1. The colors of light in the Balmer series are produced when electrons move from a high energy state to a lower energy state.

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Laurence Wilkinson
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1 Regents Review Workbook: Answers to Short Answer Questions Chapter 1: Introduction 1. The wax melts (solid liquid). or The candle cooled afterwards. 2. A flame is present. The candle is burning g. has 3 sig. figs. 4. % water = mass of water x 100 = 0.76 g. x 100 = mass of crystals 2.13g. 5. The mass will continue to decrease as long as water is lost. Heating until the mass is constant ensures that as much water as possible will be evaporated. Chapter 2: Atoms, Molecules, and Ions 1. The colors of light in the Balmer series are produced when electrons move from a high energy state to a lower energy state. 2. Anything with 13 electrons that is not The first level cannot have more than 2. The second level cannot have more than 8. So examples would include: or or , (Silicon-29 has an atomic mass of 29. Subtract the number of protons (atomic number) of 14.) 4. Show numerical setup meaning you do not need to find an answer amu x.9222 = amu x.0469 = amu x.0309 = + Or (27.98 amu x.9222) + (28.98 amu x.0469) + (29.97 amu x.0309) = Or (27.98 amu x 92.22) + (28.98 amu x 4.69) + (29.97 amu x 3.09) = Measured = 3.29% % Error = m - a x 100 = 3.29% % x 100 = 6.5% Accepted = 3.09% a 3.09% 6. Four solids at standard temperature or STP. They are Elements E, G, L and Q. (Melting Point and Boiling Point above 0 o C.) 7. One. Forms G2O, so it must have an oxidation number of +1 or have 1 valence electron that could be transferred.

2 8. Something below 64 o C (that is the melting point of Element L) 9. Element D (will not form an oxide) Chapter 3: Chemical Formulas & Equations 1. Combination or synthesis 2. 2 Fe = g 3 O = g g (molar mass) 3. Iron (III) oxide 4. AQ2 5. Covalent Bond (between 2 nonmetals) 6. The total mass of the reactants should equal the total mass of the products. Chapter 4: Mole & Stoichiometry 1. 2 Na = 46.0 g. 1 C = 12.0 g. % Carbon = part x 100 = 1 C x 100 = 12.0 g. x O = g. whole Na2CO g g. 2. Decomposition moles NaHCO3 x 1 mole CO2 = 3.5 mole CO2 2 mole NaHCO3 4. Mass of 1 mole of Ne = 20.2 grams Density = M = 20.2 grams = g Volume of 1 mole Ne = 22.4 Liters V 22.4 Liters L Chapter 5: Bonding 1. Total number electrons, positions of all electrons, number of protons, number of neutrons (number of protons and neutrons) 3. Usually only the valence electrons are involved in bonding.

4 4. The difference in electronegativity in a P-Cl bond is 1.0 (Cl is 3.2 and P is 2.2), the difference in a P-S bond is 0.4 (S is 2.6 and P is 2.2). The greater difference makes the P-Cl bond more polar. 5. About 124 o C (between 122 o C and 126 o C) 6. As the molar mass increases the intermolecular forces increase (hence the higher boiling point it is harder to separate the molecules) Chapter 6: Thermochemistry & Thermodynamics 1. Mass of water = g Q = mcδt Specific Heat of water = 4.18 J/g- o C = (100.0g)(4.18 J/g- o C)(35.0 o C) Change intemp = 35.0 o C = J [3sig figs] 2. Measured = 25.9 kj/g % error = meas. acc. x 100 Accepted = 30.2 kj/g acc. = 25.9 kj/g kj/g x kj/g = 14.2 % kj is a product 4. 2H 2 + O 2 2H 2O 5. The entropy (disorder) decreases because there are 2 different reactants that become1 product Or The reactants are gases that turn into a liquid o C 7. (solid) 8. 3 min. (from 4 to 7 mins) kj of heat is required to melt the sample (15 kj of heat per minute for 2 minutes)

