Chem second semester final

Flashcards by kayceecrist11, updated more than 1 year ago
Created by kayceecrist11 about 8 years ago


High School/Secondary Exams Honors chemistry Flashcards on Chem second semester final, created by kayceecrist11 on 25/05/2013.

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Question Answer
photoelectronic effect movment of electrons in some metals at specific wavelengths of light creating a weak current flow
electron cloud image produced due to the movment of an electron around the nucleus of an atom
orbital specific shape of an electron cloud
energy and frequency equation E = h x frequecy where E is energy and h is planks constant which equals 6.626 x 10 to the negative 34 J x second
sublevel a region in space where an electron will most likley be found , it is the region around the nucleus where the electron will most likley be seen if it could be seen
d orbital
s level orbital
p orbital
p orbital
f orbital
f orbital
aubauf principal electrons enter orbitals of lowest energy first
pauline exclusion principal orbitals contain a maximum ampunt of two electrons. the first electon has a positive spin and the second electron has a negative spin
pauline exclusion principal orbitals contain a maximum ampunt of two electrons. the first electon has a positive spin and the second electron has a negative spin
pauline exclusion principal orbitals contain a maximum ampunt of two electrons. the first electon has a positive spin and the second electron has a negative spin
hunds level in sublevels that contain more then one orbital so pretty much the P D and F the e- will enter each orbital one at a time with a positive spin before doubling up
energy leavel diagram representation fo the elecrons relative to their energy
principal quantum number states the energy region of the electrons . the principal quantum number representation is n = n1 = s level only n2= s and p levels only n3= s , p and d levels only n4 = s,p,d, and f levels only
electron configuration representation of the orbital, sublevelm and number of electrons in the sublevel 2He= 1S2 1 is the quantum number , s is the orbital,and then two is number of electrons in the sublevel. when writing out these the principal quantum numbers must be in order so if 4s is before 3d then the 3d and 4 s must be switched
placment of reaction electrons reaction electrons are located in the highest quantum number, not on the highest sublebe
promotions s-d levels electrons will prmote to allow greater stability they do this 1. half filling their outer s and d levels 2. totally filling the d level and half filling the s. it is the s level electrons that are always promoted
reasons why the condensed electron figuration is needed - elements in a noble gas arrangment are non reactive in a normal reaction system -this configuration presenst the outermost electrons which will most likley be invilved in reactions , reactions always happen in the highest quantum level
condensed shorthand notation 3Li = [ 2He] 2s1 27 [CO] 4s2 3d7
Orbital overlays Look at notes for these they are pretty straight forward
quantum address way of writing the location of an electron. 20Ca (4,0,0-1/2) 41Nb (4,2,0,1/2)
1st number of the address principal quantum number , gives the main level of an e-. They will be the principal quantum numbers 1 for s subleve. 2 for p sublevel. 3 for d subleve. 4 for f sublevel.
2nd number of the address is the angualr movment , it gives the type of orbital 1. 0 for s sublevel 1 for p sublevel 2 for d sublevel 3 for f subleve
third level of the address maximum quantum number this the orientation of the orbital . is is a Px , a DXy, or and F... to assign these you do an energy level row for this sublevel start at the middle go positive numbers to the right and negative to the left
third level of the address maximum quantum number this the orientation of the orbital . is is a Px , a DXy, or and F... to assign these you do an energy level row for this sublevel start at the middle go positive numbers to the right and negative to the left
4th level of the address magnetic spin number- is eithe a positive 1/2 for the first electron in or a -1/2 for the second electron in
stability rule 1 noble gas configuration. it wants its outer electrons to be liek that of a noble gas meaning outermost s and p levels
stability level 2 pseudo noble gas configuration filled outtermost s, p and d levels
stability rule 3 outermost filled s orbital
stability rule 4 half filled outtermost s,p,d,f
stability rule 5 none of the above.
