CHEM131 Exam 1 Study guide
CHEM131 Exam 1 Study guide CHEM131
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This 9 page Study Guide was uploaded by Cathryn Tsu on Monday February 29, 2016. The Study Guide belongs to CHEM131 at University of Maryland taught by Soumya Rastogi in Spring 2016. Since its upload, it has received 208 views. For similar materials see General Chemistry I in Chemistry at University of Maryland.
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Date Created: 02/29/16
02/29/2016 CHEM131 Exam 1 (Chapters 15) Date of exam: Thursday, 03/03/16 ____________________________________________________________________________ Chapter 1: Matter Matter anything that has mass & takes up space Solid (fixed volume, unyielding shape, particles pack closely and don’t move) Liquid (fixed volume, yielding shape, weak attraction b/t packed particles, some movement) Gas (yielding volume and shape, compressible, large spaces b/t particles) Pure substance (consists of one particle) Mixture (consists of 2 or more) can either be aelement or aompound (it can either bomogeneous ( consistent can be broken down) composition) o eterogeneous (composition varies throughout) Law of Conservation of Mass Matter can neither be created nor destroyed (Lavoiser) Law of Definite Proportions All samples of a given compound regardless of their source or preparation have the same proportions of their constituent elements (Proust) Law of Multiple Proportions When 2 elements (A & B) create 2 different compounds, the mass of element B combined w/ 1 g of element A can be expressed as a whole number ratio (Dalton) 1 g of C2 (mass)2.67 ex. 1 g of CO (mass) 1.33 = 2 Atomic Theory (Dalton) all matter is composed of miniscule particles called atoms 1) Each element is made up of tiny, indestructible particles called atoms 2) Atoms of an element have a specific mass and properties that are different from another element 3) Atoms combine in whole number ratios to form compounds 4) An element’s atoms can’t change. The only way they could change is in being bound w/other atoms in a chemical reaction Experiments Thomson & Cathode Rays : particles are composed of negative electrical charges which is a significant atomic property that results in forces that attract and repel one another. Milikan & the Oil Drop Experim: discovered the negative charge of electrons. Thomson’s Plum Pudding Model: stated that electrons were surrounded and scattered in a sphere of positive charge. Rutherford’s Gold Foil Experime: disproved Thomson’s model and instead stated that the mass and positive charge of an atom was concentrated within a space smaller than the atom. This resulted in tNuclear Theory: 1) Majority of atom’s mass (now known aseutronswhich have a neutral charge) and (+) charge particleprotons) was contained in teucleus,he small core 02/29/2016 2) Surrounding the nucleus is empty space, with () charge particles (electrons) are scattered 3) The number of protons and electrons is equal which makes the atom electrically neutral Elements on the Periodic Table Themass number is represented by A, it is the # of protons + the # of neutrons X is the element The atomic number (the # of protons and electrons) is represented by “Z” (Mass number =/ Atomic mass) Isotopes atoms that have the same amount of protons but different number of neutrons − 0 ex. Ne21 p+ e n 10 10 11 Natural abundance the percentages of each different isotope in a naturally occurring sample of a given element *natural abundances add up to 100% Ion when an atom either gains an electron (anion) and becomes (+) charged or loses an electron (cation) and becomes () charged ex. Li →Li : cation and Cl → Cl : anion How to calculate average atomic mass * amu (atomic mass unit) Atomic mass the average mass of isotopes within an element Atomic mass = ∑ (fraction of isotope 1) × (mass of isotope 1) + (fraction of isotope 2) × (mass of isotope 2) and so on n ex. “Magnesium has 3 naturally occurring isotopes with masses of 23.99 amu, 24.99 amu, and 25.98 amu and a natural abundance of 78.99%, 10%, and 11.01%, respectively. Calculate the atomic mass of magnesium” pg. 23 78.99 10.00 11.01 Atomic mass = ( 100 ×23.99 amu) + ( 100×24.99 amu) + ( 100 ×25.98 amu) = 24.31 amu ____________________________________________________________________________ Chapter 2: Measurement, Problem Solving, & the Mole Precision how similar a series of measurements are to one another and if they can be replicated Accuracy how close the measured value is to the true value Random error an equal chance that the error is too high or too low (can avg out with repeated attempts) Systematic error either too high or too low, can’t equal out (doesn’t