Semester Exam 2 Worksheet Key Notes

Significant Figures

• Determining Significant Figures:

  • 3.965 x 10^6 = 4
  • 70.07 = 4
  • 3 x 10^-3 = 1
  • 0.000900 = 3
  • 0. 0680 = 3
  • 60 = 1

• Calculations with Significant Figures:

  • Addition/Subtraction: Result should have the same number of decimal places as the number with the least decimal places.

  • Multiplication/Division: Result should have the same number of significant figures as the number with the fewest significant figures.

  • 74.12 + 58.3 + 52.345 = 184.8

  • 680 / 12.89 = 53

  • 41.33 – 26.6517 – 6.41 = 8.27

  • 0. 98 x 0.0047 x 1001 = 4.6

Scientific Notation

• Converting to Scientific Notation:

  • 0.8503 = 8.503 x 10^-1
  • 4890000 = 4.89 x 10^6
  • 90004 = 9.0004 x 10^4
  • 0. 0003210 = 3.210 x 10^-4

• Converting from Scientific Notation (Expanded Form):

  • 3.114 x 10^-2 = 0.03114
  • 9. 983 x 10^8 = 998,300,000
  • 4.68 x 10^4 = 46,800
  • 6.43 x 10^-7 = 0.000000643

Temperature Conversions

• Temperature Conversion Formulas:

  • Fahrenheit to Celsius: $C = (F - 32) \times \frac{5}{9}$
  • Celsius to Fahrenheit: $F = C \times \frac{9}{5} + 32$
  • Celsius to Kelvin: $K = C + 273.15$
  • Kelvin to Celsius: $C = K - 273.15$

• Temperature Conversions Examples:

  • -10 °F = -23 °C
  • 0. 0 °C = 32 °F
  • 160 °F = 71.1 °C
  • 160 °C = 433 K
  • 239 K = -34 °C
  • 98. 6 °C = 371.6 K

Measurement Conversions

• Measurement Conversions Examples:
  • 2 J = 0.002 kJ
  • 22 L = 22000 mL
  • 354 cm = 3.54 m
  • 593 mm = 0.593 m
  • 83 kJ = 83000 J
  • 921 mg = 0.921 g

Density

• Density Calculation: Density = Mass / Volume

  • $Density = \frac{Mass}{Volume}$

• Density Problems and Solutions:

  • Calculate the density of a material with a mass of 22.47 g and a volume of 12.5 mL: Density = 1.80 g/mL
  • The density of silver is 10.49 g/mL. A sample of pure silver has a volume of 14.183 mL, what is the mass? Mass = 148.8 g
  • Acetone has a density of 0.79 g/mL at 20 °C. What is the volume of 120. g of acetone at 20 °C? Volume = 152 mL

States of Matter

• The three states of matter are solid, liquid, and gas.

Classification of Matter

• Classification of Matter:

  • Element: A pure substance consisting of only one type of atom.
  • Compound: A substance consisting of two or more elements chemically bonded in a fixed ratio.
  • Homogeneous Mixture: A mixture with uniform composition throughout.
  • Heterogeneous Mixture: A mixture with non-uniform composition.

• Classification Examples:

  • Table Salt - compound
  • Air – homogeneous mixture
  • Neon - element
  • Soil - heterogeneous mixture
  • Concrete - heterogeneous mixture

Periodic Table

• Periodic Families:

  • Halogens (e.g., I)
  • Alkali Metals (e.g., Li)
  • Alkaline Earth Metals (e.g., Ca)
  • Noble Gases (e.g., Kr)

• Element Types:

  • Metals (e.g., K, Tc)
  • Nonmetals (e.g., Br, Xe)
  • Metalloids (e.g., Si, As)

Ions

• Ions:
  • Cation: A positively charged ion (e.g., $Mo^{+3}$, $Na^{+1}$).
  • Anion: A negatively charged ion (e.g., $Br^{-1}$, $Se^{-2}$).
  • Neither: Neutral atoms (e.g., Xe, Ar).

Polyatomic Ions

• Polyatomic Ions:
  • $NH_4^{+1}$ – ammonium
  • $OH^{-1}$ – hydroxide
  • $NO_3^{-1}$ – nitrate
  • $CO_3^{-2}$ – carbonate
  • $SO_4^{-2}$ – sulfate
  • $PO_4^{-3}$ – phosphate

Naming Compounds

• Naming Compounds:
  • KCl – potassium chloride
  • $FeSO_4$ – iron (II) sulfate
  • $PdS_2$ – palladium (IV) sulfide
  • $SiF_4$ – silicon tetrafluoride
  • $Sr(OH)_2$ – strontium hydroxide
  • $BaCl_2$ – barium chloride
  • $N<em>2O</em>3$ – dinitrogen trioxide
  • $H<em>2CO</em>3$ – carbonic acid
  • $Tc<em>3N</em>4$ – technetium (IV) nitride
  • $Pb<em>3N</em>2$ – lead (II) nitride
  • $Zn(NO<em>3)</em>2$ – zinc (II) nitrate

