining below concentration of reactants and 1 mark for correct point when equilibrium is reached. Deduct 1 mark if equilibrium concentrations are shown as being equal. (d) The reaction above is involved in the industrial preparation of hydrogen starti ...
... showing products starting from zero concn and remaining below concentration of reactants and 1 mark for correct point when equilibrium is reached. Deduct 1 mark if equilibrium concentrations are shown as being equal. (d) The reaction above is involved in the industrial preparation of hydrogen starti ...
Exam 4 - Chemistry Courses
... B. The equilibrium partial pressure of Br2(g) will be greater than 1.00 atm. C. At equilibrium, the total pressure in the vessel will be less than the initial total pressure. D. The equilibrium partial pressure of BrCl(g) will be greater than 2.00 atm. E. The reaction will go to completion since the ...
... B. The equilibrium partial pressure of Br2(g) will be greater than 1.00 atm. C. At equilibrium, the total pressure in the vessel will be less than the initial total pressure. D. The equilibrium partial pressure of BrCl(g) will be greater than 2.00 atm. E. The reaction will go to completion since the ...
intermediate chemistry may 2011 marking scheme
... point of magnesium is much higher than that of sodium. Both are metals and have mobile electrons in a delocalized state which can conduct current. (2) For Mg, each atom contributes 2 electrons to the electron sea not 1 as for Na and hence the bonding between metal ion and electron sea is greater (3) ...
... point of magnesium is much higher than that of sodium. Both are metals and have mobile electrons in a delocalized state which can conduct current. (2) For Mg, each atom contributes 2 electrons to the electron sea not 1 as for Na and hence the bonding between metal ion and electron sea is greater (3) ...
8th Grade Ch. 7 Chemical Reactions Study guide
... ____ 31. Each substance written to the right of the arrow in a chemical equation is a ____. A. reactant B. product C. precipitate D. catalyst ____ 32. According to the law of conservation of mass, how does the mass of the products in a chemical reaction compare to the mass of the reactants? A. There ...
... ____ 31. Each substance written to the right of the arrow in a chemical equation is a ____. A. reactant B. product C. precipitate D. catalyst ____ 32. According to the law of conservation of mass, how does the mass of the products in a chemical reaction compare to the mass of the reactants? A. There ...
Chemical Equilibrium Review Ch 13-14 2015
... 3. Hydrogen iodide, HI, decomposes at moderate temperatures according to the equation: 2HI(g) 2(g) + I2(g) The amount of I2 in the reaction mixture can be determined from the intensity of the violet color of I2…the more intense the color, the more I2 in the reaction vessel. When 4.00mol HI was pla ...
... 3. Hydrogen iodide, HI, decomposes at moderate temperatures according to the equation: 2HI(g) 2(g) + I2(g) The amount of I2 in the reaction mixture can be determined from the intensity of the violet color of I2…the more intense the color, the more I2 in the reaction vessel. When 4.00mol HI was pla ...
Answer Key Quiz 6 Prep 40 Questions About the Group V through
... 1) Which of the following statements about nitrogen's abundance are true? (I) Nitrogen is the most abundant element in the Earth’s crust. (II) Nitrogen is the halogen of greatest abundance in the Earth’s crust. (III) Nitrogen is the third most abundant gas in the atmosphere. (IV) Nitrogen is the sec ...
... 1) Which of the following statements about nitrogen's abundance are true? (I) Nitrogen is the most abundant element in the Earth’s crust. (II) Nitrogen is the halogen of greatest abundance in the Earth’s crust. (III) Nitrogen is the third most abundant gas in the atmosphere. (IV) Nitrogen is the sec ...
Acids-bases and Organic Review
... 1. They are Arrhenius acids 3. They are Arrhenius bases and they turn blue litmus red. and they turn blue litmus red. 2. They are Arrhenius acids 4. They are Arrhenius bases and they turn red litmus blue. and they turn red litmus blue. ...
... 1. They are Arrhenius acids 3. They are Arrhenius bases and they turn blue litmus red. and they turn blue litmus red. 2. They are Arrhenius acids 4. They are Arrhenius bases and they turn red litmus blue. and they turn red litmus blue. ...
Project Advance Chemistry 106 Sample Questions
... an increase in the fraction of molecules that have enough energy to react. an increase in the rate constant. an increase in the kinetic energy of the molecules. an increase in the average potential energy of the molecules. ...
... an increase in the fraction of molecules that have enough energy to react. an increase in the rate constant. an increase in the kinetic energy of the molecules. an increase in the average potential energy of the molecules. ...
Electrolysis of water
Electrolysis of water is the decomposition of water (H2O) into oxygen (O2) and hydrogen gas (H2) due to an electric current being passed through the water.This technique can be used to make hydrogen fuel (hydrogen gas) and breathable oxygen; though currently most industrial methods make hydrogen fuel from natural gas instead.