VSEPR

? of protons in Copper-65 ⁶⁵ ...Cu ₂₉ atomic mass= p⁺ + n⁰ ...................................X atomic # = #p⁺

#p = defines the element

Dalton's Law of Partial Pressure

* Can't use true combination law P Total=∑P initial ex. in Lab P H₂ + P H₂O

ideal gas law

** Use this equation when nothing changes i.e. there are initial measurements BUT NOT FINAL MEASUREMENTS PV=nRT

stoich

-------------- mass A→mol A→mol B→mass B avo # 6.022 x 10 ²³ g↓.......↑ How many? .....mol

net ionic equation

1) molecular eq= normal neutral equation 2) complete ionic eq= coefficient, ion charge & state *** only separate (aq) (g) (l) (s)= don't break apart 3) net ionic eq = remove non∆'s ions= leave only 1's that ∆'d

Empirical formula

1) ∆% = g 2) g÷mm 3) ÷ that ans for each element by smallest # of mols 4) if necessary to get whole no., mult. results by 2, 3,4... II. molecular formula 1) get mm of Emp formula via calculations 2) mm of True formula = given 3. n= mm True formula/ Emp formula 4. n (Empirical formula)= true formula

boyles law

1/p inverse proportional 1increase, other decrease

1atm= 760 mmHg = 760 torr

1atm= 760 mmHg = 760 torr; 1mmHg=1Torr; 1 atm=101kPA

Because pH is the negative log of the H3O+ concentration, a higher pH corresponds to a lower [H3O+], and each unit of pH represents a tenfold change in concentration.

A solution has a pH of 5. What is the pOH of the solution? Since the pH is 5, the pOH = 14 − 5 = 9.

In which solution is [OH−] equal to 0.100 M?

Ammonia (NH3) is a weak base and therefore only partially ionizes water. Its [OH-] in this solution is < 0.100 M. NaOH is a strong base. Strong bases dissociate completely in solution to form stoichiometric amounts of OH─. A weak base only partially ionizes water. Barium hydroxide (Ba(OH)2) is a strong base and therefore completely dissociates in solution. The [OH-] in this solution is 0.200 M.

monoprotic acid

Any acid that one contains one hydrogen atom in its formula is monoprotic. donates only one proton or hydrogen atom per molecule to an aqueous solution. This is in contrast to acids capable of donating more than one proton or hydrogen, which are called polyprotic acids

conjugate acid pair

Any two substances related to each other by the transfer of a proton can be considered a conjugate acid-base pair. In an acid-base reaction, a base accepts a proton and becomes a conjugate acid. An acid donates a proton and becomes a conjugate base

In a titration, we react a substance in a solution of known concentration with another substance in a solution of unknown concentration. For example, consider the acid-base reaction between hydrochloric acid and sodium hydroxide: HCl(aq)+NaOH(aq)→H2O(l)+NaCl(aq)

As we add the OH−, it reacts with and neutralizes the H+, forming water. At the equivalence point—the point in the titration when the number of moles of OH− added equals the number of moles of H+ originally in solution—the titration is complete. The equivalence point is usually signaled by an indicator, a dye whose color depends on the acidity of the solution ex pink solution

most important BASE RXNS are those in which a base neutralizes an acid (see the beginning of this section). The only other kind of base reaction that we cover in this book is the reaction of sodium hydroxide with aluminum and water.

BASE REACTIONS

In which solution is [OH−] equal to 0.100 M

Barium hydroxide (Ba(OH)2) is a strong base and therefore completely dissociates in solution. The [OH-] in this solution is 0.200 M. Correct! NaOH is a strong base. Strong bases dissociate completely in solution to form stoichiometric amounts of OH─. A weak base only partially ionizes water. Ammonia (NH3) is a weak base and therefore only partially ionizes water. Its [OH-] in this solution is < 0.100 M.--------------------------------------------------- weak bases :the most common weak bases produce OH− by accepting a proton from water, ionizing water to form OH−. B(aq)+H2O(l)⇌BH+(aq)+OH−(aq) ---

A buffer contains HCHO2(aq) and KCHO2(aq). Which statement correctly summarizes the action of this buffer?