stability rule one is the greatest and 5 is the least. atoms want their electrons to be in the greatest stability arrangment
chemical tendency the gaining or losing of electrong during the formation of an ion
metal chemical tendency tends to lose e-
nonmetals chemical tendency tends to gain e- from anions
periodic patterns traits that exist as you move across or down sections of the periodic table
atomic radius (size of radius) the measure of hald the distance between atoms lath are next to eachother in a crystal. the pattern ACROSS THE PERIOD - size of the atoms radius's decrease DOWN THE GROUo - size of the atoms radius increase
ionization energy amount of energy needed to remove an atom from an atom ... look in notes for examples. PATTERN : ACROSS THE PERIOD - ionization increases DOWN THE GROUP - ionization energy decreases this is due to the shielding effect
shielding effect the outter e- are hidden from the nucleus and easier to remove theen the inner ones because they are hidden from the nucleus
Determining is elements follow the inonization energy you find out what happens when you remove the last electron from the last level of the energy level diagram then find out where it ranks in terms of chemical stability and chemical tendency . if it its more beneficial to ionize then it wil. use the periodic pattern to find out if it follows the pattern
doing second ioniozation energy same procedure as finding the first ionization energy but you take out 2 electrons . it follows the second IE pattern if it takes on the same characteristics of the 1st IE
electron affinity measures the chane in energy when an e- is added to an atom
periodic pattern for electron affinity Across the period excluding nnoble gases the EA increases down a group it decreases due to the shielding effect
figuring out if elements follow the periodic pattern is gone about the same way as ionization energy . if an atom gains greater stability by gainig an electron then it then it needs less energy if it is hurt then it needs more energy .
electro magnetivity measures teh ability of an atom to pull electrons toward itself in a bond
periodic tend of electromagnetivity across a period - electromagnetivity increases down a group- electromagnetivity decreases
electromangetivity differences 0.0-0.4 nonpolar covalent bond 0.4 - 1.0 slightly polar convalent bond 1.0 -2.0 sliglthy polar covalent bond 2.0 and greater ionic bond
finding electromandetivity of two elements look at the chart on the back of the periodic table and find the numbers for each element and subtract them
heterogeneous mixture in which you can see the seperate particels contained in it
homogenious you can not see the seperate parts of the mixture
electrolyte a solution that conducts electricity because the substances dissolved in it allow it to conduct an electrical current because they are charged particles
nonelectrolytes a substance that dissolves in water to give the solution that does not conduct an electrical current .
process of solvation chemical making of a solution. 1. breaking solute -solute bonds 2. breaking solute-solvent bonds *** both above are endothermic 3. Making solute solvent bonds *** it is an exothermic reaction
solution smallest particles of combined matericals they do not refect light beacues the particles are so slow that the light just passes over them
colloid they are large and they refelct light and give a hazy look they are combinations of particles, alloys are a type of these
suspensions mixtures where partices are suspended in the solution then sink back down to the bottom
solute substance being dissolved
solvent substance doin the dissolving
solvation of an ionic crystal look in the notes
solvation of a molecular crystal look in the notes
rate of solution how fast we can get a substance in solution
wasy to speed up rate of sultion of a solid into a liquid 1. agitate it 2. crush the soluton 3. increase the temperature of the mixture
increasing rate of solution of a gas into a liquid 1. decrease the temp of the mixture 2. increase the pressure of a mixture
saturated solution maximum amount of solute into soultuon
ubsaturated solution more solute will be able to be dissolved
supersaturated solution more solution in solvent then is normally possible
reading solubility graphs look in notes
molarity mass solute / l of solutiom
how to make a mixture use a volumetric flask. fill it half way with distilled H20 . add the G of the solute which will be found using the molarity. then swirl it until fully dissolved then fill up to the etched line with distilled H20
