Download High School Knowledge Exam – Study Guide

High School Knowledge Exam
Precision vs. Accuracy
-Precision: The degree to which repeated measurements under unchanged conditions
show the same results (reproducibility/repeatability)
-Accuracy: The degree of closeness of measurements of a quantity to that quantity's
actual (true) value
Chemical vs. Physical Changes
-A physical change is reversible while a chemical change is not
-A physical change is a change in which no new substance is formed (no new molecules
formed)
-A chemical change results in the formation of one or more new substances (new
molecules)
Physical Change examples: Phase change, physically altering something (cutting it up,
etc.), dissolving salt in water (reversible through evaporation)
Chemical Change examples: Reactions between chemicals, burning (fire reacts with
something), color change (caused by reaction b/w chemicals)
Dalton’s Atomic Theory
1) All matter is made up of very small, discrete particles called atoms
2) All atoms of a given element are identical, and the atoms of a given element are
different from those of any other element (Incorrect – all atoms of an element are NOT
identical due to presence of isotopes)
3) Atoms cannot be subdivided, created or destroyed (Incorrect – atoms CAN be
subdivided due to subatomic particles such as neutrons, protons, electrons)
4) Atoms of different elements combine in simple whole-number ratios to form chemical
compounds
5) In chemical reactions, atoms are combined, separated or rearranged
Basic Atomic Structure (electrons, neutrons, protons)
-Protons (+ charge) and neutrons (no charge) in nucleus
-Electrons (- charge) outside nucleus
-Electrons have about 1/837th the mass of protons or neutrons
-Mass of atom mostly in nucleus
-Nucleus of atom much smaller and denser than space occupied by electrons
Mass Number and Atomic Number
Atomic Number: # of protons
Mass Number: protons + neutrons
Isotopes
-Atoms of the same element containing different numbers of neutrons and therefore
having different masses
-EX: C-14 is a common isotope of carbon used for radioactive dating
Organization of the Periodic Table (Metals, nonmetals, etc. and basic covalent vs.
ionic bonding)
-Left side = metallic (ionic bonds with nonmetals), base-forming properties
-Right side = non-metallic (covalent bonds with other nonmetals
-Atomic radii decrease left to right (due to increasing number of protons to attract
electrons yet no change in number of energy levels)
-Atomic radii increase top to bottom (due to increased number of energy levels, thus
increasing distance of outer electrons from nucleus, resulting in less attraction between
nucleus and electrons)
-Electronegativity decreases down group (shielding effect) and increases across period
(increasing non-metallic properties, and easier to complete valence shell that way)
-Ionization energies increase left to right (due to stronger attraction of nucleus to
electrons left to right, which is a result of the increase in protons left to right and no
change in energy levels)
-Base-forming properties decrease left to right
-Acid-forming properties increase left to right
-Metallic atoms form smaller ionic radii (due to losing electrons to form their ions –
cations)
-Nonmetal atoms form larger ionic radii (due to gaining electrons to form their ions –
anions)
Simple equation balancing
-Mass and charge must be conserved (same number atoms both sides and same net charge
both sides)
Example Reaction: H2 + O2
Balanced Reaction: 2H2 + O2
H2O
2H2O
Calculating the mass % of an element in a compound
(Mass of element / Mass of compound) * 100
Example Question: Calculate the mass % of oxygen in 1 mole of NaOH
Answer:
Na = 23g
O = 16g
H = 1g
NaOH = 23g + 16g + 1g = 40g
O/NaOH = 16g/40g = 4/10 = 40%
Calculating molecular mass from a chemical formula
Example Question: Calculate molecular mass for C12H22O11
Answer: 12(12.01 amu) + 22(1.008 amu) + 11(16.00 amu) = 342.30 amu
The Ideal Gas Law
PV = nRT, where the following are true:
P = Pressure in atm
V = Volume in liters
n = # of moles
R = 0.08201 (L * atm)/(mol * K)
T = Temperature in Kelvin
A gas behaves like an ideal gas at High Temperatures and Low Pressure (HTLP)
Standard Temperature and Pressure (STP)
-Following conditions exist:
Temperature = 0°C
Pressure = 1 atm
1 mole occupies 22.4L
Density
D = m/v
-FW: g/mol
-Density of gas at STP = m/v = FW/22.4
Reaction Stoichiometry
Density and Molar Mass
Example Question: Find the molar mass of a gas when the density is given as 1.25g/L
Answer:
Assume STP
1.25g/L * (22.4L/mol) = 28g/mol
Limiting Reagent Problems
Example Question: If you are given 65g of zinc and 65g of HCl, how many grams of
hydrogen gas can be produced? Which reactant will be left over? How many grams of
this reactant will not be consumed?
