No Slide Title
... • The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula. • The molar mass is the same number, but expressed in grams instead of AMU. It is the mass of one mole of the compound. ...
... • The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula. • The molar mass is the same number, but expressed in grams instead of AMU. It is the mass of one mole of the compound. ...
Lecture Notes for Chemistry 543, Part III
... As in any branch of spectroscopy, we need to determine (1) the eigenvalues of the Hamiltonian in order to know the frequencies of the possible transitions, (2) the selection rules to determine what transitions actually occur, (3) the transition dipole moments to determine the strengths of the transi ...
... As in any branch of spectroscopy, we need to determine (1) the eigenvalues of the Hamiltonian in order to know the frequencies of the possible transitions, (2) the selection rules to determine what transitions actually occur, (3) the transition dipole moments to determine the strengths of the transi ...
Regan-lecture2
... • What are the (idealised) excitation energy signatures for quadrupole collective motion (in even-even nuclei) ? – (extreme) theoretical limits Perfect, quadrupole (ellipsoidal), axially symmetric quantum rotor with a constant moment of inertia (I) has rotational energies given by (from Eclass(rotor ...
... • What are the (idealised) excitation energy signatures for quadrupole collective motion (in even-even nuclei) ? – (extreme) theoretical limits Perfect, quadrupole (ellipsoidal), axially symmetric quantum rotor with a constant moment of inertia (I) has rotational energies given by (from Eclass(rotor ...
Chapter 7: The Mole and Chemical Composition
... symbol on the periodic table). Here is how it is done. Let’s say you have 0.50 moles of Carbon. How many grams is that? 12.01 grams Carbon 0.50 mol Carbon X --------------------------- = 6.0 grams of Carbon 1 mol Carbon This method only works if you label the units of the numbers you are working wit ...
... symbol on the periodic table). Here is how it is done. Let’s say you have 0.50 moles of Carbon. How many grams is that? 12.01 grams Carbon 0.50 mol Carbon X --------------------------- = 6.0 grams of Carbon 1 mol Carbon This method only works if you label the units of the numbers you are working wit ...
Chapter 3 PowerPoint
... • There are two isotopes of carbon 12C with a mass of 12.00000 amu (98.892%), and 13C with a mass of 13.00335 amu (1.108%). What is the atomic wt. of Carbon ? • There are two isotopes of nitrogen , one with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is the percent abundan ...
... • There are two isotopes of carbon 12C with a mass of 12.00000 amu (98.892%), and 13C with a mass of 13.00335 amu (1.108%). What is the atomic wt. of Carbon ? • There are two isotopes of nitrogen , one with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is the percent abundan ...
Torque Rotational Dynamics
... • Torque is the product of force and lever arm. • The rotational inertia depends not only on the mass of an object but also on the way its mass is distributed around the axis of rotation. • The angular acceleration is proportional to the torque and inversely proportional to the rotational inertia. ...
... • Torque is the product of force and lever arm. • The rotational inertia depends not only on the mass of an object but also on the way its mass is distributed around the axis of rotation. • The angular acceleration is proportional to the torque and inversely proportional to the rotational inertia. ...
Test 7 Review - Evan`s Chemistry Corner
... equation shows conservation of mass. For example, AgNO3(aq) + NaCl(aq) ÿ NaNO3(aq) + AgCl(s). SILVER NITRATE AgNO3 Ag = 1 × 108 = N = 1 × 14 = O = 3 × 16 = ...
... equation shows conservation of mass. For example, AgNO3(aq) + NaCl(aq) ÿ NaNO3(aq) + AgCl(s). SILVER NITRATE AgNO3 Ag = 1 × 108 = N = 1 × 14 = O = 3 × 16 = ...
Molar Mass of a Compound
... Molar Mass of a Compound • The molar mass of a compound is the mass of a mole of the representative particles of the compound. • Because each representative particle is composed of two or more atoms, the molar mass of the compound is found by adding the molar masses of all of the atoms in the repre ...
... Molar Mass of a Compound • The molar mass of a compound is the mass of a mole of the representative particles of the compound. • Because each representative particle is composed of two or more atoms, the molar mass of the compound is found by adding the molar masses of all of the atoms in the repre ...
6. NaF
... Rule 2. If both elements are in the same group, the element with the higher period number is written first in the name. Rule 3. The second element in the name is named as if it were an anion, i.e., by adding the suffix -ide to the name of the element. Rule 4. Greek prefixes (see the table given to y ...
... Rule 2. If both elements are in the same group, the element with the higher period number is written first in the name. Rule 3. The second element in the name is named as if it were an anion, i.e., by adding the suffix -ide to the name of the element. Rule 4. Greek prefixes (see the table given to y ...
Chemical Quantities PPT
... Calculate the formula mass of calcium chloride Write the formula from the name given Ca2+ (from group II) and Cl- (from group VII) Formula is CaCl2 due to charge balance Formula mass: Sum of the atomic masses of atoms in the formula (1 Ca atom + 2 Cl atoms) 40.08 amu = 40.08 amu 1 Ca atom 35.45 amu ...
