Sec. 12.3: Molecular Composition of Gases 1) Boyle`s Law: a
... a) Avogadro’s law shows that the __________ __________ of two gases at the same temperature and pressure is the same as the _____________ ratio of the two gases. 19) For example: 3 H2 (g) + N2 (g) 2 NH3 (g) a) 3 L of H2 react with 1 L of N2 to form 2 L of NH3 with no H2 or N2 left over b) If we know ...
... a) Avogadro’s law shows that the __________ __________ of two gases at the same temperature and pressure is the same as the _____________ ratio of the two gases. 19) For example: 3 H2 (g) + N2 (g) 2 NH3 (g) a) 3 L of H2 react with 1 L of N2 to form 2 L of NH3 with no H2 or N2 left over b) If we know ...
chem equation Pkt Student2
... 1) Which side of the yields arrow do you find reactants? ______________________________ 2) Which side of the yields arrow do you find products? _______________________________ 3) In a chemical equation, what do the coefficients represent? ______________________________ 4) In a chemical equation, wha ...
... 1) Which side of the yields arrow do you find reactants? ______________________________ 2) Which side of the yields arrow do you find products? _______________________________ 3) In a chemical equation, what do the coefficients represent? ______________________________ 4) In a chemical equation, wha ...
mass
... When we perform a reaction, we often fail to collect all of the product that could have been produced because of difficulties with separation of the product from excess reactants or other products. The amount of product we should produce based on the starting amounts of reactants is called the theor ...
... When we perform a reaction, we often fail to collect all of the product that could have been produced because of difficulties with separation of the product from excess reactants or other products. The amount of product we should produce based on the starting amounts of reactants is called the theor ...
Chapter 2 MEASUREMENTS AND MOLES
... numbers of atoms of each element, using the smallest whole numbers of atoms. Empirical formula for glucose is CH2O tells us that carbon, hydrogen and oxygen are present in the ratio of 1:2:1.The molecular formula for glucose is C6H12O6. ...
... numbers of atoms of each element, using the smallest whole numbers of atoms. Empirical formula for glucose is CH2O tells us that carbon, hydrogen and oxygen are present in the ratio of 1:2:1.The molecular formula for glucose is C6H12O6. ...
Course __Chemistry Sept Oct Nov Dec Jan Feb March April May June
... atoms in molecules such as H2, CH4, NH3, H2CCH2, N2, Cl2 and many large biological molecules are covalent. B4. Chemical and physical properties of matter result from the ability of atoms to form bonds from electrostatic forces between electrons and protons and between atoms and molecules ...
... atoms in molecules such as H2, CH4, NH3, H2CCH2, N2, Cl2 and many large biological molecules are covalent. B4. Chemical and physical properties of matter result from the ability of atoms to form bonds from electrostatic forces between electrons and protons and between atoms and molecules ...
PRACTICE EXAM 1-C
... You are given a mixture of barium hydroxide, Ba(OH)2, and strontium hydroxide, Sr(OH)2. You dissolve this mixture in water and titrate it with hydrochloric acid. Complete neutralization of the mixture requires 96.0 mL of 1.50 M HCl. (Note that Sr(OH)2 is a strong base that reacts with HCl in the sam ...
... You are given a mixture of barium hydroxide, Ba(OH)2, and strontium hydroxide, Sr(OH)2. You dissolve this mixture in water and titrate it with hydrochloric acid. Complete neutralization of the mixture requires 96.0 mL of 1.50 M HCl. (Note that Sr(OH)2 is a strong base that reacts with HCl in the sam ...
Properties and Changes in Matter
... Matter that has collapsed due to gravity to leave nothing but a pile of neutrons touching each other with no space in between. (about 78,000 tons per teaspoon) ...
... Matter that has collapsed due to gravity to leave nothing but a pile of neutrons touching each other with no space in between. (about 78,000 tons per teaspoon) ...
EXPERIMENT 11 (2 Weeks)!
