MIDTERM EXAM – JANUARY, 2003
... The shape of a p orbital is The maximum number of electrons allowed in each of the d orbitals is Write an example of an incorrect designation for an atomic orbital? A nitrogen atom needs to gain ______________ electrons to achieve a noble gas configuration. 45. A calcium atom needs to lose _________ ...
... The shape of a p orbital is The maximum number of electrons allowed in each of the d orbitals is Write an example of an incorrect designation for an atomic orbital? A nitrogen atom needs to gain ______________ electrons to achieve a noble gas configuration. 45. A calcium atom needs to lose _________ ...
The Modern Atomic Model
... Bohr Model of the Atom (review) •Energy levels contain electrons. •Electrons travel around the nucleus. •Different orbitals varied by different quantum (energy). •Gaps between energy levels were not equal. ...
... Bohr Model of the Atom (review) •Energy levels contain electrons. •Electrons travel around the nucleus. •Different orbitals varied by different quantum (energy). •Gaps between energy levels were not equal. ...
Recitation Activity 6 (Chem 121) Chapter 6
... photons of electromagnetic radiation with wavelengths depicted by waves 1-3. a. An electron relaxing from n = 2 to n = 1 (transition A) b. An electron relaxing from n = 4 to n = 2 (transition B) c. An electron relaxing from n = 4 to n = 3 (transition C) ...
... photons of electromagnetic radiation with wavelengths depicted by waves 1-3. a. An electron relaxing from n = 2 to n = 1 (transition A) b. An electron relaxing from n = 4 to n = 2 (transition B) c. An electron relaxing from n = 4 to n = 3 (transition C) ...
- BUGS McGill
... h (Planck’s constant)………...6.626 x 10-34 Js e (electronic charge)……..…..1.6027 x 10-19 C No (Avogadro’s #)…………...6.022 x 1023 mol-1 Me (electron rest mass)…...….9.1094 x 10-31 kg Mp (proton rest mass)……..…1.673 x 10-27 kg c(speed of light)……………...2.997 x 108 ms-1, 2.997 x 1010 cm s-1 Å(angstrom)………… ...
... h (Planck’s constant)………...6.626 x 10-34 Js e (electronic charge)……..…..1.6027 x 10-19 C No (Avogadro’s #)…………...6.022 x 1023 mol-1 Me (electron rest mass)…...….9.1094 x 10-31 kg Mp (proton rest mass)……..…1.673 x 10-27 kg c(speed of light)……………...2.997 x 108 ms-1, 2.997 x 1010 cm s-1 Å(angstrom)………… ...
1) Which of the following concepts was discussed in Chapter 1
... 1) Increase the momentum of the particle 2) Decrease the momentum of the particle 3) Decrease the well width 4) Increase the well depth 5) Decrease the well depth ...
... 1) Increase the momentum of the particle 2) Decrease the momentum of the particle 3) Decrease the well width 4) Increase the well depth 5) Decrease the well depth ...
Atom (A) or Ion
... 51. Know the basics of the major contributions of the following individuals to the development of the atomic theory: John Dalton, J.J. Thompson, Ernest Rutherford, and Neils Bohr. 52. Name the ion formed by each. Indicate whether an anion or cation: ...
... 51. Know the basics of the major contributions of the following individuals to the development of the atomic theory: John Dalton, J.J. Thompson, Ernest Rutherford, and Neils Bohr. 52. Name the ion formed by each. Indicate whether an anion or cation: ...
Chapter 4 Section 2
... Answered Rutherford’s ?—electrons in a particular path have a fixed energy, they do NOT lose energy and fall into the nucleus Energy level—region around nucleus where it is likely to be moving, similar to rungs on a ladder but not equally spaced Quantum—amount of energy needed to move an electron fr ...
... Answered Rutherford’s ?—electrons in a particular path have a fixed energy, they do NOT lose energy and fall into the nucleus Energy level—region around nucleus where it is likely to be moving, similar to rungs on a ladder but not equally spaced Quantum—amount of energy needed to move an electron fr ...
Atom (A) or Ion (I)
... 51. Know the basics of the major contributions of the following individuals to the development of the atomic theory: John Dalton, J.J. Thompson, Ernest Rutherford, and Neils Bohr. 52. Name the ion formed by each. Indicate whether an anion or cation: ...
... 51. Know the basics of the major contributions of the following individuals to the development of the atomic theory: John Dalton, J.J. Thompson, Ernest Rutherford, and Neils Bohr. 52. Name the ion formed by each. Indicate whether an anion or cation: ...
Electrons in the Atom
... So scientists agreed to limit these calculations to locations where there was at least a 90% chance of finding an electron. Think of orbitals as sort of a "border” for spaces around the nucleus inside which electrons are allowed. No more than 2 electrons can ever be in 1 orbital. The orbital j ...
... So scientists agreed to limit these calculations to locations where there was at least a 90% chance of finding an electron. Think of orbitals as sort of a "border” for spaces around the nucleus inside which electrons are allowed. No more than 2 electrons can ever be in 1 orbital. The orbital j ...
