EK Chem Ch1 Gen Chem Flashcards

Question 1

Q

isotopes

Answer

A

two or more atoms of the same element that contain different number of neutrons Hydrogen= H1 = protium, H2 is deuterium and H3 is tritium. 99.98% of naturally occurring H is protium Carbon isotopes= C12, C13, C14. each of carbon’s isotopes contain 6 protons with varying numbers of neutrons. 6 protons are what define carbon, if number of protons changed it would no longer be carbon C12= 6 adn 6 C13= 6 protons and 7 neutrons C14= 6 protons and 8 neutrons

Question 2

Q

cation

Answer

A

-ion with fewer electrons than protons -positive charge size -when a neutral atom loses an e to become a cation, it gets smaller -atom still has same amount of protons but there are now more protons than electrons, so as a result that positive charge of the nucleus exerts a greater attractive force on each valence e, pulling them closer to nucleus -loss of e also reduces the repulsive forces between the electrons, further contributing to dec in size -net effect removing e makes ionic radius smaller than atomic radius

Question 3

Q

anion

Answer

A

-ion with more electrons than protons -negatively charged -gain e it becomes larger, makes the ionic radius larger than atomic radius - atom now has more e than protons, so the positive charge of the nucleus pulls less strongly on each individual valence electron. The addition of an e also inc the repulsive forces btw the e, pushing them farther away from each other. - net effect is that anion is larger than the neutral atom

Question 4

Q

ion

Answer

A

when the number of electrons in an atom does not equal the number fo protons, the atom carries a charge and is called an ion

Question 5

Q

salt

Answer

A

neutral compound* - made up of positive and negative ion together

Question 6

Q

metal

Answer

A

large atoms that tend to lose e to form positive ions and positive oxidation states - left hand side of table - atom in a sea of electrons, which emphasizes their loose hold on their electrons and the fluid like nature of their vlence electrons, easy movement of e within metals is what gives them their metallic character. -ductility (easily stretched) mallealbility (easily hammered into thin strips) -thermal + electrical conductivity -lusterous? - all metals besides mercury exist at room temperature -electrons move easily from one metal atom to the next, transferring energy or charge in the form fo heat or electricity

Question 7

Q

nonmetals

Answer

A

form covalent bonds, vs metals who form ionic -generally have lower melting points than metals and tend to form anions, which commonly react with metal cations to form ionic compounds ex. NaCl

Question 8

Q

group 11 transition metals ions…

Answer

A

makes 1+ ions, Cu+, Ag +, and Au+, all others in this group are Au3+ and Cu2+

Question 9

Q

there are five 3+ ion transition metals….

Answer

A

Cr3+, Fe3+, Au3+, Al3+, and Bi3+ all other ions if not mentioned above as 1+ are 2+

Question 10

Q

valence e

Answer

A

electrons in the outer shell, elements in the same group on the periodic table have similar chemical properties because they have the same number of valence e

Question 11

Q

diatomic molcules

Answer

A

-elements that tend to exist as diatomic molecules are H, O, N and the halogens -so when these are every discussed assume that they are in their diatomic form unless otherwise stated -so when they say “nitrogen is nonreactive” refers to N2 not N - Fl2 and Cl2 are gases at room temp, br2 diatomic liquid, I2 a diatomic solid**

Question 12

Q

halogens

Answer

A

Fl2 and Cl2 are gases at room temp, br2 diatomic liquid, I2 a diatomic solid** - react with H to form gaseous hydrogen halides, which are soluble in water, forming hydrohalic acids

Question 13

Q

periodic trends

Answer

A

-atomic radius, ionization energy, electronegativity and electron affinity - atomic radius inc as goes down and to the left, everything else inc up and to the right**

Question 14

Q

atomic radius

Answer

A

distance from center of the nucleus to the outermost electron -corresponds to the size of the atom -moving across period radius dec b/c each subsequent e has an additional proton, which pulls more strongly on the surrounding electron -moving down a group, new shells of e are dded. these outer es are “shielded” from the attraction of the protons in the inner nucleus, so atomic radius inc as move down group

Question 15

Q

Zeff

Answer

A

-the effective nuclear charge is the charge felt by the most recently added e. in perfect shielding, each e added to an atom would be completely shielded from the attractive force of all protons except for the last proton added and the Zeff would be 1 eV for each electron -without shielding, each e added would feel the full attractive force of all the protons in the nucleus, and the Zeff would simply be equal to Z for each electron

Question 16

Q

ionization energy

Answer

A

energy needed to detach an e from an atom -inc along periodic table from left to right and from bottom to top -when an e is more strongly attached to teh nucleus, more energy is required to detach it

Question 17

Q

first ionization energy

Answer

A

energy necessary to remove an e from a neutral atom in its gaseous state to form a 1+ cation -is largest for the noble gas within a given period because the electron removed was completing a stable octet configuration