By means of Brønsted-Lowry acid-base reactions, the acid component of the buffer (HClO2) neutralizes added base and the base component of the buffer (ClO2─ or KClO2) neutralizes added acid.

A buffer contains HCHO2(aq) and KCHO2(aq). Which statement correctly summarizes the action of this buffer? Ans: HCHO2(aq) neutralizes added base, and KCHO2(aq) neutralizes added acid

By means of Brønsted-Lowry acid-base reactions, the acid component of the buffer (HClO2) neutralizes added base and the base component of the buffer (ClO2─ or KClO2) neutralizes added acid.

What is the conjugate base of the acid HClO4?

Correct! ClO4 A conjugate base has one fewer H than its corresponding conjugate acid. A conjugate base is the corresponding acid with one fewer proton. HClO4-> ClO4

What is [H3O+] in a solution with [OH−] = 2.5 × 10-4 M?

Correct! In an aqueous solution there is a balance between [H3O+] and [OH ─] that is quantitatively expressed by the Kw expression. You can use this expression to convert between these concentrations. [H3O+]= 1.0 × 10-14 M/2.5 × 10-4 M= =4.0 × 10-11 M

Hydrogen bonding is an electrostatic attraction between two polar groups. Dipole-dipole interaction is an interaction between the positive end of one molecule with negative end of another molecule. Dispersion forces are the forces in which electrons in two adjacent atom behaves like dipoles and attract toward each other.

Dispersion forces are present in all atoms and molecules. Hydrogen bonding exists between molecules that have an H atom bonded directly to an F, O, or N atom. Dipole-dipole forces exist between molecules that are polar. Kr is an atom and does not contain H, F, O, or N. . Therefore, only dispersion forces are present in Kr

E= h∨ ∨= c/λ E= h*c .....___ .......λ

E= h∨ ∨= c/λ h=6.626 x 10⁻³⁴ J*s c= 3.0 x 10⁸ ¹/s

atomic size decreases as go up and right across Periodic Table

EN increases as as go up and right across Periodic Table

4 exceptions to Octet rule

H, He, =2 ve's B= 6 Ve's Be=4Ve's

Question 2: Identify the Brønsted-Lowry base in the reaction. HClO2(aq) + H2O(aq) ⇌ H3O+(aq) + ClO2−(aq)

H2OCorrect! Brønsted-Lowry bases are proton acceptors. Such a base has one more H in the products than the reactants. In this reaction,ClO2 is a product; a Brønsted-Lowry base is a proton acceptor (in the forward reaction). In this reaction, HClO2 is a proton donor; a Brønsted-Lowry base is a proton acceptor (in the forward reaction). In this reaction, H3O+ is a product. A Brønsted-Lowry base is a proton acceptor (in the forward reaction).

buffers

HF and NaF HC2H3O2 and KC2H3O2

Question 9: What is [H3O+] in a solution with [OH−] = 2.5 × 10-4 M?

In an aqueous solution there is a balance between [H3O+] and [OH ─] that is quantitatively expressed by the Kw expression [H3O+] [OH ─] =1.0*10^-14. You can use this expression to convert between these concentrations

dispersion forces

O2 is a nonpolar molecule because its atoms are identical, and does not contain an H atom bonded directly to an F, O, or N atom. Therefore, only dispersion forces are present in O2 NO is a polar molecule and does not contain an H atom bonded directly to an F, O, or N atom. Therefore, dispersion forces and dipole-dipole forces are present in NO. HF is a polar molecule and contains an H atom bonded directly to an F atom. Therefore, dispersion forces, dipole-dipole forces, and a hydrogen bonding are present in HF.