Quantum2
... that the probability of a particle being found at a particular point can be calculated from the wavefunction. •Okay, we can’t calculate the position (or other position dependent variables) precisely but given a large number of events, can we predict what the average value will be? (If you roll a dic ...
... that the probability of a particle being found at a particular point can be calculated from the wavefunction. •Okay, we can’t calculate the position (or other position dependent variables) precisely but given a large number of events, can we predict what the average value will be? (If you roll a dic ...
Chapter 28: Quantum Physics
... Example (text problem 28.52): In a ruby laser, laser light of wavelength 694.3 nm is emitted. The ruby crystal is 6.00 cm long, and the index of refraction of the ruby is 1.75. Think of the light in the ruby crystal as a standing wave along the length of the crystal. How many wavelengths fit in the ...
... Example (text problem 28.52): In a ruby laser, laser light of wavelength 694.3 nm is emitted. The ruby crystal is 6.00 cm long, and the index of refraction of the ruby is 1.75. Think of the light in the ruby crystal as a standing wave along the length of the crystal. How many wavelengths fit in the ...
File - Mr. Holz`s Website
... Ionic Bond – Transfer of electrons to create a bond between two ions that are attracted by opposite charges Covalent Bond – Bond that forms when electrons are shared between atoms Ion – Charged atoms that form from ionic bonds; atoms in which the number of electrons does not equal the number of prot ...
... Ionic Bond – Transfer of electrons to create a bond between two ions that are attracted by opposite charges Covalent Bond – Bond that forms when electrons are shared between atoms Ion – Charged atoms that form from ionic bonds; atoms in which the number of electrons does not equal the number of prot ...
File
... 29. How can you find the number of energy levels needed for an element? You can find it by looking at the element’s period number or row number ...
... 29. How can you find the number of energy levels needed for an element? You can find it by looking at the element’s period number or row number ...
Unit 2 Review for Test
... 2. Does the chemical composition of water (H2O) change when it boils? 3. List the following in order of complexity (simple to complex> ______ compound ______ element ______ atom 4. What information does this formula provide: C12H22O11? 5. Define: compound. 6. Over 96% of all living matter is made up ...
... 2. Does the chemical composition of water (H2O) change when it boils? 3. List the following in order of complexity (simple to complex> ______ compound ______ element ______ atom 4. What information does this formula provide: C12H22O11? 5. Define: compound. 6. Over 96% of all living matter is made up ...
File - Mrs. Hille`s FunZone
... If the light is not white By heating a gas with electricity we can get it to give off colors. Passing this light through a prism does something different. ...
... If the light is not white By heating a gas with electricity we can get it to give off colors. Passing this light through a prism does something different. ...
X-ray photoelectron spectroscopy
X-ray photoelectron spectroscopy (XPS) is a surface-sensitive quantitative spectroscopic technique that measures the elemental composition at the parts per thousand range, empirical formula, chemical state and electronic state of the elements that exist within a material. XPS spectra are obtained by irradiating a material with a beam of X-rays while simultaneously measuring the kinetic energy and number of electrons that escape from the top 0 to 10 nm of the material being analyzed. XPS requires high vacuum (P ~ 10−8 millibar) or ultra-high vacuum (UHV; P < 10−9 millibar) conditions, although a current area of development is ambient-pressure XPS, in which samples are analyzed at pressures of a few tens of millibar.XPS is a surface chemical analysis technique that can be used to analyze the surface chemistry of a material in its as-received state, or after some treatment, for example: fracturing, cutting or scraping in air or UHV to expose the bulk chemistry, ion beam etching to clean off some or all of the surface contamination (with mild ion etching) or to intentionally expose deeper layers of the sample (with more extensive ion etching) in depth-profiling XPS, exposure to heat to study the changes due to heating, exposure to reactive gases or solutions, exposure to ion beam implant, exposure to ultraviolet light.XPS is also known as ESCA (Electron Spectroscopy for Chemical Analysis), an abbreviation introduced by Kai Siegbahn's research group to emphasize the chemical (rather than merely elemental) information that the technique provides.In principle XPS detects all elements. In practice, using typical laboratory-scale X-ray sources, XPS detects all elements with an atomic number (Z) of 3 (lithium) and above. It cannot easily detect hydrogen (Z = 1) or helium (Z = 2).Detection limits for most of the elements (on a modern instrument) are in the parts per thousand range. Detection limits of parts per million (ppm) are possible, but require special conditions: concentration at top surface or very long collection time (overnight).XPS is routinely used to analyze inorganic compounds, metal alloys, semiconductors, polymers, elements, catalysts, glasses, ceramics, paints, papers, inks, woods, plant parts, make-up, teeth, bones, medical implants, bio-materials, viscous oils, glues, ion-modified materials and many others.XPS is less routinely used to analyze the hydrated forms of some of the above materials by freezing the samples in their hydrated state in an ultra pure environment, and allowing or causing multilayers of ice to sublime away prior to analysis. Such hydrated XPS analysis allows hydrated sample structures, which may be different from vacuum-dehydrated sample structures, to be studied in their more relevant as-used hydrated structure. Many bio-materials such as hydrogels are examples of such samples.