Chemical Formulas

• Chemical Formulas:
  • beryllium sulfate – $BeSO_4$
  • chromium (VI) phosphate - $Cr(PO<em>4)</em>2$
  • hydrofluoric acid - HF
  • sodium carbonate hexahydrate - $Na<em>2CO</em>3 \times 6 H_2O$
  • dinitrogen trioxide – $N<em>2O</em>3$
  • lead (IV) hydroxide – $Pb(OH)_4$

Chemical Reactions

• Predicting Products and Balancing Equations:
  • zinc (II) and hydrochloric acid: $Zn + 2 HCl \rightarrow ZnCl<em>2 + H</em>2$
  • magnesium plus iron (II) nitride: $3 Mg + Fe<em>3N</em>2 \rightarrow Mg<em>3N</em>2 + 3 Fe$
  • $Pb(II) + 2 HNO<em>3 \rightarrow H</em>2 + Pb(NO<em>3)</em>2$

Solubility

• Predicting Solubility:
  • sodium bromide and potassium iodide: $NaBr (aq) + KI (aq) \rightarrow NaI (aq) + KBr (aq)$ NR (No Reaction)
  • sodium hydroxide and nickel (II) nitrate: $2 NaOH (aq) + Ni(NO<em>3)</em>2 (aq) \rightarrow 2 NaNO<em>3 (aq) + Ni(OH)</em>2 (s)$
  • magnesium chloride and ammonium carbonate: $MgCl<em>2 (aq) + (NH</em>4)<em>2CO</em>3 (aq) \rightarrow MgCO<em>3 (s) + 2 NH</em>4Cl (aq)$
  • strontium nitrate and potassium phosphate: $3 Sr(NO<em>3)</em>2 (aq) + 2 K<em>3PO</em>4 (aq) \rightarrow 6 KNO<em>3 (aq) + Sr</em>3(PO<em>4)</em>2 (s)$

Types of Reactions

• Identifying Reaction Types and Balancing:
  • $Na<em>3PO</em>4 + 3 KOH \rightarrow 3 NaOH + K<em>3PO</em>4$ double displacement
  • $CaCO<em>3 \rightarrow CaO + CO</em>2$ decomposition
  • $C<em>6H</em>{12} + 9 O<em>2 \rightarrow 6 CO</em>2 + 6 H_2O$ combustion
  • $MgCl<em>2 + Li</em>2CO<em>3 \rightarrow MgCO</em>3 + 2 LiCl$ double replacement
  • $2 AgNO<em>3 + Cu \rightarrow Cu(NO</em>3)_2 + 2 Ag$ single replacement
  • $C<em>3H</em>6O + 4 O<em>2 \rightarrow 3 CO</em>2 + 3 H_2O$ combustion
  • $2 C<em>5H</em>5 + Fe \rightarrow Fe(C<em>5H</em>5)_2$ synthesis
  • $SeCl<em>6 + O</em>2 \rightarrow SeO<em>2 + 3 Cl</em>2$ single replacement

Moles and Molecules

• Calculations:

  • How many molecules are there in 4.5 g lithium oxide? $4.5 g \, Li<em>2O \times \frac{1 mol}{29.8814 g} \times \frac{6.022 \times 10^{23} molecules}{mol} = 9.1 \times 10^{22} molecules \, Li</em>2O$
  • How many molecules are there in 23.5 g carbon dioxide? $23.5 g \, CO<em>2 \times \frac{1 mol}{44.0099 g} \times \frac{6.022 \times 10^{23} molecules}{mol} = 3.22 \times 10^{23} molecules \, CO</em>2$

Grams and Molecules

• Calculations:
  • How many grams are there in $3.4 \times 10^{23}$ molecules silver (I) sulfate? $3.4 \times 10^{23} molecules \, Ag<em>2SO</em>4 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{311.7936 g}{mol} = 180 g \, Ag<em>2SO</em>4$
  • How many grams are there in $7.50 \times 10^{23}$ molecules cobalt (IV) phosphate? $7.50 \times 10^{23} molecules \, Co<em>3(PO</em>4)<em>4 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{556.68504 g}{mol} = 693 g \, Co</em>3(PO<em>4)</em>4$

Percent Composition

• Calculations:
  • Calculate the percent composition of each element in copper (II) bromide ($CuBr_2$): 28.45% Cu, 71.55% Br
  • Calculate the percent composition of each element in sodium hydroxide (NaOH): 57.48% Na, 40.00% O, 2.52% H

Empirical Formula

• Calculations:

  • A compound has 10.00 g copper and 2.52 g oxygen. Calculate the empirical formula of the compound.