Question 18

Q

second ionization energy

Answer

A

-energy required to remove second e from teh same atom to form a 2+ cation - 3rd,4,5, etc are all named the same. -second ionization energy is always greater than the first because once one electron is removed the effective nuclear charge inc for the remaining e -3rd, 4th, 5th ionization energies similarly inc -largest inc in ionization energy occurs when the e to be detached was completing a stable octet configuration

Question 19

Q

Zeff useful for periodic trends

Answer

A

inc from left to right across a period, so each new e is pulled closer to the nuc and held more tightly than the previous one 1. pulling outermost e closer dec atomic radius 2. holding the outermost e more tightly inc ionization energy 3. atoms with greater Zeff will pull more strongly on e in covalent bonds, inc electronegativity across a period 4. atoms with stronger Zeff will more readily accept another electron, so electron affinity inc across a period

Question 20

Q

electronegativity trend

Answer

A

tendency of an atom to attract electrons shared in a covalent bond -when two atoms have diff electronegativities, they share electrons unequally causing polarity -relative electronegativity determines the direction of polarity within a bond and within an overall molecule -inc across a period from left to right and up

Question 21

Q

electronegativity

Answer

A

most common used measurement for this is pauling scale, which ranges from a value of 0.79 Cesium to a value of 4 for fluorine -F most electronegativity element

Question 22

Q

electronegativity of H

Answer

A

-falls between that of boron and that of carbon -when bonded with H, carbon and elements to teh right of carbon will carry a partial negative charge while H will carry a partial positive -think Ch4, boron, and elements to the left of boron will carry a partial positive charge when bonded to hydrogen, while teh hydrogen will carry a partial negative charge. Think of the hydrides H- in NaH or LiAlH3

Question 23

Q

pauling scale for electronegativity large differences

Answer

A

electronegativity values provide system for predicting which type of bond will form btw two atoms -atoms with large electronegativity 1.6 or larger* on Pauling scale as a rule of thumb, will form ionic bonds! -metals and nonmetals usually exhibit large electronegativity differences and form ionic bonds with each other

Question 24

Q

pauling scale for electronegativity moderate differences

Answer

A

-atoms with moderate differences in electronegativities (0.5-1.5 on Pauling scale) will generally form polar covalent bonds

Question 25

Q

pauling scale for electronegativity moderate differences

Answer

A

atom with v minor electronegativity differences generally form nonpolar covalent bonds, 0.4 or smaller on pauling scale

Question 26

Q

Pauli Exclusion

Answer

A

that no two electrons in the same atom can have the same four quantum numbers

Question 27

Q

electron affinity

Answer

A

willingness of an atom to accept an additional electron -more precisely, it is the energy released when an electron is added to an isolated atom -inc on periodic table from left to right and from bottom to top -the sign of e affinity values can be different for different atoms because some atoms release energy when accepting an e, and thus become more stable while others require energy input to force the addition of an electron since the additional electron dec stability WARNING: e affinity is sometimes described in terms of exothermicity, for which the energy released is given a negative sign. We can state this as follows: electron affinity is more exothermic to the right and up on the periodic table. noble gases do not follow this trend. Electron affinity values for the noble gases are endothermic, because noble gases are stable. therefore, significant amounts fo energy are required to force them to take on electrons and become less stable

Question 28

Q

fluoride (size influence)

Answer

A

small atoms hold charge in a concentrated way because they have fewer orbitals available to distribute and thereby stabilize charge - concentration of charge makes the smallest element in each group bond more readily and with greater bond strength, especially when in ionic form -Fluorine is a good ex, F- is too small to manage its full negative charge. Therefore it is generally insoluble and bonds immediately when in solution. For this reason, in toothpaste, fluoride a poison in high concentrations bonds immediately with the enamel of teeth before it can be ingested.

Question 29

Q

Aufbau principle

Answer

A

THINK GERMAN ORDER - “building up principle” - with each new proton added to create a new element, the new e added to maintain neutrality will occupy the lowest energy level available -everything being equal, lower energy state of a system should be the most stable in the system

Question 30

Q

electron configuration tricks

Answer

A

make sure total number of e in your e configuration equals the total number of e in the atom -an e can momentarily absorb energy and jump to a higher energy level, creating an atom in an excited state versus ground state which is when electrons are all at their lowest energy levels (normal form from table)

Question 31

Q

electron configurations of transition metals

Answer

A

-not same as the nearest noble gas -ions are formed from losing e from the subshell with teh highest principle quantum number FIRST** -generally this is hte s subshell, it is also important to note that there are a few expectations to the electron configuration rules in the transition metals -Half-filled and filled subshells offer greater stability -elements in groups 6 and 11 are expected to have nearly half-filled or nearly filled d subshells, instead they borrow one e from the highest s subshell so they end up with a half filled s sub shell and a half-filled or filled d subshell -This phenomenon is most likely to appear on MCAT with Cr and Cu, which have only one e in the 4s orbital

