Question paper - Unit A173/02 - Module C7 - Higher tier
... Oxford Cambridge and RSA Copyright Information OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of a ...
... Oxford Cambridge and RSA Copyright Information OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of a ...
Intermediate 1 Unit 2 Homework 5
... Some metals are found uncombined in the Earth’s crust. E.g. Gold, silver and copper. Most metals however are found combined with other elements. Some metals are extracted from their ore, by heating with carbon e.g. iron. Some metals are extracted from their ores by using electricity. E.g. aluminium. ...
... Some metals are found uncombined in the Earth’s crust. E.g. Gold, silver and copper. Most metals however are found combined with other elements. Some metals are extracted from their ore, by heating with carbon e.g. iron. Some metals are extracted from their ores by using electricity. E.g. aluminium. ...
de Caux - Combustion of Methane Demonstration
... 2. Combine dish soap and water to produce the soap solution within the desired container (beaker, Erlenmeyer flask, or large plastic tub). A 3% detergent solution is recommended (ex. ~9 mL of dish soap with 290 mL of water). The solution should produce bubbles readily to ensure that the natural gas ...
... 2. Combine dish soap and water to produce the soap solution within the desired container (beaker, Erlenmeyer flask, or large plastic tub). A 3% detergent solution is recommended (ex. ~9 mL of dish soap with 290 mL of water). The solution should produce bubbles readily to ensure that the natural gas ...
File
... 59. The ionization constant, Kb, of the base HONH2 is 1.1 x 108 . The pH of a 1.0 M aqueous soluton of HONH2 is closest to A) 4.0 B) 6.0 C) 8.0 D) 10.0 E) 14.0 60. Which of the following 0.10 M aqueous solutions has a pH less than 7 ? A) KI B) NH4NO3 C) K2CO3 D) NH3 E) Ca(OH)2 61. If 0.15 mol of K2 ...
... 59. The ionization constant, Kb, of the base HONH2 is 1.1 x 108 . The pH of a 1.0 M aqueous soluton of HONH2 is closest to A) 4.0 B) 6.0 C) 8.0 D) 10.0 E) 14.0 60. Which of the following 0.10 M aqueous solutions has a pH less than 7 ? A) KI B) NH4NO3 C) K2CO3 D) NH3 E) Ca(OH)2 61. If 0.15 mol of K2 ...
Final Review Sheet Answers (the 6 page packet)
... symmetry. The lone pair on the central Sb push the three chlorine atoms toward each other in the trigonal pyramidal shape. SbCl5, however, is symmetrical and therefore, does not have any separation of charges. (c) The normal boiling point of CCl4 is 77 °C, whereas that of CBr4 is 190 °C. Both molecu ...
... symmetry. The lone pair on the central Sb push the three chlorine atoms toward each other in the trigonal pyramidal shape. SbCl5, however, is symmetrical and therefore, does not have any separation of charges. (c) The normal boiling point of CCl4 is 77 °C, whereas that of CBr4 is 190 °C. Both molecu ...
SAMPLE EXAM #2
... 15. According to the kinetic molecular theory for gases, particles of a gas a. are very large particles. b. are very far apart. c. lose their valence electrons. d. move slowly. e. decrease kinetic energy as temperature increases. 16. Which relationship is INCORRECT? a. as the temperature of a gas in ...
... 15. According to the kinetic molecular theory for gases, particles of a gas a. are very large particles. b. are very far apart. c. lose their valence electrons. d. move slowly. e. decrease kinetic energy as temperature increases. 16. Which relationship is INCORRECT? a. as the temperature of a gas in ...
2 - CronScience
... 1) Assemble the correct formulas for all the reactants and products, using “+” and “→” 2) Count the number of atoms of each type appearing on both sides 3) Balance the elements one at a time by adding coefficients (the numbers in front) where you need more - save balancing the H and O until LAST! ...
... 1) Assemble the correct formulas for all the reactants and products, using “+” and “→” 2) Count the number of atoms of each type appearing on both sides 3) Balance the elements one at a time by adding coefficients (the numbers in front) where you need more - save balancing the H and O until LAST! ...
AP Chem Stoichiometry Topic#4 Questions WS Name: Date: Per
... (4) Based on the structural formula, calculate the percentage of carbon by mass present in the compound. (5) The diagram represents the collection of elements formed by a decomposition reaction. (a) If the blue spheres represent N atoms and the red ones represent O atoms, what was the empirical form ...
