Chem 1A Lecture 1
... • Understand the concepts of formula weight and the mole as a counting number for particles (atoms or molecules) • Be able to use balanced chemical equations to covert between particles/mole/mass of one reactant or product to particles/mol/mass of another • Understand the concepts of limiting reacta ...
... • Understand the concepts of formula weight and the mole as a counting number for particles (atoms or molecules) • Be able to use balanced chemical equations to covert between particles/mole/mass of one reactant or product to particles/mol/mass of another • Understand the concepts of limiting reacta ...
C5H12 + 8 O2 → 5 CO2 + 6 H2O
... 3. Some elements have the same oxidation number in almost all their compounds, and can be used as __________ to determine the oxidation numbers of ...
... 3. Some elements have the same oxidation number in almost all their compounds, and can be used as __________ to determine the oxidation numbers of ...
Stoichiometry: Calculations with Chemical Formulas and Equations
... 3. Calculate the empirical formula with the anhydrous (no water) compound as one unit and the water as the second. Stoichiometry © 2009, Prentice-Hall, Inc. ...
... 3. Calculate the empirical formula with the anhydrous (no water) compound as one unit and the water as the second. Stoichiometry © 2009, Prentice-Hall, Inc. ...
File
... H2O produced when a given quantity of the alcohol is combusted. We must determine the empirical formula for isopropyl alcohol, a task that requires us to calculate the number of moles of C, H, and O in the sample. Plan We can use the mole concept to calculate grams of C in the CO2 and grams of H in ...
... H2O produced when a given quantity of the alcohol is combusted. We must determine the empirical formula for isopropyl alcohol, a task that requires us to calculate the number of moles of C, H, and O in the sample. Plan We can use the mole concept to calculate grams of C in the CO2 and grams of H in ...
Part II - American Chemical Society
... a. Several different compounds have the formula C2H4O2. Two of these contain –CO2 groups. i. Give the structures and names of the two compounds with –CO2 groups. ii. These compounds boil at 31.5 ˚C and 118 ˚C. Assign the two boiling points to the structures in i. and account for the boiling point di ...
... a. Several different compounds have the formula C2H4O2. Two of these contain –CO2 groups. i. Give the structures and names of the two compounds with –CO2 groups. ii. These compounds boil at 31.5 ˚C and 118 ˚C. Assign the two boiling points to the structures in i. and account for the boiling point di ...
AP Chemistry Summer Assignment
... 32. Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 x 103 g of lithium hydroxide? ...
... 32. Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 x 103 g of lithium hydroxide? ...
A Study of Matter
... liquid (the sweating on a glass that is colder than it’s environment) • Sublimation point- temperature at which a solid changes directly to a gas without first changing into a liquid. (dry ice) ...
... liquid (the sweating on a glass that is colder than it’s environment) • Sublimation point- temperature at which a solid changes directly to a gas without first changing into a liquid. (dry ice) ...
Reactions in Aqueous Solution
... If 8 moles of NH3 are reacted with 14 moles of O2 in a rigid container with an initial pressure of 11 atm, what is the partial pressure of NO2 in the container when the reaction runs to completion? ( Assume constant temperature) a. 4 atm b. 6 atm c. 11 atm d. 12 atm ...
... If 8 moles of NH3 are reacted with 14 moles of O2 in a rigid container with an initial pressure of 11 atm, what is the partial pressure of NO2 in the container when the reaction runs to completion? ( Assume constant temperature) a. 4 atm b. 6 atm c. 11 atm d. 12 atm ...
3 Quantitative Chemistry Higher IL Pack
... Q1. Calcium carbonate tablets are used to treat people with calcium deficiency. ...
... Q1. Calcium carbonate tablets are used to treat people with calcium deficiency. ...
Reaction Rate Graphs C12-3
... Ineffective collisions involve particles that rebound essentially unchanged. The rate of reaction depends upon the frequency of collisions and the fractions of those collisions that are effective. Reaction Rate is measured as a decrease in the concentration of reactants per unit time or an incre ...
... Ineffective collisions involve particles that rebound essentially unchanged. The rate of reaction depends upon the frequency of collisions and the fractions of those collisions that are effective. Reaction Rate is measured as a decrease in the concentration of reactants per unit time or an incre ...
Chemistry 2nd Semester Final Review
... so we cannot use the combined gas law or the ideal gas law. The only law that will work is one that involves only volume and moles of gas (because we can convert from grams to moles). Enter Avogadro! Avogadro’s law: (V1/n1)=(V2/n2); Avogadro's Law states that for a gas at constant temperature and pr ...