Question 32

Q

Cu e configuration

Answer

A

-e configuration of Cr is [Ar] 4s^1 3d^5

Question 33

Q

Cr e configuration

Answer

A

-e configuration of Cu is [Ar] 4s^1 3d^10

Question 34

Q

Hund’s rule

Answer

A

e will not fill any orbital in the same subshell until all orbitals in that subshell contain at least one e, and that the unpaired e will have a parallel spins -remember like charges repel each other, if consider the energy of two particles with like charges, we would find that as the particles approach the mutual repulsion creates an inc in potential energy - this occurs when e approach each other so e avoid sharing an orbital when possible, spreading out amongst the orbitals of a given subshell to minimize PE

Question 35

Q

paramagnetic

Answer

A

there are unpaired e so subshell is not completely filled
spin of each unpaired e is parallel to the others, as a result electrons align with an external magnetic field

Question 36

Q

diamagnetic

Answer

A

-no unpaired e like He, so subshells are completely filled

- they are unresponsive to an external magnetic field

Question 37

Q

emission line spectrum

Answer

A

when excited e falls from a higher energy state to a lower energy state, energy is released
this energy creates this, that is characteristic of the given element

Question 38

Q

absorption line spectrum

Answer

A

-measures the radiation absorbed when e absorb energy to move to a higher energy state

Question 39

Q

photoelectric effect

Answer

A

Einstein demonstrated one to one, photon to electron collision
used one to one collision to prove that light is made up of particles
light shining on a metal may cause the emission of electrons, sometimes called photoelectrons in the context of he photoelectric effect.
Since the energy of a wave is proportional to its intensity, one might expect that when the intensity of lighting shinning on a metal is inc by inc the number of photons, the KE of an emitted electron would inc accordingly. But this is not the case, instead, the kinetic energy of the electrons increases only when intensity is inc by inc the frequency of each photon. if the frequency is less than the necessary quantum of e, no electrons at all will be emitted regardless of number of photons. This demonstrates that the electrons must be ejected by one ot one photon electron collision rather than by the combined energies of two or more photons.

Question 40

Q

work function

Answer

A

minimum amount of energy required to eject an electron is called work function, phi of the metal KE excess= Hf (energy in) - phi (electron out) hf is the energy put in by a photon, and phi is the energy required to eject the e from the atom -energy left over is the electron’s KE

Question 41

Q

hydrogen bond

Answer

A

-responsible for high bp of water -strongest type of intermolecular force, it is still much weaker than any covalent bond -strongest type of dipole-dipole interaction. H bond occurs btw a H that is covalently bonded to a F, O or N and N, F, or O atom from another molecule -F, O or N are HIGHLY electronegative, when bound to H a large dipole moment is formed, leaving H with a strong partial positive charge

Question 42

Q

intermolecular attractions

Answer

A

-attraction btw separate molecules, occur due to dipole moments -partial neg charge of one molecule is attracted to partial positive charge of another molecule -they are weak electrostatic bonds, generally about 1% as strong as covalent bonds. -Attraction btw two molecules is roughly proportional to the magnitude of their dipole moments the stronger the dipole, stronger the attraction ** these are usually insignificant in gases because the molecules are spread far apart, but these forces apply to solids and liquids hte same

Question 43

Q

london dispersion forces/ van der waals

Answer

A

-weakest dipole dipole force btw two instantaneous dipoles - these dipole dipole bonds are called London dispersion forces -all molecules exhibit London dispersion forces, even when they are capable of stronger intermolecular interactions

Question 44

Q

mega

Question 45

Q

kilo

Question 46

Q

deci

Question 47

Q

centi

Question 48

Q

milli

Question 49

Q

micro

Answer

A

u (weird u) 10^-6

Question 50

Q

nano

Question 51

Q

pico

Answer

A

p, 10^-12

Question 52

Q

femto

Answer

A

f, 10^-15

Question 53

Q

percent composition of mass

Answer

A

-multiply each element’s atomic weight by the number of atoms it contributes to the empirical formula -then divide the result by the net weight of all the atoms in the empirical formula, which yields the mass fraction of that element in the compound - multiply by 100% to get hte percent composition by mass ex. percent mass of carbon in glucose, empirical formula CH2O molecular weight of carbon/molecular weight of CH2O = 12/30= 0.4 0.4 X 100= 40 glucose is 40% carbon by mass

Question 54

Q

determine empirical formula ex

Answer

A

if we are asked to find the empirical formula of a compound that is 12.5% H, 37.5% C, 50% O by mass= from a 100 g sample= 12.5 g H/ 1 g/mol = 12 mol 12/3 = 4 37.5 g C/ 1 g/mol = 3 moles 3/3 = 1 50 g O/ 1 g/mol = 3 moles = 3/3 = 1 There is a 4:1 ratio of H to each of the other elements. Empirical formulas is H4CO

