chem112 exam

how much energy is absorbed when 355 g of acetone is vaporized at 56.1 C? (acetone boiling point is 56.1)

(355g)(1mol acetone/58.0791g acetone)(29.1 kJ/1mol aceteone)= 178 kJ

c to f

(C x 1.8) + 32

c-f

(C x 1.8) + 32

f-k

(F - 32) * 5/9 + 273.15

f to c

(F-32)/1.8

f-c

(F-32)/1.8

K-f

(K-273.15)*9/5+32

specific heat capacity

(c)energy required to raise the temperature of a specific amount of substance by one degree water per gram 2.087 j /g C (s), 4.184 j/g C (l) , 2.042 j/gC (g) per mole 47.6 j/m C (s), 75.37 J/mol C(l), 36.03 J/mol C (g)

R=

0.08206 L atm/mol K

log(2.2)=

0.34 10^.34=2.2 calculator 10^x (.34)=2.2

ln(2.2)=

0.79 ln=loge log (2.71)^2.2 e^(.79)=2.2 log(2.2)/log(2.71)= 0.79

torr to mmHg

1 torr = 1 mmHg

solve for x 81=13^x

1.71 log13 81 means what power do I need to take 13 to to get 81 13^1.71=81 in calculator log(81)/log(13)

liquid water enthalpy of vaporization

100 C 40.7 kj/mol 25 C 44.0 kj/mol

1atm= kpa?

101.3kpa = atm?

1 atm = mbarr?

1013.25 mbar = atm?

na2so4 how dissociate into how many ions?

3 ions 2 na, 1 so4

1 atm

760 mmHg

torr-atm

760torr=1atm

lnp2/lnp1 =

= lnp2 - lnp1

propanol has a normal boiling point of 97.00 C and a heat of vaporization of 41.2 kj/mol. What is the vapor pressure of propanol at 61.00 C, in mmHg?

=180mmHg T2=97.00 C --> 370.15K P2= 760 mmHg T1= 61.00 C --> 334.15 K ln p2/p1= ΔH vap/R (1/t1 - 1/t2) ---> lnP2-lnP1 ΔH vap in j/mol instead of kj/mol to consistent with R 41.2kj/mol --> 41200j/mol R=8.3145 J/k mol ln(760mmHg)-lnP1= (41,200 j/mol)/(8.3145 J/K mol) ((1/334.15K)-(1/370.15K)) ln(760mmHg)=6.633 6.633-lnP1=1.422 lnP1=6.633-1.442=5.191 P2=e^5.191=180mmHg

e

=2.71

Solvent

A liquid substance capable of dissolving other substances water (solvent) salt (solute)

polar molecule

A molecule that has electrically charged areas.

solute

A substance that is dissolved in a solution.

solute

A substance that is dissolved in a solution. salt (solute) and water (solvent)

Ion

An atom or group of atoms that has a positive or negative charge. unequal amount of protons and electrons

Adhesion

An attraction between molecules of different substances concave meniscus in container if more attracted to container

electrolytes

Aqueous ionic compuounds, and strong acids acids (metal or ammonium ions) (nh4+) 100% ionization al2(so4)3 = 5 ions (2 al3+, and 3 (so4)2-) more particles = higher boiling point

Cohesion

Attraction between molecules of the same substance convex meniscus in container if more attracted to itself

c-k

C + 273.15

henrys law

C=kP

Allotropes

Different forms of the same element

exceptions to the octet rule

Hydrogen (2), Boron (3-6), Aluminum(sometimes 6, but 3), Phosphorus(10), Sulfur(up to 12).

k-c

K-273.15

Buckyball

Molecule composed of 60 carbon atoms arranged in a sphere. c60

ln P2/P1= ΔH vap/R (1/t1 - 1/t2)

Pvap(torr) 450 720 T(K) 270 370 lnPvap 6.11 6.58 1/T (K^-1) 0.0037 0.0027 ln P2/P1= ΔH vap/R (1/t1 - 1/t2) ln 720torr/450torr=ΔH vap/ 8.3145 J/k mol (1/270 - 1/370) 0.470=ΔH vap/8.3145 j/K mol (0.0010 K^-1) ΔH vap=8.3145 J/K mol (0.470) / 0.0010 K =3907.815 j/mol ==3.9k kj/mol

lnPvap=-ΔH vap/R (1/T) +lnB

Pvap(torr) 450 720 T(K) 270 370 lnPvap 6.11 6.58 1/T (K^-1) 0.0037 0.0027 lnPvap=-ΔH vap/R (1/T) +lnB y=mx+b m=-ΔH vap/R m(slope)= is y2-y1/x2-x1 (6.58-6.11)/(0.0027-0.0037)=-470 K slope=-ΔH vap/R ---> ΔH vap=-slopexR -(-470K) R --> - (-470K) (8.3145 J/k mol) 3907.815 j/mol -- 3.9 kj/mol

