South Pasadena Chemistry Name Period Date / / S T A T I O N 1 H E A T For water, Cp,ice = 2.10 J/ g C, Cp,water = 4.18 J/ g C, Cp,steam = 2.08 J/g C, Hfus = 333 J/g, Hvap = 2260 J/g 50.0 g of water at 0 C is heated to 50 C. 50.0 g of water at 0 C is completely frozen to ice. This process is [ endothermic exothermic ] because energy is [ absorbed by released from ] the system. The temperature [ increases remains unchanged decreases] & molecular attractions are [ strengthened about the same weakened ]. The value of Q is [ negative positive ]. Calculate Q for this process. This process is [ endothermic exothermic ] because energy is [ absorbed by released from ] the system. The temperature [ increases remains unchanged decreases] & molecular attractions are [ strengthened about the same weakened ]. The value of Q is [ negative positive ]. Calculate Q for this process. Q = m T Cp = (50g)(50 C)(4.18 J/ g C) = 10450 J Q = m Hfus = (50g)(333 J/ g) = 16650 J Circle the portion that corresponds to this process: Circle the portion that corresponds to this process: Temp Temp Heat Heat S T A T I O N 2 N U C L E A R C H E M I S T R Y Consider the isotope K +. Find the following: Number of protons: 19 Number of neutrons: 21 Number of electrons: 18 Atomic number: 19 Mass number: Net Charge: +1 K can undergo beta decay, electron capture, and positron decay. Write the equations for each of these nuclear reactions. Beta decay: Electron capture: Positron decay: 19 K 0 1 + 20 Ca K + 0 19 1 e Ar 18 K 0 19 +1 + + Ar 18 After 3.75 10 9 years, only 12.5% of a sample of K remains. What is the half-life of K? Time 0 3.75 10 9 yrs 3h = 3.75 10 9 years Half-lives 0 h 1 h 2 h 3 h h = 1.25 10 9 years Amount (%) 100% 50% 25% 12.5%
S T A T I O N 3 E L E C T R O N C O N F I G U R A T I O N Write the long form ground state electron configurations for the following species. Ar 1s 2 2s 2 2p 6 3s 2 3p 6 S 2 1s 2 2s 2 2p 6 3s 2 3p 6 (isoelectronic with Ar) Fe 3+ 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 (Remove 4s electrons before 3d ebecause they re on a higher floor) Ga 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 1 Write the short form ground state electron configurations for the following species. Sn [Kr] 5s 2 4d 10 5p 2 Ba [Xe] 6s 2 Au [Xe] 6s 2 4f 14 5d 9 (remember the f orbitals) Cd 2+ [Kr] 4d 10 (remove 5s electrons before 4d because they re on a higher floor) S T A T I O N 4 P E R I O D I C T R E N D S Indicate the trends across the periodic table for each of the following (write Increases or Decreases): Atomic Size Ionization Energy Electronegativity Across period: Decr Incr Incr Down family: Incr Decr Decr Higher Ionization Energy: Si P because: P has more protons; greater nuclear charge Larger atomic radius: Cl Br because: Br has more layers of electrons Lower Ionization Energy: Ca Sr because: Sr has more layers of electrons Smaller atomic radius: O F because: F has more protons; greater nuclear charge Order from smallest to largest size: S 2, Cl, Cl, Ar Ar < Cl < Cl < S 2 Ar has more p +, so Ar < Cl. Cl has fewer e, so Cl < Cl. Cl has more p + so Cl < S 2.