Question 55

Q

Aluminum

Question 56

Q

Ammonium

Question 57

Q

Barium

Question 58

Q

Calcium

Question 59

Q

Chromium II, Chromous

Question 60

Q

Chromium III, Chromic

Question 61

Q

Copper I, cuprous

Question 62

Q

Copper II, cupric

Question 63

Q

Hydrogen, hydronium

Question 64

Q

Iron II, ferrous

Answer 49

A

** remember pancreas releases this in digestion HCO3-

Answer 50

A

(CrO4) 2-

Answer 51

A

(Cr2O7) 2-

Answer 52

A

(C2O4) ^-2

Answer 53

A

(HC2O4) ^-

Answer 54

A

(MnO4) ^-

Answer 55

A

(PO4)^ 3-

Answer 56

A

(HPO4)^ 2-

Answer 57

A

(H2PO4)^ -

Answer 58

A

(SO4) ^2-

Answer 59

A

grams/ atomic or molecular weight 6.022 X 10^23 amu= 1 gram

Answer 60

A

concerns atoms that sponateously break apart - all atoms other than Hydrogen are subject to some type of spontaneous decay - the mass and identity of an atom is located within its nucelus -the rate at which decay occurs varies dramatically - atoms with a relatively high decay rate are said to be radioactive

Answer 61

A

-length of time necessary for one half of a given amount of a substance to decay. A given sample has a predictable rate of decay, usually given in terms of a half-life. -radioactive decay follows first order kinetics, and the amount of atoms that remain after decay can be expressed as follows= At= Aoe^ (-kt) or ln(At/Ao)= -kt where At is the amount at time t, Ao is the original amount, k is the rate constant and t is time - notice that radioactive decay describes a constant e being raised to an exponent kt, EXPONENTIAL DECAY*** for radioactive decay

Answer 62

A

there are only 4 possible variables 1. the initial amount of substance 2. the final amount of substance 3. the length of the half-life 4. the number of half-lives (often given as a time period, in which case you simply divide by the length of a half-life *Any mcat half-life question will provide you with three of these variables in some form and will ask you to find the fourth. This o answer a half-life question count the number of half-lives on your fingers. For instance, if 12.5% of a substance remains after 5 years, what is the half life? The initial amount is of course 100%. The final amount is 12.5%, the number of half-lives is found by dividing the initial amount by 2 until you arrive at the final amount. 50% is once, 25% is twice, 12.5% is three times. so three half-lives

Answer 63

A

The initial amount equals 500, final amount equals 62, half life is 2 years. Divide the initial amount by 2 until you arrive at the final amount. Keep track of the number of half-lives on your fingers, 250 is one, 125 is two, 62.5 is three. so 3 half lives or 6 years

Answer 64

A

-loss of an alpha particle! -an alpha particle is a helium nucleus, meaning it contains 2 protons and 2 neutrons. An example of alpha decay is : 238^ U 92 –> -4^alpha 2 (look on page 34) + 234 ^ Th 90 = CHANGE IN MASS NUMBER= -4 - change in atomic number, -2 new element name= 2 to the left on periodic table

Answer 65

A

this is the breakdown of a neutron into a proton and electron and the expulsion of the newly created electron - since a neutron is destroyed but a proton is created, the mass number stays the same, but the atomic number increases by one - positron emission and electron capture are types of beta decay ex. 234 ^Th (90) –> 234 ^Pa (91) + 0^ e (-1) - neutron becomes proton, electron emitted or positron absorbed - no change in mass number - change in atomic number is +1 - new element name is 1 to the right on the periodic table

Answer 66

A

a neutrino is also emitted during beta decay - is a virtually massless particle, it is typically represented by greek letter nu (v)

Answer 67

A

the emission of a positron when a proton becomes a neutron, this is considered to be a type of beta decay. A positron can be though of as e with a positive charge - both e and positrons are consdiered to be beta particles - in positron emission, a proton is transformed into a neutron adn a positron is emitted 22^ Na (11) –> 0 ^ e (1) + 22^ Ne (10) - proton becomes neutron, positron is emitted! - no change in mass number, change in atomic number is -1 - new element name is 1 to the left on the peridodic table

Answer 68

A

capture of an e and merging of hat e with a proton to create neutron - proton is destroyed and a neutron is created - no change in mass number, change in atomic number is -1 - new name element is 1 to the left o n periodic table ex. 201^ Hg (80) + 0 ^e (-1) –> 201 ^ Au (79) + 0 ^y 0

Answer 69

A

emit high energy gamma ray, this is a high frequency proton -it has no mass or charge, and does not change the identity pf the atom from which it is given off - gamma ray emission often accompanies the other types of radioactive decay can occur when an e and positron collide -1 e + 1 e –> 0 ^ y0 + 0^ y0 - no change in mass number, no change in atomic number, no change in element name

Answer 70

A

-this is matter-antimatter collision - mass is destroyed converted to energy in the form of gamma rays

Answer 71

A

electrons in covalent bond between two different atoms ( H and Cl in this case) are not equally shared by the atoms. This is due to the electronegativity difference btw the two atoms, the more electronegative atom (Cl) has greater share of the electrons than the less electronegative atom (H). Consequently, the atom that has the greater share of the bonding electrons bears a partial negative charge and the other atom automatically beras a partial positive charge of equal magnitude

Answer 72

A

often occurs between atoms that are the same

electronegativity difference between bonded atoms is small (<0.5 Pauling units)

electrons are shared equally between atoms

Answer 73

A

always occurs between different atoms

electronegativity difference between bonded atoms is moderate (0.5 and 1.9 Pauling units)

electrons are not shared equally between atoms

Answer 74

A

Methane (CH4) is an example of a compound where non-polar covalent bonds are formed between two different atoms. One carbon atom forms four covalent bonds with four hydrogen atoms by sharing a pair of electrons between itself and each hydrogen (H) atom. The electronegativity value for carbon (C) and hydrogen (H) is 2.55 and 2.1 respectively, so the difference in their electronegativity values is only 0.45 (<0.5 criteria); the electrons are thus equally shared between carbon and hydrogen. So we can conveniently say that a molecule of methane has a total of four non-polar covalent bonds.

