Chemistry [QEE-R 2012]
-
Upload
vanny-rose-martin -
Category
Technology
-
view
717 -
download
5
Transcript of Chemistry [QEE-R 2012]
Matter: A Quick Review
Dimensional Analysis and Measurement
Modern Periodic Table and Electronic Structure of Atoms
Mole: Its Concepts and ApplicationsChemical Language: Reactions and
EquationsSolutions and Concentration Units
Gas Laws
Matter
Pure Substance
Element
monoatomic
polyatomic (molecule)
Compound ionic
molecule
Impure Substance (Mixture)
Homogeneous (i.e. solutions)
Heterogeneous (i.e. colloids and
suspensions)
approach used in problem solving
provides a systematic way of solving numerical problems in science and other disciplines as well as checking numerical solutions for possible error/s
units must be carried through all
calculations
correct use of
conversion factor/s to change one unit
to another
Given: Qty1 Find: Qty2
Solution:
factor/s
Conversion ×
quantity
Given = Unkown
( )1
2
12Qty
Qty × Qty = Qty
22Qty = Qty
Given: Qty1 Find: Qty2
Solution:
factor/s
Conversion ×
quantity
Given = Unkown
( )2
1
12Qty
Qty × Qty = Qty
( )
2
2
1
2Qty
Qty ≠ Qty
A. Base Quantities and Units
Quantity SI Metric English
length m
m, km, cm,
mm, m,
nm
mi, ft, yd, in
mass kgkg, Tg, mg,
cg, g, ngslug, oz
time ss, ms, hr,
min
s, hr, min,
day, wk,
mo, yr
A. Base Quantities and Units
Quantity SI Metric English
temperatur
eK C F
amount of
substancemol - -
electric
currentA - -
luminosity cd - -
B. Some Derived Quantities and Units
Quantity SI Metric English
area m2 m2, km2,
cm2, mm2
mi2, ft2, yd2,
in2
volume m3 km3, cm3,
mm3; mL
mi3, ft3, yd3,
in3; gal; qt
speed m/sm/min;
km/hr; cm/s
ft/s, mi/hr,
mi/min, in/s
density kg/m3
Mg/m3;
g/cm3;
g/mL
oz/qt;
oz/cm3
An aluminum foil is found to be 8.0 10-5
cm thick. What is it thickness in
micrometers?
m 10
μm 1 ×
cm 1
m 10 × cm 10 × 8.0 = thickness
6-
-2
5-
μm
μm 0.80or μm 10 × 8.0 = thickness-1
μm
A gas at 25 C exactly fills a container previously
to have a volume of 1.05 103 cm3. The container
plus the gas are weighed and found to have a mass
of 837.6 g. The container when emptied of all gas,
has a mass of 836.2 g. What is the density of the
gas at 25 C?
volume
mass = density
gas
gas
gas
( )
cm 10 × 1.05
g 836.2 - g 837.6 = density
33gas cm
g10 × 1.3 = density
3
3-
gas
most significant tool in ORGANIZING and SYSTEMATIC REMEM-BERING of chemical facts
list the important characteristics of all elements in increasing atomic no.
Isotopic Notation
13Al26.98
Atomic No. = # of p = # of e (neutral atom)
Mass No. = p + n
Isotopic
Notation
Atomic
No.
Mass
No.p n e
Electronic
structure
13Al2713 27 13 14 13
[10Ne] 3s2,
3p1
40Ar1818 40 18 22 18
[10Ne] 3s2,
3p6
13Al+3 13 27 13 14 10[2He] 2s2,
2p6
the mass, in grams, per mole of a substance (atom, ion, molecule)
how to find: just add the atomic weights of that substance (compound or molecules)
thus, in g/mol
MM= amu (monoatomic element)MM= FW (ions)MM= MW (molecules)
Molar Mass
atomic mass
unit/atomic weight
formula
mass/weight
Molecular
mass/weight
1 mol Fe weighs 55.85
g
MMFe = 55.85 g/mol
1 mol Fe2O3
weighs 159.70 g
MMFe2O3 =
159.70 g/mol
1 mol O2 weighs
32.00 g
MMO2 = 32.00
g/mol
1 mol O weighs 16.00 g
MMO = 16.00 g/mol
1 mol NaCl weighs
58.44 g
MMNaCl = 58.44
g/mol
1 mol H2O
weighs 18.02 g
MMwater =
18.02 g/mol
interconversion of mass of a substance, in grams, and the no. of particles of that substance (atoms, ions, or molecules)
the no. of moles of substance is central to stoichiometry
grams molesunit
particlesuse MM use
6.022 x 1023
Calculate the no. of moles of glucose,
C6H12O6 (MM = 180.0 g/mol), in 5.380 g of
this substance.