... (4) Based on the structural formula, calculate the percentage of carbon by mass present in the compound. (5) The diagram represents the collection of elements formed by a decomposition reaction. (a) If the blue spheres represent N atoms and the red ones represent O atoms, what was the empirical form ...
Chemical reactions
... reaction. They are written in the left term of the equation. Reaction products = substances formed in a chemical reaction. They are written in the right term of the equation Because in a chemical reaction, the nature of atoms of the substances is not changed, the chemical equations are equalized so ...
... reaction. They are written in the left term of the equation. Reaction products = substances formed in a chemical reaction. They are written in the right term of the equation Because in a chemical reaction, the nature of atoms of the substances is not changed, the chemical equations are equalized so ...
Chapter 3 Part 2 Review
... human body contains C, H, O and N. In one experiment, the complete combustion of 2.175 g of lysine gave 3.94 g CO2 and 1.89 g H2O. In a separate experiment, 1.873 g of lysine gave 0.436 g NH3. a) Calculate the empirical formula. b) If the molar mass is about 150 g, calculate the molecular formula. ...
... human body contains C, H, O and N. In one experiment, the complete combustion of 2.175 g of lysine gave 3.94 g CO2 and 1.89 g H2O. In a separate experiment, 1.873 g of lysine gave 0.436 g NH3. a) Calculate the empirical formula. b) If the molar mass is about 150 g, calculate the molecular formula. ...
Chemistry Midterm Review 2006
... 7. Define density. What is the equation? 8. Ice floats because it is more or less dense than water? 9. Put the 3 states of matter in order of increasing density. 10. A copper penny has a mass of 3.1 g and a volume of .35 cm3. What is the density? 11. A plastic ball has a volume of 19.7 cm3 and a den ...
... 7. Define density. What is the equation? 8. Ice floats because it is more or less dense than water? 9. Put the 3 states of matter in order of increasing density. 10. A copper penny has a mass of 3.1 g and a volume of .35 cm3. What is the density? 11. A plastic ball has a volume of 19.7 cm3 and a den ...
MSTA WOW Chemistry
... 5. Now allow students to suggest which liquid they would like to go into the graduated cylinder. Pour each additional 10 mL of liquid in the cylinder by tilting the cylinder and allowing each to slowly run down the sides of the container. The slow pour is essential to prevent mixing of the liquids. ...
... 5. Now allow students to suggest which liquid they would like to go into the graduated cylinder. Pour each additional 10 mL of liquid in the cylinder by tilting the cylinder and allowing each to slowly run down the sides of the container. The slow pour is essential to prevent mixing of the liquids. ...
PERIODIC TABLE
... 40- When (C8H16) is burned in oxygen atmosphere, we obtain (CO2) and (H2O) according to the following equation: a C8H16 + b O2 → c CO2 + d H2O In a balanced equation, the factors a, b, c, and d have the values: a- (a = 1, b = 1, c = 1, d = 1) b- (a = 1, b = 12, c = 8, d = 16) c- (a = 1, b = 12, c = ...
... 40- When (C8H16) is burned in oxygen atmosphere, we obtain (CO2) and (H2O) according to the following equation: a C8H16 + b O2 → c CO2 + d H2O In a balanced equation, the factors a, b, c, and d have the values: a- (a = 1, b = 1, c = 1, d = 1) b- (a = 1, b = 12, c = 8, d = 16) c- (a = 1, b = 12, c = ...
Chapter 14…Kinetic Theory
... Placing a solute in a solvent will (increase/decrease) the boiling point and (increase/decrease) the freezing point of a solution. (Ionic/molecular) compounds cause greater change of boiling/freezing points in solutions. Aluminum chloride (AlCl3) will dissociate into _____(how many) ions and will ca ...
... Placing a solute in a solvent will (increase/decrease) the boiling point and (increase/decrease) the freezing point of a solution. (Ionic/molecular) compounds cause greater change of boiling/freezing points in solutions. Aluminum chloride (AlCl3) will dissociate into _____(how many) ions and will ca ...
Word Equations • a summary
... The number of each kind of atom does not change, the atoms are simply rearranged ...
... The number of each kind of atom does not change, the atoms are simply rearranged ...
Name ……………………………..………...… …….. Index No
... 7. In an experiment to determine the molar heat of neutralization of hydrochloric acid with sodium hydroride, students of Kassu Secondary school reacted 100cm3 of 1M hydrochloric acid with 50cm3 of 2M sodium hydroxide solution. They obtained the following results. Initial temperature of acid = 25.00 ...