In which solution is [H3O+] less than 0.100 M? One of these solutions has a hydronium ion concentration that is less than 0.100 M. Recall that strong acids completely ionize while weak acids do not.

One of these solutions has a hydronium ion concentration that is less than 0.100 M. Recall that strong acids completely ionize while weak acids do not. Sulfuric acid (H2SO4) is a diprotic acid that is strong in its first ionization and weak in its second; therefore, its [H3O+] is greater than 0.100 M. Perchloric acid (HClO4) is a strong acid and has an [H3O+] = 0.100 M. Correct! Formic acid (HCHO2) is the only weak acid among the choices, so its [H3O+] is less than 0.100 M. ---------------------------------------------------------------------------------------Since HCl is a strong acid, it completely ionizes. The concentration of H3O+ is 1.5 M. [H3O+]=1.5 M Since HC2H3O2 is a weak acid, it partially ionizes. The calculation of the exact concentration of H3O+ is beyond the scope of this text, but you know that it is less than 3.0 M. [H3O+]<3.0 M

Ideal gas law constant

R=.08206 L * atm /mol*K

color of visible light with shortest wavelength λ = violet

ROYGBIV= visible light *low energy, low frequency ( grt wavelength)←R →V high energy, high frequency ( short wavelength) ( most energetic photons)

imf The strength of the intermolecular forces between the molecules or atoms that compose a substance determines the state—solid, liquid, or gas—of the substance at room temperature. Strong intermolecular forces tend to result in liquids and solids (with high melting and boiling points). Weak intermolecular forces tend to result in gases (with low melting and boiling points). In this book, we focus on four fundamental types of intermolecular forces. In order of increasing strength, they are the (London) dispersion force, the dipole-dipole force, the hydrogen bond, and the ion-dipole force.

Remember that all molecules (including polar ones) have dispersion forces. In addition, polar molecules have dipole-dipole forces. These additional attractive forces raise their melting and boiling points relative to nonpolar molecules of similar molar mass. For example, consider the compound

Stand Temperature & Pressure

STP 0°; 1 atm ; 22.4L/mol 22.4L/mol ---> this is the molar vol of gas

Formic acid (HCHO2) is the only weak acid among the choices, so its [H3O+] is less than 0.100 M.

Sulfuric acid (H2SO4) is a diprotic acid that is strong in its first ionization and weak in its second; therefore, its [H3O+] is greater than 0.100 M. Perchloric acid (HClO4) is a strong acid and has an [H3O+] = 0.100 M.

diatomic molecules

S₈ P₄ H₂ N₂ O₂ F₂ Cl₂ Br₂ I₂

A 25.00-mL sample of an HNO3 solution is titrated with 0.102 M NaOH. The titration requires 28.52 mL to reach the equivalence point. What is the concentration of the HNO3 solution?

The concentrations of the two solutions would only be the same if the volume of NaOH required to reach the endpoint were exactly the same as the volume of the sample being titrated (which is not the case here). Determine the number of moles of NaOH required to reach the equivalence point. Then determine the number of moles of HNO3 in the sample and divide by the sample volume to get the concentration. The problem says that the HNO3 solution has a volume of 25.00 mL. The volume of the 0.102 M NaOH solution required to meet the equivalence point is 25.82 mL.

pOH

The pOH is related to [OH─] by the defining equation pOH = ─ log[OH─].

Acid + Metal→H2 gas +salt

The reaction between an acid and a metal usually produces hydrogen gas and a dissolved salt containing the metal as a cation.

BRØNSTED-LOWRY DEFINITION Acid—An acid is a proton (H+ ions) donor. Base—A base is a proton (H+ ion) acceptor.

This definition more clearly accounts for what happens to the H+ ion from an acid: It associates with a water molecule to form H3O+ (a hydronium ion). The Brønsted-Lowry definition also works well with bases (such as NH3) that do not contain OH− ions but that still produce OH− ions in solution. NH3 is a Brønsted-Lowry base because it accepts a proton from water:

Which of the following has the highest boiling point?