Answer 75

A

1, like C-O= 2.5-3.5= 0 so polar covalent bond

0= non covalent bond

if difference greater than 0.5, polar covalent, less than 0.5 nonpolar covalent bond!

difference btw polar covalent and ionic, higher than 1.7 mostly considered to be an ionic bond, lower than 1.7 is considered to be polar covalent range

around 1.5-2.1 is difference btw polar and ionic bond

Answer 76

A

So, the result of this exercise is that we have six towels attached to each other through thread and Velcro. Now if I ask you to pull this assembly from both ends, what do you think will happen? The Velcro junctions will fall apart while the sewed junctions will stay as is. The attachment created by Velcro is much weaker than the attachment created by the thread that we used to sew the pairs of towels together. A slight force applied to either end of the towels can easily bring apart the Velcro junctions without tearing apart the sewed junctions.

Exactly the same situation exists in molecules. Just imagine the towels to be real atoms, such as hydrogen and chlorine. These two atoms are bound to each other through a polar covalent bond—analogous to the thread. Each hydrogen chloride molecule in turn is bonded to the neighboring hydrogen chloride molecule through a dipole-dipole attraction—analogous to Velcro. We’ll talk about dipole-dipole interactions in detail a bit later. The polar covalent bond is much stronger in strength than the dipole-dipole interaction. The former is termed an intramolecular attraction while the latter is termed an intermolecular attraction.

Answer 77

A

Intramolecular forces are the forces that hold atoms together within a molecule.

Answer 78

A

Intermolecular forces are forces that exist between molecules.

Answer 79

A

A polar covalent bond is formed when atoms of slightly different electronegativities share electrons. The difference in electronegativity between bonded atoms is between 0.5 and 1.9. Hydrogen chloride, \text{HCl}HClstart text, H, C, l, end text; the \text{O}-{H}O−Hstart text, O, end text, minus, H bonds in water, \text{H}_{2}\text{O}H2Ostart text, H, end text, start subscript, 2, end subscript, start text, O, end text; and hydrogen fluoride, \text{HF}HFstart text, H, F, end text, are all examples of polar covalent bonds.

Answer 80

A

This type of covalent bonding specifically occurs between atoms of metals, in which the valence electrons are free to move through the lattice. This bond is formed via the attraction of the mobile electrons—referred to as sea of electrons—and the fixed positively charged metal ions. Metallic bonds are present in samples of pure elemental metals, such as gold or aluminum, or alloys, like brass or bronze.

The freely moving electrons in metals are responsible for their a reflecting property—freely moving electrons oscillate and give off photons of light—and their ability to effectively conduct heat and electricity.

Answer 81

A

These forces occur when the partially positively charged part of a molecule interacts with the partially negatively charged part of the neighboring molecule. The prerequisite for this type of attraction to exist is partially charged ions—for example, the case of polar covalent bonds such as hydrogen chloride. Dipole-dipole interactions are the strongest intermolecular force of attraction.

Answer 82

A

This is a special kind of dipole-dipole interaction that occurs specifically between a hydrogen atom bonded to either an oxygen, nitrogen, or fluorine atom.

Hydrogen bonding is a relatively strong force of attraction between molecules, and considerable energy is required to break hydrogen bonds. This explains the exceptionally high boiling points and melting points of compounds like water, and hydrogen fluoride***

Hydrogen bonding plays an important role in biology; for example, hydrogen bonds are responsible for holding nucleotide bases together DNA and RNA

Answer 83

A

These are the weakest of the intermolecular forces and exist between all types of molecules, whether ionic or covalent—polar or nonpolar.

The more electrons a molecule has, the stronger the London dispersion forces are. For example, Br2, has more electrons than Clr, so bromine will have stronger London dispersion forces than chlorine, resulting in a higher boiling point for bromine, 59 C compared to chlorine -35 C.

Also the breaking of London dispersion forces doesnt require that much energy, which explains why nonpolar covalent compounds like methane- CH4 oxygen and nitrogen which only have London dispersion forces of attraction btw the molecules freeze at VERY LOW TEMPERATURES***

Answer 84

A

Polar covalent compounds—like hydrogen chloride, HCl, hydrogen iodide HI- have dipole-dipole interactions between partially charged ions and London dispersion forces between moelcules.