... 7. In an experiment to determine the molar heat of neutralization of hydrochloric acid with sodium hydroride, students of Kassu Secondary school reacted 100cm3 of 1M hydrochloric acid with 50cm3 of 2M sodium hydroxide solution. They obtained the following results. Initial temperature of acid = 25.00 ...
2 KClO 3
... Wheels + Pedals + Handlebar ---» Bicycle Unbalanced: a list of ingredients & results ...
... Wheels + Pedals + Handlebar ---» Bicycle Unbalanced: a list of ingredients & results ...
CHEM 13 NEWS EXAM 1998 - University of Waterloo
... 22. Two flexible containers for gases are at the same temperature and pressure. One holds 0.50 grams of hydrogen and the other holds 8.0 grams of oxygen. Which one of the following statements regarding these gas samples is false? (The relative atomic mass of oxygen is 16.0 and that of hydrogen is 1. ...
... 22. Two flexible containers for gases are at the same temperature and pressure. One holds 0.50 grams of hydrogen and the other holds 8.0 grams of oxygen. Which one of the following statements regarding these gas samples is false? (The relative atomic mass of oxygen is 16.0 and that of hydrogen is 1. ...
Chemistry Final Exam Review 2006-2007
... what would be the partial pressure of nitrogen? c) If a sample of gas occupies 15.9 L at 34 C, what will its volume be at 27 C if the pressure does not change? d) The volume of a sample of oxygen gas is 300.0 ml when the pressure is 1.00 atm and the temperature is 27.0 C. At what temperature would t ...
... what would be the partial pressure of nitrogen? c) If a sample of gas occupies 15.9 L at 34 C, what will its volume be at 27 C if the pressure does not change? d) The volume of a sample of oxygen gas is 300.0 ml when the pressure is 1.00 atm and the temperature is 27.0 C. At what temperature would t ...
Chemistry: the study of composition, structure, and properties of
... exactly the same properties and came composition. – Pure water is always 11.2% hydrogen and 88.8% oxygen. ...
... exactly the same properties and came composition. – Pure water is always 11.2% hydrogen and 88.8% oxygen. ...
Alcohols Oxidation by oxygen O2 in presence of
... aldehyde without the inter mention of carbon- carbon double bond, and it indicates that the method mentioned above is to be an appropriate method for the oxidation of functional group OH in the presence of functional groups such as C=C. Under this oxidation, the hetero aromatic alcohols having atoms ...
... aldehyde without the inter mention of carbon- carbon double bond, and it indicates that the method mentioned above is to be an appropriate method for the oxidation of functional group OH in the presence of functional groups such as C=C. Under this oxidation, the hetero aromatic alcohols having atoms ...
Slide 1 - MrCard.Org
... into 2 or more simpler substances, always just a single reactant in a decomposition • Example – breakdown of calcium carbonate upon ...
... into 2 or more simpler substances, always just a single reactant in a decomposition • Example – breakdown of calcium carbonate upon ...
Catalytic reforming
Catalytic reforming is a chemical process used to convert petroleum refinery naphthas distilled from crude oil (typically having low octane ratings) into high-octane liquid products called reformates, which are premium blending stocks for high-octane gasoline. The process converts low-octane linear hydrocarbons (paraffins) into branched alkanes (isoparaffins) and cyclic naphthenes, which are then partially dehydrogenated to produce high-octane aromatic hydrocarbons. The dehydrogenation also produces significant amounts of byproduct hydrogen gas, which is fed into other refinery processes such as hydrocracking. A side reaction is hydrogenolysis, which produces light hydrocarbons of lower value, such as methane, ethane, propane and butanes.In addition to a gasoline blending stock, reformate is the main source of aromatic bulk chemicals such as benzene, toluene, xylene and ethylbenzene which have diverse uses, most importantly as raw materials for conversion into plastics. However, the benzene content of reformate makes it carcinogenic, which has led to governmental regulations effectively requiring further processing to reduce its benzene content.This process is quite different from and not to be confused with the catalytic steam reforming process used industrially to produce products such as hydrogen, ammonia, and methanol from natural gas, naphtha or other petroleum-derived feedstocks. Nor is this process to be confused with various other catalytic reforming processes that use methanol or biomass-derived feedstocks to produce hydrogen for fuel cells or other uses.