avg out) 02/29/2016 Kinetic energy total energy of an object Potential energy energy involved with an object’s motion Density mass m d = V olume or d = V ex. “A chemist needs 35.0 g of concentrated sulfuric acid for an experiment. The density at room temperature is 1.84 g/mL. What volume of acid is required?” Discussion worksheet 1, #10 m m 35.0 g d = so V= = 19.0 mL V d 1.84 g/mL 23 Mole he amount of material consistin6.02214 ×10 particles 23 This number is known asAvagadro’s number ( mol = 6.022 × 10 toms) 22 ex. “An aluminum sphere contains 8.55 ×10 aluminum atoms. What is the sphere’s 3 radius in cm? The density is 2.70 g/cm ” pg. 55, example 2.12 3 22 Given: d= 2.70 g/cm and 8.55 ×10 Al atoms 23 Relationships: 1 mol = 6.022 × 1 toms 26.98 g Al = 1 mol 22 1 mol Al 26.98 g Al 1 cm 3 1) 8.55 ×10 l atoms× 6.022 ×10 Al atoms1 mol Al× 2.70 g/cm 1.4187 cm 4 3 2) V= 3πr 3 r =√ 3V 3 3(1.4187 =.697 cm 4π √ 4π Molar mass mass of 1 mol of an element’s atoms; numerically equal to element’s atomic mass in amu ex. Calculate the mole of 127.08 g of Copper. Discussion worksheet 2, #6a 1 mol Cu = 63.5 g Cu 1 mol 127.08 g Cu ×63.5 g Cu 2.00 mol Cu Significant Figures 1) All nonzeros are significant 2) Zeros b/t nonzeros are significex. 2.040 3) Leading zeros aren’t significx. 0.00 4) Trailing zeros are significant when they’re: 02/29/2016 a) after the decimalx. 65.0) b) after a nonzeros and before a decimalx. 350.0 In calculations: Multiplication/Division the result abides by the fewest sig figs of the numbers being multiplied/divided ex. 4.796 ×5.6 = 27 Addition/Subtraction the result abides by the fewest sig figs after the decimal place ex. 3.693 + 2.34 = 6.03 Conversions (look at your prefix multipliers sheet on ELMS, helps a t**) DESIRED UNIT start with yoGIVEN QUANTITY × GIVEN UNIT −9 ex. #8a on Discuss12n worksheet 1; Convert 9.4 × 10 m to pm 9.4 × 10 m × 10 p= 9.4 × 10 pm 1 m Going about problems from here on out (how to try to not get confused) The book really tries to help lay out how to attack problems moving from here forward, especially with problems concerning wavelength, KE, and so on. 1) Sort through the information you’re given and what they want you to find 2) Create a plan on how you might get from what you’re given to what you need to find a) Figure out the relationships you have to use and what you’ll need to convert if anything b) Think about if you’ll need to use more than one formulas to get to your answer 3) Solve the problem (harder than it sounds, I know it’s okay you can do it) 4) Check your answer ____________________________________________________________ Chapter 3: The QuantumMechanical Model of the Atom Electromagnetic light spectrum Waves can be characterized by their: a) Amplitude vertical height of the crest/depth of a trough that determines light intensity. b) Wavelength ( λ ) distance b/t adjacent crests; determines its color 02/29/2016 c) Frequency (υ) # of cycles that pass through a stationary point during a certain time (measured Hz / s )−1 i) related to wave speed ii) υis inversely proportional to its λ (↑ λ= ↓ υand vice versa) C (light) 8 Frequency = λ (wavelength) c = .00 × 10 m/s (photo creds: http://www.cyberphysics.co.uk/topics/light/emspect.htm) Wave vs. particle experiments Interference when waves cancel/build upon each other depending on their alighment Constructive interference if 2 waves with the same amplitude aligh w/overlapping crests, a wave w/double the amplitude results Defraction when waves come into contact w/ an obstacle or slit, it bends around it Interference from 2 slits: the resulting waves interfere w/ one another usually shown by a pattern of dark/bright lines Photoelectric effect many metals emit electrons when light shines on them In the beginning, it was believed that the energy transfer from light to the metal’s electron but that didn’t make so much sense. this meant that the transferred energy would have to exceed the electron’s bindning energy b/c the light intensity is low, there should be a lag time however Einstein found that this wasn’t true. light is quantized; electron emission from metal depends on wheter there is enough energy to remove 1 electron threshold frequency condition: ϕ= hυ ϕ= binding energy of the electron υ> υ o( the threshold frequency, υ oh as to be greater than the light frequency) h = 6.626 ×10 −3 J (Planck’s constant) Bohr atom & calculations The Bohr atom model electrons travel around the nucleus in orbits that occurred at fixed distances from the nucleus, reflective of the electron’s wave nature radiation is released/absorbed when it “jumps” from one stationary state to the next Wave nature of matter replaced the Bohr model, can be seen by