Answer:
Zn + 2HCl ZnCl2 + H2
65g Zn * (1 mol Zn / 65g) = 1 mol Zn
65g HCl * (1 mol HCl / 36.5g) = 1.78 mol HCl
Need 2 moles HCl for each mole Zn, but only 1.78 mol HCl while 1 mol Zn, so Zn is in
excess, and HCl is the limiting reagent (zinc will be left over)
1.78 mol HCl * (1 mol H2 / 2 mol HCl) = 0.89 mol H2
0.89 mol H2 * (2g H2 / 1 mol H2) = 1.78g H2 produced
1.78 mol HCl * (1 mol Zn / 2 mol HCl) = 0.89 mol Zn used
1 – 0.89 mol Zn = 0.11 unused mol Zn
0.11 mol Zn * (65g Zn / 1 mol Zn) = 7.15g unused Zn
Percent Yield Problems
% Yield = (Actual Yield / Theoretical Yield) * 100
Example Question: Quicklime, CaO, is to be prepared by roasting 2 * 103g of limestone,
CaCO3, according to the following reaction: CaCO3 CaO + CO2. However, the actual
yield of CaO is 1.05 * 103g. What is the percent of yield?
Answer:
2 * 103g CaCO3 * (1 mol CaCO3 / 100g CaCO3) = 20 mol CaCO3
20 mol CaCO3 * (1 mol CaO / 1 mol CaCO3) = 20 mol CaO
20 mol CaO * (56g CaO / 1 mol CaO) = 1,120g CaO
% Yield = [(1.05 * 103g CaO) / (1,120g CaO)] * 100 = 93.7%
Celsius to Kelvin temperature conversions
K = °C + 273.15
Significant Figures and Scientific Notation
Significant Figure Rules/Examples:
1) Zeros between nonzero digits are significant (1,005 has 4 significant figures, and 1.03
has 3)
2) Zeros at the beginning of a number are never significant (0.02 only has 1 significant
figure)
3) Zeros at the end of a number are significant if the number contains a decimal point
(0.0200 has 3 significant figures)
Significant Figure Rules/Examples for Calculations:
1) For addition and subtraction, the result has the same number of decimal places as the
measurement with the fewest decimal places.
Example:
20.42 (2 decimal places)
+
1.322 (3 decimal places)
+
83.1 (1 decimal place)
=
104.842
=
104.8 (Answer rounded to 1 decimal place)
2) For multiplication and division, the result contains the same number of significant
figures as the measurement with the fewest significant figures.
Example:
6.221 (4 significant figures)
*
5.2 (2 significant figures)
=
32.3492
=
32 (Answer rounded to 2 significant figures)
Scientific Notation:
-For large numbers, move the decimal point to the left until only 1 digit remains left, and
then indicate the number of moves of the decimal point as the exponent of 10 (3,630 3.630 3.630 * 103)
-For small numbers, move the decimal point to the right until only 1 digit remains left,
and then indicate the number of moves of the decimal point as the negative exponent of
10 (0.00123 0001.23 1.23 * 10-3)
Scientific Notation in Calculations:
-For multiplication (A * 10x)(B * 10y) = (AB)(10x + y); alter the result as needed to make
sure scientific notation rules are followed
-EX: (2.3 * 105)(5 * 10-12) = 11.5 * 10-7 = 1.15 * 10-6
-For division (A * 10x) / (B * 10y) = (A / B)(10x - y) ; alter the result as needed to make
sure scientific notation rules are followed
-EX: (1.5 * 10-3) / (5 * 10-2) = 0.3 * 10-1 = 3 * 10-2
Scientific Notation in Addition and Subtraction:
-For addition, make the exponents the same through algebraic manipulation, factor out
the exponent, and add normally
-EX: (4.2 * 104) + (7.9 * 103) = (4.2 * 104) + (0.79 * 104) =104 (4.2 + 0.79) = 4.99 * 104
-For subtraction, make the exponents the same through algebraic manipulation, factor out
the exponent, and add normally
-EX: (6.02 * 10-3) - (2.41 * 10-4) = (6.02 * 10-3) - (0.241 * 10-3) = 10-3 (6.02 - 0.241) =
5.779 * 10-3
Unit Conversions
Tera = 1012 = 1,000,000,000,000
Giga = 109 = 1,000,000,000
Mega = 106 = 1,000,000
Kilo = 103 = 1,000
Hecto = 102 = 100
Deka = 101 = 10
- = 100 = 1
Deci = 10-1 = 0.1
Centi = 10-2 = 0.01
Milli = 10-3 = 0.001
Micro = 10-6 = 0.000001
Nano = 10-9 = 0.000000001
Pico = 10-12 = 0.000000000001