Which of the following has the highest boiling point? : They are all nonpolar. CH4 CH3CH3 CH3CH2CH3 CH3CH2CH2CH3 All of these molecules are non polar. So, they will have dispersion forces as the only intermolecular force. The greater the molar mass of molecule, greater would be the dispersion force and hence greater will be the boiling point. Boiling point will be largest for CH3CH2CH2CH3 Answer: CH3CH2CH2CH3

Kw= ion product constant for H2O =1.0x10⁻¹⁴ ** no units ** do not round pH?

[H₃O⁻][OH⁻] =(1.0x10⁻⁷M)(1.0x10⁻⁷M) ==1.0x10⁻¹⁴ [H₃O⁺]=1.0x10⁻⁷M NEUTRAL SOLUTION

exothermic rxn

aA +bB↔cC + heat HEre: add Heat= shift away (L) remove heat= twd removal=go twds heat ~shift twd-(R) kJ =heat ** temp affects ENDOTHERMIC rxns

HCl(aq)+NaHCO3(aq)→H2O(l)+CO2(g)+NaCl(aq)

acid reacting with carbonates or bicarbonates (compounds containing CO32− or HCO3−). This type of gaseous-neutralization reaction produces water, gaseous carbon dioxide, and a salt.

rxn rate

amt produced that forms in a given amt of time

Brønsted-Lowry bases

are proton acceptors. Such a base has one more H in the products than the reactants. h20→H3o⁺ A conjugate base has one fewer H than its corresponding conjugate acid. ex. HCLO4→CLO4

amphoteric In Brønsted-Lowry terminology, able to act as either an acid or a base.

arrhenius base A substance that produces OH− ions in aqueous solution.

4 e groups, 2 lone pairs bond angles <109.5

bent .... °° H- S -H ....°°

2 outer atoms 3 e groups, 1 lone pair bond angle 120deg<

bent .....°° O- N -O

Electrolyte solutions

contain dissolved ions (charged particles) and therefore conduct electricity.

Non electrolyte solutions

contain dissolved molecules (neutral particles) and therefore do not conduct electricity.

Buffer: How buffers resist pH change A buffer contains significant amounts of a weak acid and its conjugate base. The acid consumes any added base, and the base consumes any added acid. In this way, a buffer resists pH change. blood Like all buffers, blood contains significant amounts of both a weak acid and its conjugate base. When additional base is added to blood, the weak acid reacts with the base, neutralizing it. When additional acid is added to blood, the conjugate base reacts with the acid, neutralizing it. In this way, blood maintains a constant pH.

contains significant amounts of both a weak acid and its conjugate base. Buffers can also be composed of a weak base and its conjugate acid.

form of energy atoms absorb= HEAT

form energy atoms EMIT= EM RADIATION

supersaturated solution

holds more than the normal maximum amount of solute. The solute will normally precipitate from (come out of) a supersaturated solution.

saturated solution

holds the maximum amount of solute under the solution conditions. If additional solute is added to a saturated solution, it will not dissolve

le chatliers principle

if a system rxn is at equilibrium *& a stress added; (like a ∆ in concentration, temp, vol, or pressure) the sys will respond by removing that stress & which thereby re-est. equilibrium

catalyst

like in oxy la affects spd rxn; doesn't mess with equilibrium *enzymes= biological catalyst that's like lock n key model= ur making proteins & gets particle A, Chem rxn & from grabs particles & chem rxn SNAPS & makes protein

2 outer atoms , 0 lone pairs 180 deg

linear

which transition would produce a lower energy photon

n=3 to n=2 or n=7 to n=6 distance between orbitals is greatest in smaller n, those closest to nucleus= orbitals furthest from nucleus have closesr distances to each other and thus moving down or the electron tranistion would produce lower energy going down from something close together

equivalence point,

neither reactant is present in excess, and both are limiting. The number of moles of the reactants are related by the reaction stoichiometry (se

solubility

of a compound is defined as the amount of the compound, usually in grams, that dissolves in a certain amount of liquid.- *increases with higher temp------------------------------------------------------------------------ For example, the solubility of sodium chloride in water at 25 °C is 36 g NaCl per 100 g water, while the solubility of calcium carbonate in water is close to zero.

oxidation state

or oxidation number—for each element based on the number of electrons assigned to it.

