1 UNIT 9 CHEMICAL KINETICS 1. (c) mole 1–1 sec–1 The rate law for a reaction A + B →products‚ is rate = k [A]1[B]2. 7. Then‚ which one of the following statements is false ? (a) If [B] is held constant while [A] is doubled‚ the reaction will proceed twice as fast. (b) If [A] is held constant while [B] is reduced to one quarter‚ the rate will be halved (c) If [A] and [B] are both doubled‚ the reaction will proceed 8 times as fast. 2. Fro a first order reaction‚ a straight
Premium Chemical kinetics Rate equation Chemical reaction
Static and Kinetic Friction Introduction The amount of friction force between two surfaces in contact depends on the type of the surfaces in contact and the amount of compression between the surfaces. Static friction is the force that is acting against your force before the object begins to move. If you exert a small push on the box‚ the box will not move because static friction is directly opposite to the push. If you apply a greater force than the static friction force‚ the friction increases
Premium Friction Force
Kinetic Theory Objectives • Describe how the kinetic-molecular theory is used to explain how gases behave at different temperatures. (Exploration 1) • Analyze data that shows how gas particle mass affects that gas’s behavior. (Exploration 2) • Describe the Maxwell-Boltzmann Distribution. (Explorations 1 and 2) Description of Activity The kinetic-molecular theory states that a collection of gas molecules’ average kinetic energy has a specific value at any given temperature
Free Gas Temperature Oxygen
signals the breakdown of fats to provide energy in a process called gluconeogenesis. This breakdown of fats produces ketones such as acetone‚ these molecules are acidic. A build-up of acetone and acetoacetate in Roberta’s blood lowers the pH of her blood. This is known as ketoacidosis. She will begin to rapidly breathe in order to compensate for the high levels of acetone and low pH of her blood. 2. She feels very thirsty despite drinking lots of water. This is a common symptom of diabetes‚ another
Premium Blood Heart Oxygen
uniform disk of radius R = 0.25 m has a string wrapped around it‚ and a m = 3 kg weight is hanging on the string. The system of the weight and disk is released from rest. a) When the 3 kg weight is moving with a speed of 2.2 m/s‚ what is the kinetic energy of the entire system? KETOT = KEwheel+KEweight = (1/2)(I)(w2)+(1/2)(m*v2) =(0.5* v2)(m+1/2M) =0.5*(2.2^2)*(3+(.5*15)) J b) If the system started from rest‚ how far has the weight fallen?
Premium Mass Classical mechanics Energy
Kinetics of Hydrogen Peroxide February 22‚ 2007 Chem. 1130 TA: Ms. Babcock Room 1830 Chemistry Annex PURPOSE OF THE EXPERIMENT Kinetics of Hydrogen Peroxide The major purpose of this experiment is to determine the rate law constant for the reaction of hydrogen peroxide and potassium iodide. In this experiment‚ the goal will be to try to measure the rate law constant at low acidity‚ since at low acidity‚ anything less than 1.0 x 10-3M‚ the effect of the hydrogen ion is negligible. To calculate
Premium Rate equation Reaction rate Chemical kinetics
Kinetics Introduction Nivaldo J. Tro describes kinetics as the study of how changes that occur in chemical reactions take place over time‚ and because of its vast utilization in a multitude of industries‚ it may be one of the most significant and fascinating aspects in the entire chemical world. One application of the study of kinetics can be applied to the determination of the rate of a chemical reaction involving a certain selection of chemicals (FD&C Blue #1 and sodium hypochlorite). The
Premium Chemical reaction Reaction rate Chemistry
puck of mass m initially at speed u collides head-on (without rotation) with a stationary puck of mass M. Find the velocities of both puck after the collision if: i) the collision is fully elastic ii) the collision if fully inelastic. i) momentum: kinetic energy: mu = mv+MV (+ve in direction of initial u) 1 /2 m u2 = 1/2 m v2 + 1/2 M V2 2 eqns in 2 unknowns: V = (u - v) m/M substitute in K eqn: u2 = v2 + (M/m) V2 = v2 + (M/m) (u - v)2 (m/M)2 = v2 + (u - v)2 (m/M) let ρ = (m/M) ⇒ v2 (1 + ρ) - 2ρ
Premium Mass Kinetic energy Classical mechanics
Ah Seung Chong Molecular Biology CTW: Enzyme Kinetic Dr. Cruz 07/22/2010 Enzyme kinetics Introduction Enzymes are biological catalysts or assistants‚ without enzyme many of important processes of life could not happen. Most of enzymes are proteins that help speed up chemical reactions by lowering amount of activation energy needed for the reaction1. Enzymes are usually highly selective‚ only bind to specific substrate and convert it to product at a particular rate1. The rate of the reaction
Premium Enzyme Chemical reaction Catalysis
Biochemistry Unit The Kinetics of Alkali Phosphatase Inhibition 1. OVERVIEW This practical builds on the enzymology lab skills you learned in the Acid Phosphatase practical. Again‚ you will measure the initial reaction velocity (V 0) of an enzyme reaction‚ but this time in the absence and then presence of an inhibitor. Last time you used Acid Phosphatase (Prac 1)‚ but this time you will use the enzyme Alkali Phosphatase. These enzymes have different primary (and hence tertiary) structures
Premium Enzyme PH