its diffraction as it isn’t caused by interference b/t electron pairs but by electron’s selfinterference faster electron movement, ↑KE, shorter λ de Brogile reaction: λ= h mv 02/29/2016 Energy things: 1 2 KE = 2v hυ = ϕ+ KE Mass of e = 9.11 × 10 −31kg The Uncertainty Principle the unobserved electron can occupy 2 states but the act of observation forces it into one state (either a particle or a wave) Quantum numbers Orbital is specified by 3 connected uantum numbers: 1) N, the principal quantum # 2) L, determines the orbital’s shape (can be any integer up to n 1) 3) m 1 l to +l s defines the electron’s spin (can be either ½ or +½ ) Orbital shapes Orbital a probability distribution map where the electron will most likely be located E n 2.18 ×10 −18 J (12 → determines the total size & energy of the orbital n Principal level orbitals with the same value of n Sublevel orbitals with the same value of n & l Energy difference b/t energy levels (*has to be +) : ΔE = E − E → ΔE =− 2.18 × ( −2) 12 final initial nf ni J l = 0 s l = 1 p l = 2 d l = 3 f S orbitals have a spherical shape and are symmetrical P orbitalshave 2 lobes on both sides of the nucleus and aren’t symmetrical D orbitals have a cloverleaf shape ____________________________________________________________ Chapter 4: Periodic Properties of the Elements Periodic Coulomb’s law Electron shielding Penetration 02/29/2016 Electron configuration − Orbital diagrams symbolizes the e as an arrow reflective of its spin and the orbital as a box Pauli’s Exclusion Principle no 2 electrons in an atom can have the same 4 quantum #s; each orbital can only have a max of 2 e with opposing spins Coulomb’s Law the potential energy of 2 charged particles relies on their charges & their separation; like charges repel Shielding (the repulsion of one electron by another) the atom experiences anffective nuclear charge if there was a 3rd electron charge Penetration (when the 3rd electron enters the atom) the atom experiences a ↑in nuclear charge and a ↓in energy Effective nuclear charge = eff= Z − S Z = actual nuclear charge S = the charge of shielded electrons Aufbau’s Rule e lectrons will fill the lowest available energy level before filling higher energy levels Hund’s Rule e lectrons will fill separate orbitals individually before pairing in order to create a more stable atom (photocreds: http://davidjohnewart.com/Chemistry/chemtheft/09.html) *Remember* s 1 orbital ⇒2 e − p 3 orbitals ⇒6 e d 5 orbitals ⇒10 e f 7 orbitals ⇒ 14 e Core & Valence electrons Core e in complete principal energy levels and d & f sublevels − Valence e that are in the outermost principal energy level ex. Si 1s s 2p s 3p Inner electron configuration(the preceding noble gas) can be used with the uter electron configuration(electron configuration beyond the noble gas) to shorten the electron configuration Periodic table First 2 families on the left ⇒sblock Last 6 families on the right ⇒ pblock Lanthanides/ Actinides ⇒ dblock 02/29/2016 Noble gases Alkali metals (Group Alkali earth metals Halogens (Group 7A 1A elements) (Group 2A elements) elements) filled valence have an outer 1 have an outer 2 have an outer 2 electrons and outer configuration of ns configuration of ns configuration of ns levels np unreactive easily ease their lose their 2 electrons gain 1 electron to electrons to form ions to form ions w/a 2+ form ions with a 1 w/a 1+ charge charge charge Atomic radius set of average bonding radii taken from measurement on a large number of elements & compounds Ion formation The charge of main group elements that form cations = the group # The charge of main group elements that form anions = the group # 8 Transition elements differ* its cation the highest nvalue orbitals are removed first, regardless of reverse order of filling Ionic radii Cations < Neutral atoms < Anions Ionization energy ( the energy needed to remove an electron from the atom/ion in the gaseous state) Electron affinity energy charge associated w/gaining electrons; usually negative because of energy released 02/29/2016 trends aren’t as consistent as atomic radius or ionization energy but EA becomes more () as it moves right ___________________________________________________________ Chapter 5: Molecules & Compounds Remember common polyatomic ions ***** Ionic compounds (charge neutral) can either be simple or ransitional Simple Transitional Prefixes within nomenclature only apply to covalent bonds Hydrated ionic compounds contain a certain amount of water molecules associated with each formula unit ex. MgSO 4 7H 2O ⇒Magnesium sulfate heptahydrate ____________________________________________________________
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