Nonpolar covalent compounds- like methane CH4 and nitrogen gas, N2 only have london dispersion forces between molecules. The rule of thumb is= ****= the stronger the intermoelcular forces of attraction, the more energy is required to break those forces!!***

This translates into ionic and polar covalent compounds having higher boiling and melting points, higher enthalpy of fusion and higher enthalpy of vaporization than covalent compounds

Boiling and melting points of compounds depend on the type and strength of the intermolecular forces present, as tabulated below:

Answer 85

A

london dispersion forces, v small difference in electronegativity btw C and H so canceld out, nonpolar as a result of that
no H bonding, no dipole-dipole
boiling point
think about electrons in these bonds moving in these orbitals! v small attraction btw two molecules, v weak only thing holding together these methane molecules
since it is weak expect boiling point of methane to be extremely low= -164 degrees celsius, since room temp around 20-25 C methane already boiled and turned into a gas, gas at room temperature adn pressure!!! if you increase the number of carbons, you will inc the number of attrctive forces possible, if inc carbons actually inc boiling point of other hydrogen carbons dramatically, if larger molecules adn sum up all of those extra forces turns out to be signficiant if workign with larger molecules!

Answer 86

A

sigma bonds + # of lone pairs

Answer 87

A

E= h V

long wavelength loewr frequency, smallernumebr of crests passing point per minute or second, carries less energy, slow rollign waves vs. smaller wavelength and higher frequency, that is higher energy

Answer 88

A

c= wavelength X frequency

shorter wavelengths are more harmful

long wavelength loewr frequency, smallernumebr of crests passing point per minute or second, carries less energy, slow rollign waves vs. smaller wavelength and higher frequency, that is higher energy

Answer 89

A

a constant

c= 3.00 X 10^8 m/s

important equation= E= hc/wavelength

Answer 90

A

Na compared to Na+ , parent is bigger Na> Na +
if comparing Na to Na-, you are bulking up anion when add extra electron, so Na- > Na*
when comparing isoelectronic (meaning same number of electrons) atoms and ions really empty space, greaer positive chrge in middle will pull in e more* = so rule for isoelectronic ions, the one with more protons is smaller** so comparing things with same number of electrons more protons willl squeeze in e more and make radius smaller**

Answer 91

A

the energy change that occurs when an element becomes its anion!

the more exothermic that reaction is, the larger and more negative the electron affinity is the happier this element is to become its anion* elements who like to form anions liek F- favorable electron affinity!

so like f lives to become F- so electron affinity for that would be a negative number, negative electron affinity has pretty large magnitude to signify favorable process for becoming an anion

some of these trends run parallel to each other, but definition of electorn affinity is energy change when element becomes an anon*** electronegativity is informative but not exactly what they are askign for, its when this element is participating in a bond how much does it pull electron desnity toward itself**

Answer 92

A

energy required to make cation*** energy needed to take away electron and make cation* opposite of electron affinity whch is adding e to make anion

Answer 93

A

proxy for reactive, things more metallic usually more reactive!
metals on left, nonmetals on right so if nonmetals on right as go towards right becomes less metallic, zone from metals to nometals
if throw Na in water reacts get a poof, K in water bigger explosion, if through Rb into water your table starts shakign, if put Cs in water wear a hazmat suit!
EXCEPTION: in general when go down a group have an inc in metallic character! so K has greater metallic character and is more reactive than Na, by time get down to Cs very reactive
BUT there is an exception to that if look at periodic table the first row of the transition metals are more metallic then the second or the third row of transition metals* that is just a fact to memorize*

Answer 94

A

colored because of energy diff btw 4s adn 3d orbital wavelenvght of that photon corresponds to light in the visible range**

Answer 95

A

also relates to fact that 4s and 3d orbitals are very close in energy to eachother* specific to 4s and 3d* so when doing electron configuration add electrons that are 4s before you would add 3d
BUT when makign ion take out 4s before hte 3d** which is weird but energies of those orbitals are so close to eachother weird flexibility or variability that occurs
ex. Fe = [Ar] 4s^2 3d6

but Fe2+= [Ar] 3d6, so take out 4s electrons first* an EXCEPTION to usual rule!!**

Answer 96

A

if high 2nd IE, cost removign 2nd electorn with 2 IE is so highw ill stay at plus 1

so sodium and potassium like to form plus one, then general trend for ionization energy so in terms of second highest ionization energy inc when go up a group**

as atoms become bigger valence e less controlled by nucleus, shielding, so outer electrons easier to rip off** so Na would have a highest second ionization energy*

= Na

Answer 97

A

Cl- larger radius than Cl, where as Na+ would have smaller radius than Na

In general anions have much larger radii across the board compared to cations

Cl-> then K+, Na and Ca2+ becuase they are cations

Answer 98

A

visible portion of electromagnetic spectrum

Answer 99

A

No lone pair on aluminum, can technically have 6 electrons like Be*

lewis acid can take another pair of electrons, part of aromatic electrophilic subsitition* refresher on where you have seen alcl3 before*

Answer 100

A

CH IS NOT POLAR almost anything else will be a little bit polar enough to rgister here