... so we cannot use the combined gas law or the ideal gas law. The only law that will work is one that involves only volume and moles of gas (because we can convert from grams to moles). Enter Avogadro! Avogadro’s law: (V1/n1)=(V2/n2); Avogadro's Law states that for a gas at constant temperature and pr ...
Unit 3
... • Increasing pressure favours the side with the smaller volume of gas. Consider: N2O4(g) 2NO2(g) 1 mole 2 moles 1 volume 2 volumes • If we increase the pressure we favour the forward reaction, so more N2O4 is formed. ...
... • Increasing pressure favours the side with the smaller volume of gas. Consider: N2O4(g) 2NO2(g) 1 mole 2 moles 1 volume 2 volumes • If we increase the pressure we favour the forward reaction, so more N2O4 is formed. ...
4 - College of Arts and Sciences
... Empirical / Molecular Formulas Determine the empirical formula of a compound that has (by mass) 21.7% C ; 9.6% O ; and 68.7% F. • answer C3OF6 Determine the molecular formula of the compound if its molar mass = 166 • Answer: same as empirical formula ...
... Empirical / Molecular Formulas Determine the empirical formula of a compound that has (by mass) 21.7% C ; 9.6% O ; and 68.7% F. • answer C3OF6 Determine the molecular formula of the compound if its molar mass = 166 • Answer: same as empirical formula ...
PHYSICAL SETTING CHEMISTRY
... A separate answer sheet for Part A and Part B–1 has been provided to you. Follow the instructions from the proctor for completing the student information on your answer sheet. Record your answers to the Part A and Part B–1 multiple-choice questions on this separate answer sheet. Record your answers ...
... A separate answer sheet for Part A and Part B–1 has been provided to you. Follow the instructions from the proctor for completing the student information on your answer sheet. Record your answers to the Part A and Part B–1 multiple-choice questions on this separate answer sheet. Record your answers ...
June 01, 2008
... If a mass of 8.75 g of K2Cr2O7 was added to Solution A and the volume increased by 3.27 ml, what would the concentration of the new solution be in g/ml? ...
... If a mass of 8.75 g of K2Cr2O7 was added to Solution A and the volume increased by 3.27 ml, what would the concentration of the new solution be in g/ml? ...
Question 2
... Carbon tetrachloride, CCl4, can be produced in the reaction below. CH4 + 4Cl2 CCl4 + 4HCl a) What mass of CH4 is needed to exactly combine with 3.4 g Cl2? b) How many grams of Cl2 are required to produce 91 g CCl4, assuming excess CH4? c) What mass of CH4 must have reacted, if 2 mg HCl is liberate ...
... Carbon tetrachloride, CCl4, can be produced in the reaction below. CH4 + 4Cl2 CCl4 + 4HCl a) What mass of CH4 is needed to exactly combine with 3.4 g Cl2? b) How many grams of Cl2 are required to produce 91 g CCl4, assuming excess CH4? c) What mass of CH4 must have reacted, if 2 mg HCl is liberate ...
Practice Exam-Final Fall 2016 W-Ans
... 16. How many hydrogen atoms are there in 48.0 g of CH4? (a) 1.81x1023 (b) 7.22x1024 (c) 6.02x1023 (d) 1.20x1025 (e) 4.70x1025 Hint: According to the chemical formula, one mole of CH4 contains 1 mole of C atoms and 4 moles of hydrogen atoms. Thus, the mole of H = 4 x {mass of CH4/molar mass of CH4}. ...
... 16. How many hydrogen atoms are there in 48.0 g of CH4? (a) 1.81x1023 (b) 7.22x1024 (c) 6.02x1023 (d) 1.20x1025 (e) 4.70x1025 Hint: According to the chemical formula, one mole of CH4 contains 1 mole of C atoms and 4 moles of hydrogen atoms. Thus, the mole of H = 4 x {mass of CH4/molar mass of CH4}. ...
chemical reaction
... reactants causes a fast rate of reaction. Concentration is a measure of the amount of one substance when it is dissolved in another substance. • When concentration is high, there are many reactant particles in a given volume. So, there is little distance between particles and the particles collide m ...
... reactants causes a fast rate of reaction. Concentration is a measure of the amount of one substance when it is dissolved in another substance. • When concentration is high, there are many reactant particles in a given volume. So, there is little distance between particles and the particles collide m ...
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 ...
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.