moalrity with stiochemisty how many grams of Cl2 are produced when 87.2 ml of 42MHCL reacts with excess phophorous. You will be ginvent the balaenced equation and you must use the molarity relationship ANSWER: 1.30 x 10 to the 1 grams of Cl2
molality moles solute / Kg solvent
using molality in math how many grams NaOh are neede to reproduce 7.6 M NaOH using 665 g H20 ANSWER: 2.02 x 10 to the second g NaOH
dissociation formation of ions in aqueous solution from ionic substances
ionization formation of ions in aqueous solutions from molecular substances
colligative properties relationship influenced by the number of particles in a sample
freezing point depression delta t f = Kf . m . number of formula particle delta t f is the number of degrees the freezing point will change Kf is the molar Fp constant m is the moalr concentration
finding the freezing point depression determint the frezing point depression of 9.81 m KCl aqueous solution ANSWER: -36.5 degrees celcius
boliing point elevation delta T =Kb . m. number of formula particles. delta K b = change of degrees for boiling m = molal concentration kb= molar boliling point constant
finding boiling point elevation calculate the boiling point of 5.26 m nonelectrolyte ether solution ANSWER: 45.2 degrees celcius
molar dilution equation M1V1=M2V2
finding molar dilutions what amount of 4.7 m H2SO4 solution is needed to prepare 150 ml of a 1.5 molar H2SO4 solution ANSWER; 4.8 x 10 to the 1 ml
normality H+ or OH- mol equivelenace/ l solution 3.3 M HCL = 3.3 normal HCl 3.3 M H2SO4 = 6.6 normal H2SO4 4.4 normal H2Cl = 2.2 H2SO4
organic chemistry compounds composed of covalently bonded carbon, hydrogem and smaller amounts of elements such as nitrogen, oxygen, halgoens ect
alkane single bonds in the carbon chain
prefixes for expressing number of carbons in a compound 1. meth 2 eth 3 prop 4 but 5. pen 6. hex 7. hept 8. oct 9. non 10. dec
ane ending saying there is only one bond
substituents additions to the main chaing of carobns
substituent the halogens Florine- floro Chlorie- Chloro Iodine - Iodo Bromine- Bromo
substituents the methyl group Ch3-methyl Ch2, Ch3 - ethyl CH2,CH2,CH3- propyl
substituents the methyl group Ch3-methyl Ch2, Ch3 - ethyl CH2,CH2,CH3- propyl
other substituents amino - NH2 nitro - NO2
nomenclature rules 1. locate the longest carbon chaing , this is assigned the family name 2. the substituents are assigned carbon positions so they will dispay the lowest number of carbon positions 3. substituents are arrened in aplhabetical order are prefixes are used to describe the number of atoms
condensed formula read notes for this . you will always start at the first carbon and wirte it out from there
alkene family says theri are double bonds in the carbon chain
alkyne family says their are triple bonds in the main carbon chain
alkene and alkyne nomenclature the multiple bonds are assigned the smallest numb carbon in the chain 2. the substituents are then assigned the next smalles carbon numbers 3 alphabetical order is next used
deriving R in the ideal gas law equation You first need to convert your PV, n and T to the appropriate thing like atm, mL, Mmoles and Kalvins. then plug the values in the equation and you will find R
Using the Ideal gas law equation determine the volume of a sample of chlorine weighing 12.4g at 25.6 degrees celcius and 3.95 atmospheres R= 8.21 atmL / molesK ANSWER : 1.09 x 10 to the 1st L CL2
Graph of the ideal Gas Law equation so the graph is extrapalated. P and V are on the y axsis and n and T are on the other. there will be 2 slopes the slope of A and the slope of B and they are equal to eachother
Molar conversions 1 mole = molar mass 1 mole = 6.02 x 10 to the 23rd particles 1 mole = 22.4 L at STP *** for the ideal gas laws you can not use this conversion
Using the ideal gas law and stiochemetry C5H10 reacts with O2 at 25.0 degrees celcius at 842 torr. determine the volume in ml of CO2 produced if the pentine measures 18.9 g. Ok so your gonna do the converstions for the ideal gas law equations. then your gonna do the mapping and go from g C5H10 to moles C5H10 to moles of the CO2 at this point you cant go from moles to l because because its not at STP so you use the ideal gas law equation to get to liters and the go from liters to Ml CO2 . ANSWER: 2.98 x 10 to the 4th mL CO2
Daltons law of partial pressures the total pressure of a mixture of gases equals the patrial pressures of all the gases that are mixed into it
Daltons law of partial pressures equation PT= P1+P2+P3....
Daltons law of partial pressures equation PT= P1+P2+P3....