Dipole attractions

The attraction of the δ- end of one polar molecule to the δ + end of another polar molecule. medium intermolecular force CO - CO aka polar interactions

van der Waals forces

a slight attraction that develops between the oppositely charged regions of nearby molecules aka london dispersion forces weakest intermolecular force only intermolecular force that exists between nonpolar molecules or atoms

molecular solid

a solid that consists of atoms or molecules held together by intermolecular forces

capillary action

adhesion + cohesion of liquid

bond polarity

arrow is pointing toward more electronegative atom 2.5 C <----+ H 2.1 (np)

hydrogen bonds

attractive forces in which a hydrogen covalently bonded to a very electronegative atom is also weakly bonded to an unshared electron pair of another electronegative atom H-- N, O, F strongest intermolecular force a type of dipole attraction h cov bond to n,o,f --- n,o,f lone pair hf hbond is stronger than h2o hbond

cis/trans isomer

cis has 2 groups on same side trans has 2 groups on different side c2h2cl2

coal, diamond, pencils

coal (amorphous carbon) diamond (crystalline form carbon) pencils (graphite, type of carbon) allotropes of carbon

lattice energy

decreases as size of ion increases

Electronegativity

desire to gain an electron nonmetals have high electronegativity (F-) metals have low (Na+) increases toward fluorine

electrolytes

dissociate into ions salts, strong acids

non electrolyte

do not dissociate into ions

Covalent Bond

electrons are shared H2O polar and non polar polar covalent bonds, electrons shared unevenly electronegativity difference less than .5 is considered nonpolar greater polar (there is no true agreement) greater than 1.8 is ionic rather than covalent

ionic bond

electrons are transferred metal and non metal NaCl

Enthalpy (ΔH)

energy it takes to phase change ΔHfus=334 j/g ΔHvap= 2260 j/g

ΔHsub

energy required for sublimation of 1 mol of a solid ΔHsub= ΔHfus+ΔHvap ΔHsub= 6.01 kj/mol + 40.7kj/mol = 46.7 kj/mol

surface tension

energy required to increase the surface area of a liquid/force required to increase the length of a liquid surface by a given amount caused by unbalanced net inward attraction of the molecules along the surface of a liquid mercury has a very high surface tension tendecy of a liquid to reduce its surface area

ΔHfus

energy required to melt 1 mol of a solid ΔHfus of water is 6.01 kj/mol

ΔHvap

energy required to vaporize 1 mol of liquid ΔHvap of water is 40.7kJ/mol in kj/mol or j/mol

i-factor

equal to number of particles in solution when 1 mole of substance dissolves in water NaCl= Na+, Cl- = i=2 HNO3 = H+, NO3- = i =2

intermolecular forces

forces of attraction between molecules liquid and solid state weak and temporary melting and boiling point increase as intermolecular forces increase ion-dipole>hydrogen bonding>dipole-dipole>dispersion

R

gas law constant 8.3145 J/mol K

deposition

gas---->solid

enthalpy of vaporization

j/mol energy required for the vaporization of 1 mol of a liquid

methanol has a normal boiling point of 64.60 c and a heat of vaporization of 32.5 kj/mol. What is the vp of methanol at 25.00C in mmHg

ln p2 - ln p1= Hvap/R (1/T1 - 1/T2) ln(760 mmHg) - ln p1= 35200 J/mol / 8.3145 j/k mol (1/298.15k - 1/337.75k) reciprocals before subtracting 6.633 -lnp1= 1.66 lnp1= 6.633- 1.66 = 4.97 P2=e^4.97=144mmHg

ln(x/y)=

lnx-lny=

ln x

log e x

electron pair repulsion

lone pair- lone pair >lone pair - bonding pair> bonding pair-bonding pair lone pair> triple bond > double bond> single bond

vapor pressure, hvap

low vp, high hvap hvap is energy required to vaporize 1mol of substance

molar mass

molar mass =g/mole moles=g/molar mass g=molar mass x moles

molality

moles of solute/kg of solvent 0.05 mole NaCl/1kg H2O

molarity

moles of solute/liters of solution

surface area effects

more contanct between molecules=stronger dispersion forces (isomers)

crystalline solids

nacl crystal lattice (repeating pattern)-unit cell defined by position of atoms in unit cell simple, body centered, face centered uniform = one specific melting point

amorphous solids

no orderly internal structure no specific melting point (gradually soften) (butter), rubber, glass imfs not evenly distributed

nonvolatile

not easily vaporized

bond type electronegativity

pure covalent <0.4 polar covalent 04.-1.8 ionic >1.8

ΔH fus, vap, sub in j/g

q=m(ΔH) m=mass q=j

ΔH fus, vap, sub in kj/mol

q=n(ΔH) q=kj n=moles

Clausius-Clapeyron equation

relationship between vapor pressure, temperature, and enthalpy of vaporization lnPvap= -ΔHvap/R (1/T) +ln b pvap= vapor prsssure in atmospheres R= 8.3145 J/mol K T temperature in kelvins b is a constant specific for each liquid ln p2/p1= ΔH vap/R (1/t1 - 1/t2) in order for hvap to be positive in this equation 2 has to be divided by 1 in p and minus 1 in t, if they are both first hvap is negative in this equation lnp2/p1=-hvap/r(1/t2-1/t1)

Viscosity

resistance of a liquid to flow depends on intermolecular forces, size of molecules, and temperature of the liquid intermolecular forces up viscocity up size/surface area up viscocity up temperature up viscosicty down

crystalline solid

solid made of repeating pattern

phase changes

solid to liquid = melting liquid to gas = vaporization gas to liquid= condensation liquid to solid = freezing solid to gas= sublimation gas to solid= deposition

amorphous solid

solid with random arranged does not have definite melting point

Sublimation

solid--->gas

covalent network solids

solids in which the units that make up the three-dimensional network are joined by covalent bonds

metallic solids

solids that are composed of metal atoms

ionic solids

solids whose composite units are ions; they generally have high melting points

capilary action

tendency of water to rise in a thin tube

ion-dipole

the charge of an ion is attracted to the partial charge on a polar molecule Na+ and H2O

solvent

the substance in which the solute dissolves

volatile

vaporize easily

formal charge

ve-(b+d)

boiling point

vp=1atm