  $10.00 g \, Cu \times \frac{1 mol}{63.546 g} = 0.157366 mol$

  $2.52 g \, O \times \frac{1 mol}{15.999 g} = 0.157506 mol$

  • CuO

Empirical and Molecular Formulas

• Calculations:

  • A compound contains hydrogen and oxygen. The percent of hydrogen was found to be 5.9265 %. The molar mass of the compound is 34.01468 g/mol. Determine the empirical and molecular formulas of this compound.

  $5.9265 g \, H \times \frac{1 mol}{1.0079 g} = 5.879814 mol$

  $94.0735 g \, O \times \frac{1 mol}{15.999 g} = 5.879814 mol$

  • Empirical Formula: HO
  • Molecular Formula: H2O2

Molecular Formula

• Calculations:

  • A compound with the empirical formula $C<em>{11}H</em>{12}N<em>2O</em>2$ was found to have a molar mass of approximately 204.25 g/mol. What is the molecular formula of the compound?

  $\frac{204.25}{204.25} = 1$

  • $C<em>{11}H</em>{12}N<em>2O</em>2$

Limiting Reactant and Theoretical Yield

• Calculations:

  • copper (II) chloride + sodium nitrate yields copper (II) nitrate + sodium chloride
  • Balanced chemical equation: $CuCl<em>2 + 2 NaNO</em>3 \rightarrow Cu(NO<em>3)</em>2 + 2 NaCl$
  • If you have 54.0 g of each reactant, identify the limiting reactant and determine the theoretical yield of copper (II) nitrate. Calculate the percent yield of product if 57.3 g were produced.

  $54.0 g \, CuCl<em>2 \times \frac{1 mol \, CuCl</em>2}{134.452 g \, CuCl<em>2} \times \frac{1 mol \, Cu(NO</em>3)<em>2}{1 mol \, CuCl</em>2} = 0.401630322 mol \, Cu(NO<em>3)</em>2$

  $54.0 g \, NaNO<em>3 \times \frac{1 mol \, NaNO</em>3}{84.994 g \, NaNO<em>3} \times \frac{1 mol \, Cu(NO</em>3)<em>2}{2 mol \, NaNO</em>3} = 0.317669483 mol \, Cu(NO<em>3)</em>2$

  $0.317669483 mol \, Cu(NO<em>3)</em>2 \times \frac{187.554 g \, Cu(NO<em>3)</em>2}{mol \, Cu(NO<em>3)</em>2} = 59.6 g \, Cu(NO<em>3)</em>2$

  $\frac{57.3 g}{59.6 g} \times 100\% = 96.1\%$

  • Limiting Reactant: $NaNO_3$
  • Theoretical Yield: 59.6 g $Cu(NO<em>3)</em>2$
  • Percent Yield: 96.1%

Balancing Equations and Stoichiometry

• Calculations:

  • $Al<em>2(SO</em>3)<em>3 + 6 NaOH \rightarrow 3 Na</em>2SO<em>3 + 2 Al(OH)</em>3$
  • If you have 69.5 g of each reactant, identify the limiting reactant and determine the theoretical yield of the solid product. Calculate the percentage yield of solid product if 33.3 g were produced.

  $69.5 g \, Al<em>2(SO</em>3)<em>3 \times \frac{1 mol \, Al</em>2(SO<em>3)</em>3}{294.153 g \, Al<em>2(SO</em>3)<em>3} \times \frac{2 mol \, Al(OH)</em>3}{1 mol \, Al<em>2(SO</em>3)<em>3} = 0.4725432 mol \, Al(OH)</em>3$

  $69.5 g \, NaOH \times \frac{1 mol \, NaOH}{39.997 g \, NaOH} \times \frac{2 mol \, Al(OH)<em>3}{6 mol \, NaOH} = 0.579210107 mol \, Al(OH)</em>3$

  $0.4725432 mol \, Al(OH)<em>3 \times \frac{78.003 g \, Al(OH)</em>3}{mol \, Al(OH)<em>3} = 36.9 g \, Al(OH)</em>3$

  $\frac{33.3 g}{36.9 g} \times 100\% = 90.24\%$

  • Limiting Reactant: $Al<em>2(SO</em>3)_3$
  • Theoretical Yield: 36.9 g $Al(OH)_3$
  • Percent Yield: 90.24%

Matter Chart

• Matter Chart Elements:
  • element name
  • element symbol
  • atomic number
  • mass number
  • protons
  • electrons
  • neutrons