A molecule is polar if it has :

at least one polar bond
no overall symmetry- (into this becuase if you have even SO3

Answer 101

A

assume 100 g so %–> gram amounts, so if 58.8% Ba equals 58.8 g
convert g to moles for each element ex. 58.8 g Ba (1 mol Ba/ 137 g) = .43 mol Ba
we care about ratio of these numbers! so divide mol amounts by smallest to largest**
use whole number ratio to get subscripts in the formula* for compound

Answer 102

A

find empirical formula, ex. 58.8% C, 9.8% H, and 31.4 % O; assume 100 g use 58.8 g C X (1 mol. 12 g) = 4.9 mol of C divided by smallest in this case it was 1.96 so 4.9 /1.96
if not whole number ratios, multiply by smallest factor possible that would give you a whole numner; whatever do to one have to do to all of them, so empirical formula C5, H10, O2
part 2- to find molecular formula, so need molar mass of compound is equal to 204 grams/mol given information; mass of empirical formula ( 5 x mass of C, 10 X 1 for H, plus 2 X 16 for O)= 204 g mol; tells us that the empirical formula is not the same thing as the molecular formula, so we know molecular formula has to be a multiple of empirical formula, know mass of empirical formula unit 102 g/ mol, so if see at a mass of 102 and target is a mass of 204 so multiple by 2, tells us should multiply empirical formula subscripts* by 2
204/ 102 g/mol = 2 so C₁₀H₂₀O₄

Answer 103

A

AB + X –> XB + A

usually metal in compound coming in and replacing, booting out hte other

Answer 104

A

A + B —> AB

AB–> A + B

Answer 105

A

AB + XY–> AY + XB

Answer 106

A

by definition means rapid combination with oxygen- KNOW PRODUCTS ARE CO2 and H2O** so hydrocarbon can have methanol, propane, octane, gasoline in cars, recognizing what the formula is*

hydrocarbon + O2–> CO2 + H2O

* the less stable a compound is the more energy it releases upon combustion**less stable higher energy, so when reacting them can release more energy*

Answer 107

A

CA(NO3)2 aq + 2 KOH (aq) –> Ca(OH)2 (s) + 2 KNO3 (aq)

when asked to find net ionic equation, what htis is really asking is can you simplify or boil down this process to say in a more concise way what actual change is taking place! what is the transformation that is the heart of this reaction; to simplify the change taking place** to boil it down, so what that means is first makign what is called a complete ioni equation* so for things present as ions in the solution to write them as ions* so starting with Calcium nitrate* that is aqueous so that exists as ions, Ca2+, so of these 2NO3- ionic form actually exists in solution

Ca²⁺ (aq) + 2NO₃- (aq) + 2K⁺(aq) + 2 OH- (aq) —> Ca(OH)₂ (s-exists like this) + 2K⁺(aq) + 2NO₃- (aq)

cancel out anything that appears exactly the same way on the left and right side of this equation; any ions uncharged are referred to as spectator ions, not participating in any change here just along for the ride
left with= Ca²⁺ + 2 (OH-) –> Ca(OH)2 is net ionic equation, just includes things that participate in some kind of change, so these ions find eachother and create solid that perticipates from solution* thats net ionic equation*

Answer 108

A

Electrons within an atom are usually understood as occupying the lowest-energy possible orbitals, which is referred to as the “ground state” for a particular electron. However, when an atom absorbs energy, an electron can be promoted to a higher-energy orbital, entering what is known as an “excited state.” When an electron drops from a higher-energy orbital back to its ground state, it releases energy as a photon (itself a form of electromagnetic radiation).

The amount of energy required for an electron to transition between orbitals is specific to each element. Thus, the emission or absorption of electromagnetic radiation results in a characteristic atomic emission spectrum and atomic absorption spectrum that are unique to each element. For hydrogen, the amount of energy gained or lost in an electron transition is given by the Rydberg formula, ΔE = R(1/nf2 – 1/ni2), where R is the Rydberg constant (2.18 × 10-18 J) and nf and ni represent the final and initial energy states of the electron, respectively.

If enough energy is absorbed by an atom, an electron can be expelled from the atom entirely. The minimum amount of energy necessary to do so is known as the work function of a substance. The fact that this energy can be delivered by an incident photon is known as the photoelectric effect.

Answer 109

A

Radioactive decay is the spontaneous transformation of one atomic nucleus into another. Often, this involves a change from one element into a different element. The only way this transformation can happen is by changing the number of protons (and often neutrons) in the nucleus, since atomic identity is defined by the number of protons.

There are a number of ways that this can happen, and when it does, the atom is forever changed. There is no going back – the process is irreversible. There are four primary types of decay: alpha decay, beta decay (β+ and β-), gamma decay, and electron capture. In alpha decay, an alpha particle, containing two protons, two neutrons, and a +2 charge, is emitted. In beta-minus decay, a neutron is converted into a proton in the nucleus, and a β- particle (an electron) is ejected to maintain charge balance. In beta-plus decay, a proton is converted into a neutron, and a β+ particle (a positron) is emitted to preserve charge. Gamma decay involves the emission of a gamma ray, which is a high-energy photon, from an excited nucleus. Finally, in electron capture, a nucleus “grabs” an electron, which changes a proton into a neutron.