6126
6126
6126glucoseOHC g 180.0
OHC mol 1 × OHC g .3805 = mol
6126glucoseOHC mol .029890 = mol
How many copper atoms, Cu (MM = 63.5
g/mol), are there in a traditional copper penny
weighing 3 g. Assume the penny to be 100%
copper.
Cu mol 1
atoms Cu 10 × 6.022 ×
Cu g 63.5
Cu mol 1 × Cu g 3 = atoms Cu
23
atoms Cu 10 × 3 = atoms Cu22
Consider the combustion of butane, C4H10, the
fuel in disposable cigarette lighters.
(l)22(g)2(g)10(l)4OH + CO → O + HC
(l)22(g)2(g)10(l)4O10H + 8CO 13O + H2C →
2 mol
C4H10
13 mol
O2
8 mol
CO2
10 mol
H2O
2
2
104
2
1042CO mol 1
CO g 01.44 ×
HC mol 2
CO 8 × HC mol .341 = CO g
CO g 342 = CO g22
Consider the combustion of butane, C4H10, the fuel
in disposable cigarette lighters.
(l)22(g)2(g)10(l)4O10H + 8CO 13O + H2C →
Determine the mass in grams of CO2 (MM = 44.01
g/mol) formed when 1.34 mol of C4H10 (MM = 58.14
g/mol) reacts.
104
104
2
104
2
2
2104HC mol 1
HC g 8.145 ×
O mol 13
HC mol 2 ×
CO g 32.00
O mol 1 × O g .006 = HC g
HC g .681 = HC g104104
Consider the combustion of butane, C4H10, the fuel
in disposable cigarette lighters.
(l)22(g)2(g)10(l)4O10H + 8CO 13O + H2C →
Determine the mass in grams of C4H10 (MM = 58.14
g/mol) required to react with 6.00 g of O2 (MM = 32.00
g/mol).
SOLUTE >> the substance to be dissolved
SOLVENT >> the dissolving medium
Concentration units:
solution of V
mol = M Molarity, 1.
L in
solute
solvent of mass
mol = m Molality, 2.
kg in
solute
100 x solution of mass total
mass = percent mass 3.
solute
solute
mol/L 0.0444or M .04440 =Mglucose
A solution used for intravenous feeding contains 4.80 g
of glucose, C6H12O6 (MM = 180.16 g/mol) in 600.0 mL
of solution. What is the molar concentration of glucose?
solution of V
mol = M
L in
glucose
glucose
nsol'mL 1000
nsol'L 1× nsol'mL 0.006
OHC g 180.16
OHC mol 1×OHC g 4.80
= M6126
6126
6126
glucose
mol/kg 0.296or m .2960 =mglucose
A solution used for intravenous feeding contains 4.80 g
of glucose, C6H12O6 (MM = 180.16 g/mol) in 90.0 g of
water. What is the molal concentration of glucose?
water of m
mol = m
kg in
glucose
glucose
water g 1000
water kg 1×water g 0.90
OHC g 180.16
OHC mol 1×OHC g 4.80
= m6126
6126
6126
glucose
mol/kg 0.296or m .2960 =mglucose
A solution used for intravenous feeding contains 4.80 g
of glucose, C6H12O6 (MM = 180.16 g/mol) in 90.0 g of
water. What is the molal concentration of glucose?
water of m
mol = m
kg in
glucose
glucose
water g 1000
water kg 1×water g 0.90
OHC g 180.16
OHC mol 1×OHC g 4.80
= m6126
6126
6126
glucose
4%1 =mass)(by %NaCl
In a solution prepared by dissolving 24 g of NaCl in
152 g of water, determine the mass percent NaCl.
100 x solution of mass
mass = )/%(
NaCl
NaClg
g
100 x g) 152 + g (24
g 24 = )/%(
NaClg
g
A 1.13 molar solution of aqueous KOH (MM = 56.11
g/mol) has a density of 1.05 g/mL. calculate its
molality or molal concentration.
water of m
mol = m
kg in
KOH
KOH
] KOH mol 1
KOH g 56.11 × KOH mol 13.1[ - ]
nsol'mL 1
nsol' g 1.05 ×
nsol'L 1
nsol'mL 1000 × nsol'L 1[
KOH mol 13.1 = m
KOH
mol/kg 1.14or m .141 =mKOH