How to use Daltos law of partial pressures What is the pressure of a gas collected by the displacemet of H2O if the total pressure is 895.6 torr ar 24.0 degrees celcius PT= p + 120+ px H20 Pressure = 22.4 torr 895 torr = 22.4+Px Px= 873.2 torr
Effusion movment of gas through a narrow opening
diffision movment of gas from a region of higher concentration to a region of lower concentration
diffision movment of gas from a region of higher concentration to a region of lower concentration
Grams Law of Effustion equation Rate of A / Rate of B = the square root of MMB / MMA **** MM stands for molar mass
using grams law of effusion equation option 1 Compare the rate of effusion between CO2 and N2 ANSWER : N2 effuses 125 times faster then CO2
using Grams law of effusion option B Co2 effuses at a rate of 1.16 times faster then and unknown gas determine the molar mass of the Gas. Anawer = 59.4 grams / mole
quantum mechanics the sturcures of an atom relating the frequency , wavelenght, and energy signatures of the atoms electrons
electro magnetic spectrum display of the relationshi of electromangnetic radiation
radiation way of categorizing energy systems
the electromagnetic spectrum Radio Waves micro waves inferred Visibe ROYGBIV ultra volate x-rays gamma rays cosmic rays the spectrum moves from left to right!
crest the top of a wave
trough the very bottom of the wave
Wavelenght the distance between corresponding parts of the wave.
lambda the greek letter that symbolizes wavelength
Frequency the number of cycles a wave goes through within a given amount of time . it is figured out cycles / minute
nu the greek letter that looks like a v and represents frequency
wavelenght and frequency relationship inversly preportional
wavelenght and the electromagnetic spectrum as you move from right to left the wavelength becomes shorter
frequency and the electromagnetic spectrum the frequency increases as you move across the electromagnetic spectrum
equation for wavelength frequency C=Wavelength X frequency C is the speed of light and is a constant number equaling 3.00 x 10 to the 8th meter / second
Energy and wavelength relationship they are inversly preportional
energy and the electromagnetic spectrum the center visible light is neutral and then going to the left it is low energy and then from the center going right is high energy
quantum expression of energy from the electromagnetic spectrum
photon quantum of light energy
particle theory of light light travels in bundles (particles) that pulse from a source. EX: a red glow from a hot metal
Pressure Conversions 1 atm - 760 terr 1 atm - 760 mg Hg 1 atm - 101.3 KPa *** all four of these measurments are equal and can be interchanges they are also conversion factors and can be used in dimensional analysis
Boyles Gas laws at a constant temperature, the volume and pressure of a gas will vary inversley . as pressure increases volume will decrease and as pressure decreases the pressure will increase.
boyles gas law equation P1 V1 = P2 V2 both the pressure and the volume must be taken at the same time.
manometer a device used to measure the cahnge in pressure of a gas.
example probelm of Boyles Gas law at a constant temperature , the volume of a gas sample is 26.9 L at 32 atm. what is the volume in mililiters if the pressure becomes 878 torr? ANSWER : 7.4 X 10to the 4th ml
Charles's Gas Laws at a constant pressure temperature and volume will vary preportionally . As temperature increases so will pressure and as pressure decreases so will temperature.
Temperature in Gas Law Caluclations Temperature in Gas laws is always calculated in Kelvin Units .
Calcuation for Kelvin Unites degrees Celcuis + 273 = Kelvin Units
Charles Gas Law Graph This graph will taper off into values unknown , when this happens the term is extrapalted, temperature will be on the Y axsis and the volume will be on the X axsis
Charles Gas law equation V1T2=V2T1
Gay-Lussiacs Gas Law at a constant volume themperature and pressure will vary preportionally . As temperature increasis so will pressure as temperature decreases so will pressure
gay Lussiacs graph this graph is also extrapalated. The temperature will be on the Y axsis and the pressure will be on the X axsis.
Heating of a Gas the kenetic energy of the Gas will increase. The number of collision wil the sides of the container the gas is in increases and the pressure increases as well.
Cooling of a Gas if you place a container of a gas in an ice bath it slows the kenetic energy of the particels down. because the energy has decreases the volume will decreases because they are exerting less force and also the pressure also decreases because their are less collisions of the gas against their container
combined gas laws relationship that allowes the volume,pressure and temperature to vary simoultaneously. In this realtionship if something is held constant then it is canciled out of the equation
combined gas law equation P1V1T2 = P2V2T1
ideal gas law equation PV=nRT n= number of moles of the Gas R= the ideal gas law constant
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