When an atom does not gain or lose any protons, and only the neutron count is changed, a new isotope is formed. An isotope is a variety of an element that is distinguished from other varieties of the same element by having a different atomic mass; however, isotopes of a given element all share the same atomic number and chemical properties. These isotopes are often used in radiolabeling techniques in the biological sciences (for example, using 2H, or deuterium (D), to track amino acid uptake in protein translation).

Answer 110

A

Another high-yield fundamental concept is half-life (t1/2), which is the time required for one-half of the parent isotopes in a sample to decay into daughter (radiogenic) isotopes. If we know the half-life of a material, then we can determine how much of a sample is lost (1- ½n) or remains (½n) at any given time, as expressed in the number of half-lives that have passed (n).

Answer 111

A

amount of time between each drop having btw concentration is a constant*

IT IS A CONSTANT****

can calculate half life by looking for any pair of concentraitons 1/2 apart however long it takes for drop to occur, just pick any two points half life parat on epxoential decay graph should give you same length of time

Answer 112

A

-radiation taht ejects an electron from an sp orbital

highest energy one that injets electron from orbital close as possible to nucelus, means tightest hold so needs tightest energy to kick that electron out, eject electron more energy to get rid of it

Answer 113

A

shine light and emits waves
shinign light or electromagnetic wave onto a material, and as a result of that we are getting electrons to pop off
electrons beign liberated from material as a result of light shinning**
how does this work? changings things about light and how they are done?
look in notebook

Answer 114

A

applies idea of quantization of energy to the atom

saying electrons in the atom can only occupy fixed energy levels*

he thinks of them as circular orbits, but saying the occupy these 6 circular orbits each associated with specific energy

b/c in fixed energy levels, it moves around by absorbing fixed /specific amount of energy that corresponds to a photon of exactly that amount*

photon HAS TO HAVE exact amount of energy, has to have difference btw first energy level adn second energy level, means has to also have a very very specific frequency and therefore have a very specific color!

yellow or green, corresponds to exact difference, super important which allows Bohr to explain line emission spectra

Answer 115

A

Energy nth level = -13.6 eV/ n²

unit electron volts, eV= an aside eV is a very small unit of energy, 1.6 x 10-19 J, they expect you on exam to be able to do this conversion

Answer 116

A

each band/line in line emission spectrum is a specific frequnecy of light, aka color

THAT means it corresponds to a specific energy photon, that line corresponds to photons of all the same energy, so therefore what those energies are are the differences btw energy levels in atom** ideal all this empty spaace becuase photons of htat energy cannot be emited, can only emite photons that correspond to exact right number btw

has frequency photon, which correspond to energy differences in atom

can only have photons correspond to differences, 10.2 eV can emit a photon with 10.2 eVs, or I could do one for 3-1 difference btw those is 12 .1, or can do 1.5 to 3.4 differences btw those is 1.9 eV

Answer 117

A

A is correct. Bohr proposed “stationary orbits,” or orbits at the specific distances at which an electron is stable. These orbits are associated with energy levels. Jumping between orbits must then require either the absorption or the emittance of a photon with a frequency that is related to the difference in energy levels between the orbits and to Planck’s constant.

B- is wrong b/c this is very close. However, the Bohr model only gives accurate results for one-electron systems and when the charged points move at speeds significantly lower than the speed of light.

Answer 118

A

Question 12

An unknown molecule, which has the ability to form multiple stable oxidation states, tends to exist in brightly-colored compounds. This species is most likely a:

A.

Noble Gas.

Noble gases do not have multiple stable oxidation states.

B.

halogen.

Some halogens are brightly colored in their atomic states. They do not, however, have multiple stable oxidation states.

C.

alkaline earth metal.

Alkaline earth metals have an oxidation state of +2 and therefore do not have multiple stable oxidation states.

D.

transition metal.

D is correct. Two of the most well-known chemical hallmarks of transition metals are their ability to possess multiple oxidation numbers and their tendency to form brightly-colored compounds. This second characteristic stems from the arrangement of their d electrons.

Answer 119

A

Question 16

Alkali metals are defined by their:

I. reactivity with water.

II. ability to act as strong reducing agents.

III. ability to form strong acids when bound to hydrogen.

IV. ground-state valence shell configuration of ns1.

A.

I only

Show Explanation

B.

I and III

III: Alkali metals form hydrides, such as NaH, when they bind to hydrogen. These compounds are actually basic, not acidic.

C.

I, II, and IV

C is correct. Alkali metals are located in the first column of the periodic table; they include sodium and potassium. These metals are highly reactive with water. Due to their extreme tendency to lose a single electron and reach a noble gas configuration, they serve as good reducing agents.

D.

I, III, and IV

III: Alkali metals form hydrides, such as NaH, when they bind to hydrogen. These compounds are actually basic, not acidic.

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