S T A T I O N 5 L E W I S S T R U C T U R E S Draw the electron dot structures for the following species, and write the number of valence electrons: C S 2 P Mg 2+ # Val e : 4 # Val e : 8 # Val e : 5 # Val e : 0 Draw the Lewis dot structures for the following compounds. Indicate the type of bond between atoms, whether the molecule is polar or non-polar, and what kind of Inter-molecular Forces are found. Compound CO2 NH3 CH4 Lewis Structure Non-Polar Covalent or Non-polar or Polar Molecules Inter-Molecular Forces Non-Polar Covalent Bonds Non-Polar Molecule Polar Molecule Non-Polar Molecule London Dispersion Forces Hydrogen Bonds London Dispersion Forces S T A T I O N 6 B O N D I N G A N D S O L I D S Identify each of the following descriptions as Ionic Bond (I), Covalent Bond (C), or Metallic Bond (M). I Between a metal and non-metal C Between two non-metal atoms M Between Zn and Cu M Between two metal atoms I Electrons transferred between atoms C Between C and O I Between Na and F C Electrons shared between atoms I Attraction between oppositely charged ions Identify each of the following descriptions as an Ionic Solid (I), Molecular Solid (Mo), Network Covalent Solid (N), or Metallic Solid (M). I Conducts electricity when dissolved in water. M Brass (s) N Graphite M Positive ions in a sea of electrons Mo Have low melting and boiling points. N Contains only covalent bonds. I MgCl2 (s) M Malleable and ductile. M Zinc (s) Mo P4O10 (s) M Conducts electricity as a solid. Mo Has London dispersion forces.
S T A T I O N 7 G A S L A W S A sample of methane gas, CH4 (g), placed in a 2.0 L container at 30 C has a pressure of 300 torr. Standard pressure is 760 torr. When the pressure is raised to the standard pressure, the volume will [ increase stay the same decrease ] because pressure and volume are [ directly not inversely ] related. (Assume that temperature remains constant.) Standard temperature is 0 C. When the temperature is lowered to the standard temperature, the volume will [ increase stay the same decrease ] because volume and temperature are [ directly not inversely ] related. (Assume that pressure remains constant.) Calculate the volume when the sample is brought to STP. P1 = 300 torr P2 = 760 torr P1 V1 P2 V2 (300 torr)(2.0 L) V1 = 2.0 L V2 =? = = T1 T2 (303 K) T1 = 303 K T2 = 273 K V2 = 0.711 L A 32.0-g sample of an unknown noble gas occupies 24.0 L at 20 C and has a pressure of 0.800 atm. Calculate the moles of the gas in the sample. (R = 0.0821 atm L mol 1 K 1 ) P = 0.800 atm PV = nrt V = 24.0 L (0.800 atm)(24.0 L) = n (0.0821)(293 K) n =? n = 0.798 mol T = 293 K Find the molar mass of the unknown noble gas and its identity. m = 32.0 g n = 0.798 mol MM = m n = 32.0 g = g/mol Ar 0.798 mol (760 torr) V2 (273 K) S T A T I O N 8 E Q U I L I B R I U M Consider the equilibrium process, 2 O3 3 O2. A sample of O3 (g) was placed in a 1.0 L container at a particular temperature. The plot of the concentrations over time was made. concentration (M) 0.80 0.60 0. 0.20 0 0 1 2 3 4 5 6 time (s) [O 2] [O 3] At what time was equilibrium established? How do we know? Equilibrium established at 4 s because at that time, the concentrations of O2 and O3 (macroscopic property) are constant. Write the expression for the Keq and find its value. Keq = [O2]3 [O3] 2 = (0.80)3 (0.20) 2 = 12.8 At equilibrium, this reaction is [ reactant-favored product-favored ] because: Keq > 1 What is equal at equilibrium? Rates of forward and reverse reactions Write the expression for the Keq for the following reaction: S4 (s) + 12 F2 (g) 4 SF6 (g) Keq = [SF6]4 [F2] 12 Remember that solids and liquids are not included in Keq.
S T A T I O N 9 L E C H A T E L I E R S P R I N C I P L E Given the equilibrium reaction below, indicate the direction of the shift and the change in concentration of each substance when each of the following stresses is placed on the system. 2 NO (g) + O2 (g) 2 NO2 (g) Given the equilibrium reaction below, list three stresses that would produce more SF6 (g). heat + S4 (s) + 12 F2 (g) 4 SF6 (g) Add NO2 (g) Shifts left Remove NO (g) Shifts left Add O2 (g) Shifts right Add Ne (g) No shift Decrease volume Shifts right When the temperature of the system is lowered, more NO (g) is produced. Is the reaction endothermic or exothermic? Why? The reaction is endothermic, so heat is a reactant, and lowering the temperature would shift the reaction to the left making more NO (g). Increase the temperature Decrease the volume / increase the pressure Add more F2 (g) NOT add S4 (s)