OILRIG

oxidation, then, is the loss of electrons, and a more fundamental definition of reduction is the gain of electrons. If one substance loses electrons (oxidation), then another substance must gain electrons (reduction)

....I P..S..Cl H..C..N..O..F

polar molecule i. lone pair on central atom ii. or if bonded atoms = diff EN values ex. F has diff EN than Cl ; all attached to central atom ............F ............| .......Cl-N-Cl

∃ products over reactants 1994 final * NOT CURRENT ONE

products ÷reactants products/reactants both raised to approp powers from ratio ka= weak acid kb= weak base ks solubility product EQUILIBIRUM EXPRESSION FOR ANY RXN products [C]∧c [D]∧d ------------------- reactants [A]^a [B]

A buffer,

resists pH change by neutralizing added acid or added base.

Covalent bond

single bond 2 e- shared dbl bond 4e- shared triple bond 6e- shared

ionization energy

smaller atoms on top right= harder to ionize = higher ionization energy= electrons held tightly to postive nucleus larger atoms bottom left= easier to ionize =ionization energy decreases as go down , left periodic table=electrons held less tightly to nucleus

weak acids

strong acids= HBr, HCl, HNO₃, H₂SO₄,HCLO₄,HCLO₃, HI

pH and pOH at 25 °C from the expression for Kw. Kw= [H3O+][OH−]=1.0×10−14

sum of pH and pOH is always equal to 14.00 at 25 °C pH + pOH =log(1.0×10−14)−14.0014.0014.00

4outer atoms 4 e groups, 0 lone pairs 109.5 degrees

tetrahedral ......H ......| H- C -H .......| ......H

pH of a solution

the negative of the logarithm of the hydronium ion concentration: pH=−log[H3O+] -- Calculating pH from [H3O+] A solution having an [H3O+]=1.5×10−7 M (acidic) has a pH of: pH====−log[H3O+]−log(1.5×10−7)−(−6.82)6.82 -- Calculating [H3O+] from pH 10^⁻pH Calculate the H3O+ concentration for a solution with a pH of 4.80. = 10^⁻pH=10^⁻4.80 -- The pOH scale is analogous to the pH scale but is defined with respect to [OH−] instead of [H3O+]. pOH=−log[OH−] Notice that p is the mathematical function: −log; thus, pX=−log X. Calculating [OH−] from pOH Calculate [OH−] for a solution with a pOH of 8.55. [OH−] =10^⁻pOH=10^-8.55=2.8*10^⁻9

3 outer atoms, 0 lone pairs , bond angles =120deg

trigonal planar .....O .....| Cl- C -Cl

3 outer atoms, 4 e groups, 1 lone pair bond angles <109.5

trigonal pyramidal ....Cl .....I Cl- B -Cl ....°°

Pi Vi = P2 V2 = R _______ ________ niTi n2 T2

true combination law Pi =pressure initial Vi =vol initial ni=moles initial Ti= temperature initial R=.08206 L * atm /mol*K

when atoms absorb energy= electrons make transitions from LOW TO HIGH ENERGY LVLS

when atoms EMIT energy= electrons make transitions from HIGH TO LOW LVLS

concentration

↑particle, ↑lvl=↑product BUT **MUST OVERCOME energy activation

Q=mc∆T....Q= heat or absorption of heat; all chemical rxns that we use for energy are exothermic= emits energy

∆T= Tf-Ti c= specific heat capacity m= mass