... spoon" from under the hood. Your instructor will put a very small amount of red phosphorus into the spoon. UNDER THE HOOD, light the phosphorus in the flame of a burner. Lower the spoon with the burning phosphorus into the bottle. CAUTION: Do not touch the bottle with the hot spoon. Remove the spoon ...
... spoon" from under the hood. Your instructor will put a very small amount of red phosphorus into the spoon. UNDER THE HOOD, light the phosphorus in the flame of a burner. Lower the spoon with the burning phosphorus into the bottle. CAUTION: Do not touch the bottle with the hot spoon. Remove the spoon ...
matter crct/final exam review
... 65. Why is the condensation point and the boiling point the same temperature for a particular substance? ...
... 65. Why is the condensation point and the boiling point the same temperature for a particular substance? ...
Chapter 3 Stoichiometry STOICHIOMETRY: The chemical arithmetic
... With a 50 % Yield, How many moles of NH3 are produced from (a) 3 grams of H2 and ½ mole of N2? ½ mole = (½ mole)x(17 g/mole) grams of NH3 (b) 3 grams of H2 and 28 grams of N2? ...
... With a 50 % Yield, How many moles of NH3 are produced from (a) 3 grams of H2 and ½ mole of N2? ½ mole = (½ mole)x(17 g/mole) grams of NH3 (b) 3 grams of H2 and 28 grams of N2? ...
Final Exam Review Sheets
... 26. List four properties of water that are influenced by its strong hydrogen bonds. a. high surface tension: Surface tension is a contractive tendency of the surface of a liquid that allows it to resist an external force caused by cohesion b. high specific heat capacity (absorbs a lot of heat as it ...
... 26. List four properties of water that are influenced by its strong hydrogen bonds. a. high surface tension: Surface tension is a contractive tendency of the surface of a liquid that allows it to resist an external force caused by cohesion b. high specific heat capacity (absorbs a lot of heat as it ...
4.1 Writing and Balancing Chemical Equations
... balancing a chemical equation. Consider as an example the reaction between one methane molecule (CH4) and two diatomic oxygen molecules (O2) to produce one carbon dioxide molecule (CO2) and two water molecules (H2O). The chemical equation representing this process is provided in the upper half of Fi ...
... balancing a chemical equation. Consider as an example the reaction between one methane molecule (CH4) and two diatomic oxygen molecules (O2) to produce one carbon dioxide molecule (CO2) and two water molecules (H2O). The chemical equation representing this process is provided in the upper half of Fi ...
The student will
... 2. Can use experimental data to determine the rate law, determine the order of the reaction, and to define proper units for the constant. 3. Can compare and contrast zero, first, and second order reactions in terms of the plot needed to give a straight line, the relationship of the rate constant to ...
... 2. Can use experimental data to determine the rate law, determine the order of the reaction, and to define proper units for the constant. 3. Can compare and contrast zero, first, and second order reactions in terms of the plot needed to give a straight line, the relationship of the rate constant to ...
Atoms in Combination: The Chemical Bond
... electron to chlorine, which is one electron shy of the “magic” number 18. The result is the ionic compound sodium chloride—ordinary table salt. In these diagrams, electrons are represented as dots in shells around a nucleus. ...
... electron to chlorine, which is one electron shy of the “magic” number 18. The result is the ionic compound sodium chloride—ordinary table salt. In these diagrams, electrons are represented as dots in shells around a nucleus. ...
this PDF file
... - x CP(A)- y CP(B)- z CP(C)- w CP(D) = u (a(E) + b(E)T+c(E)T-2+d(E)T3+e(E)T2) + v (a(F) +b(F)T+c(F)T-2+d(F)T3+e(F)T2) + p(a(G)+b(G)T+c(G)T-2+d(G)T3+e(G)T2) + q(a(H)+b(H)T+c(H)T-2+d(H)T3+e(H)T2) - x(a(A)+b(A)T+c(A)T-2+d(A)T3+e(A)T2) - y(a(B)+b(B)T+c(B)T-2+d(B)T3+e(B)T2) - z(a(C)+b(C)T+c(C)T-2+d(C)T3+ ...