... Calculate the formula mass of calcium chloride Write the formula from the name given Ca2+ (from group II) and Cl- (from group VII) Formula is CaCl2 due to charge balance Formula mass: Sum of the atomic masses of atoms in the formula (1 Ca atom + 2 Cl atoms) 40.08 amu = 40.08 amu 1 Ca atom 35.45 amu ...
7 Angular Momentum I
... This is true because J+ |µmax , νi = 0. Then, we can substitute J− J+ with Eq. (17) and get ν = j(j + 1). Applying J+ to the state with a maximal value of µ we should get 0. But we can apply J− . If we apply it k-times the final state will be proportional to |j − k, νi. This state also should satisf ...
... This is true because J+ |µmax , νi = 0. Then, we can substitute J− J+ with Eq. (17) and get ν = j(j + 1). Applying J+ to the state with a maximal value of µ we should get 0. But we can apply J− . If we apply it k-times the final state will be proportional to |j − k, νi. This state also should satisf ...
Mole - My CCSD
... 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the ...
... 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the ...
Empirical and Molecular Formulas Empirical Formula: The smallest
... Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1 4.068/3.390 = 1 mol of C 5.08/3.390 = 1.5 mol of H 3.390/3.390 = 1 mol of O Then look at the three numbers of moles and determine the lowest number they can be multiplied by to get all whole numbers. In this ...
... Step 2: Determine simplest ratio by dividing the lowest amount of moles determined in step 1 4.068/3.390 = 1 mol of C 5.08/3.390 = 1.5 mol of H 3.390/3.390 = 1 mol of O Then look at the three numbers of moles and determine the lowest number they can be multiplied by to get all whole numbers. In this ...
Aalborg Universitet
... stationary scattering objects in (4.14). In Section 5 we derive the usual Landauer-Büttiker formula from (4.14); we state our main results in Theorem 5.2. We also derive a continuity equation in subsection 5.1, which shows that the steady current is the same, no matter where we measure it on a give ...
... stationary scattering objects in (4.14). In Section 5 we derive the usual Landauer-Büttiker formula from (4.14); we state our main results in Theorem 5.2. We also derive a continuity equation in subsection 5.1, which shows that the steady current is the same, no matter where we measure it on a give ...
The Mole
... Because atoms are so small we can’t use the conventional units of grams and kilograms. Mass of a single carbon atom: 1.99 x 10-23 g 1 atomic mass unit = 1.66 x 10-24g ...
... Because atoms are so small we can’t use the conventional units of grams and kilograms. Mass of a single carbon atom: 1.99 x 10-23 g 1 atomic mass unit = 1.66 x 10-24g ...
Empirical and Molecular Formulas
... An oxide of sulfur has a percent composition of 40.05% S and 59.95% O. In 100 g of the compound, 40.05 g are S and 59.95 g are O. Next, find the amount of mol for each element. ...
... An oxide of sulfur has a percent composition of 40.05% S and 59.95% O. In 100 g of the compound, 40.05 g are S and 59.95 g are O. Next, find the amount of mol for each element. ...
CHM 4412 Physical Chemistry II - University of Illinois at
... operator, a different function (B) results. The Schrödinger equation says that the input and output functions should be the same (Ψ), apart from a constant factor (E). ...
... operator, a different function (B) results. The Schrödinger equation says that the input and output functions should be the same (Ψ), apart from a constant factor (E). ...
Lecture 21.Roational..
... wheelie,” a large acceleration causes the bike’s front wheel to leave the ground. Let M be the total mass of the bike-plus-rider system; let x and y be the horizontal and vertical distance of this system’s CM from the rear wheel’s point of contact with the ground (see figure). (a) Determine the hori ...
... wheelie,” a large acceleration causes the bike’s front wheel to leave the ground. Let M be the total mass of the bike-plus-rider system; let x and y be the horizontal and vertical distance of this system’s CM from the rear wheel’s point of contact with the ground (see figure). (a) Determine the hori ...
MODULE FOR INTERNATIONAL STANDARD CLASS
... alkanes and cycloalkanes. The object of petroleum refining is to obtain various hydrocarbon products from the mixture. In part, this is accomplished by fractional distillation of the petroleum, but the demand for certain products, particularly gasoline, is greater than what can be supplied by distil ...
... alkanes and cycloalkanes. The object of petroleum refining is to obtain various hydrocarbon products from the mixture. In part, this is accomplished by fractional distillation of the petroleum, but the demand for certain products, particularly gasoline, is greater than what can be supplied by distil ...
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. ...
Chemistry
... - The volume of a gas changes with changes in temperature and pressure - Because of this a volume of gas is usually measured at standard temperature and pressure (STP) - STP is 0 C and 101.3 kPa or 1 atmosphere - The Earth is surrounded by a blanket of air, which we call the atmosphere. It reaches o ...
... - The volume of a gas changes with changes in temperature and pressure - Because of this a volume of gas is usually measured at standard temperature and pressure (STP) - STP is 0 C and 101.3 kPa or 1 atmosphere - The Earth is surrounded by a blanket of air, which we call the atmosphere. It reaches o ...