... - x CP(A)- y CP(B)- z CP(C)- w CP(D) = u (a(E) + b(E)T+c(E)T-2+d(E)T3+e(E)T2) + v (a(F) +b(F)T+c(F)T-2+d(F)T3+e(F)T2) + p(a(G)+b(G)T+c(G)T-2+d(G)T3+e(G)T2) + q(a(H)+b(H)T+c(H)T-2+d(H)T3+e(H)T2) - x(a(A)+b(A)T+c(A)T-2+d(A)T3+e(A)T2) - y(a(B)+b(B)T+c(B)T-2+d(B)T3+e(B)T2) - z(a(C)+b(C)T+c(C)T-2+d(C)T3+ ...
LECTURE_pptnotes Fipps Stochiometry
... Limiting Reagent: The reactant that limits the amount of product that can be formed in a ...
... Limiting Reagent: The reactant that limits the amount of product that can be formed in a ...
problem 8 on 2003B exam
... Answer Question 4 below. The Section II score weighting for this question is 15 percent. 4. Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described below. Answers to more than five choices will not be graded. In all cases, a reaction occurs. Assu ...
... Answer Question 4 below. The Section II score weighting for this question is 15 percent. 4. Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described below. Answers to more than five choices will not be graded. In all cases, a reaction occurs. Assu ...
Unit 1 - Measurement Atomic Theory
... 2) Mono prefix is optional with the first nonmetal 3) Last nonmetal ending is –ide ...
... 2) Mono prefix is optional with the first nonmetal 3) Last nonmetal ending is –ide ...
Introduction to Chemistry and the Metric System
... 5 types of reactions: synthesis, decomposition, single replacement, double replacement, combustion monatomic and diatomic elements reactant, product, precipitate, know your solubility rules net ionic equation, spectator ion Problems: 1. Name the reaction type and balance: ______________ a. _ ...
... 5 types of reactions: synthesis, decomposition, single replacement, double replacement, combustion monatomic and diatomic elements reactant, product, precipitate, know your solubility rules net ionic equation, spectator ion Problems: 1. Name the reaction type and balance: ______________ a. _ ...
Chapter 2 - Cloudfront.net
... Every sample of a given substance has identical intensive properties because every sample has the same composition. ...
... Every sample of a given substance has identical intensive properties because every sample has the same composition. ...
sample paper chemistry clas xi set 3
... dioxide and chemical reaction involving hydrocarbons. (ii)It takes place during dry warm season in presence of sunlight (iii)It is oxidising in nature. ...
... dioxide and chemical reaction involving hydrocarbons. (ii)It takes place during dry warm season in presence of sunlight (iii)It is oxidising in nature. ...
Stoichiometry
Stoichiometry /ˌstɔɪkiˈɒmɨtri/ is the calculation of relative quantities of reactants and products in chemical reactions.Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of product can be empirically determined, then the amount of the other reactants can also be calculated.As seen in the image to the right, where the balanced equation is:CH4 + 2 O2 → CO2 + 2 H2O.Here, one molecule of methane reacts with two molecules of oxygen gas to yield one molecule of carbon dioxide and two molecules of water. Stoichiometry measures these quantitative relationships, and is used to determine the amount of products/reactants that are produced/needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry. In the example above, reaction stoichiometry measures the relationship between the methane and oxygen as they react to form carbon dioxide and water.Because of the well known relationship of moles to atomic weights, the ratios that are arrived at by stoichiometry can be used to determine quantities by weight in a reaction described by a balanced equation. This is called composition stoichiometry.Gas stoichiometry deals with reactions involving gases, where the gases are at a known temperature, pressure, and volume and can be assumed to be ideal gases. For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, due to the existence of isotopes, molar masses are used instead when calculating the mass ratio.