5 Chapter 7: Gases, Liquids, & Solids 1. Freezing or Solidification (sample started as a liquid and then temp remained constant) 2. Make temp. scale go from 65.0 o C to 79.0 o C (scale goes up by 2 o C, you will not have 0 o C) 3. Gram-formula mass is just molar mass (so add up the atoms) 18 Carbon = g. 36 Hydrogen = 36.0 g. 2 Oxygen = g g o C (from 801 o C to 1465 o C) 5. AB or CD (avg. kinetic energy = temp.) 6. CD (ionics conduct electricity in the liquid phase) ml x 1L = ml ml 8. P1 = 1.0 atm. V2 = P1V1 = (1.0 atm)(125.0 ml) P2 = 1.5 atm. P2 (1.5 atm) V1 = ml V2 =? (1.0 atm)(125.0 ml) = (1.5 atm)(v2) 9. Number of particles are the same (if volume, temperature and pressure are the same for a gas) Chapter 8: Solutions g NaNO3 x 1 mole NaNO3 = mole NaNO g. NaNO3 2. Approx more grams of NaNO3 (saturated solution can absorb about 87.0 grams- approx.) 3. Evaporate the water 4. Pressure has no effect on the solubility because it is not a gas. 5. Approx grams will undissolve or settle to the bottom (at 15 o C approx. 30 grams is soluble)

6. Unsaturated (under those conditions the solution should hold 10.5 mg.) 7. Oxygen is a nonpolar molecule and water is a polar molecule 8. ppm O2 = 0.0070 gram x 1,000,000 = 7.0 ppm O2 1000.

6 6. Unsaturated (under those conditions the solution should hold 10.5 mg.) 7. Oxygen is a nonpolar molecule and water is a polar molecule 8. ppm O2 = gram x 1,000,000 = 7.0 ppm O grams Chapter 9: Kinetics & Equilibrium 1. The system was initially at equilibrium because all of the concentrations were constant or non-changing. 2. The concentration of NH3 increased because the addition of H2 stressed the system and caused a shift toward the products. 3. The H2 molecules begin to collide with N2 molecules to produce NH3 molecules, therefore the number of free H2 molecules decreases. 4. The rate of the forward and reverse are equal if it is at equilibrium. 5. The concentration of H + increases (because HC2H3O2 dissociates into H + ) 6. 3 S(s) + 2 KClO3(s) 3 SO2 (g) + 2 KCl (s) + energy 7. Heat (friction) when two balls are struck together. 8. Cl When temperature increases, atoms move faster and collide more often and with more energy. Chapter 10: Acids & Bases 1. KOH + HCl HOH + KCl 2. MB = MA x VA = 1.22 M x ml ml VB 3. NaHCO3 4. Ethanoic Acid = 5 D = M so M = D x V = 1.8 g x 0.20 L = 0.36 g V L

7 6. Between x 10-7 mole per liter 8. As temp. increases, solubility of O2 (g) decreases 9. ppm O2 = 2.7 x 10-2 g x 1,000,000 = 7.1 ppm O g The DO concentration of 7.1 ppm is healthy because it is between 6 ppm and 8 ppm. 10. Methanol (CH3OH) is neither an acid nor a base, so it should be relatively neutral. For bromthymol blue to be green it must be within the ph range for a color change, between a ph of Since this is the most neutral of the 3 bottles, bottle B must be methanol. Methanol, an organic compound, does not dissociate in water, therefore it is a non-conductor (non-electrolyte) as demonstrated by Bottle B. 11. Reacts with the metal Mg, because Mg is higher on the Reactivity Chart (Table F) than H. Bromthymol blue displays the acid color of yellow. 12. Methyl orange will remain yellow with a ph that extends from 4.4 (middle of the acid range) to 14 (an extreme base) so all solutions base (NaOH), neutral (methanol =CH3OH) and acid (HCl) could be in the yellow range for methyl orange. 13. Hydrochloric Acid 14. MA = MB x VB = M x ml =.833 M ml sig figs. VA 16. Wash thoroughly any areas of skin that come in contact with the acid or base. Don t taste chemicals. Read all Safety Data Information. Etc. Chapter 11: Redox & Electrochemistry 1. 1 Cd (s) + 1 NiO2 (s) + 2 H2O (l) 1 Cd(OH)2 (s) + 1 Ni(OH)2 (s) 2. Anode = Cd (from reading) Cd 0 Cd Cd loses electrons to Ni. Therefore Cd is more likely to be oxidized then Ni.

8 4. 3 Cu Al 3 Cu + 2 Al Solid aluminum atoms lose electrons and become aqueous ions that are dissolved in the solution. Therefore the solid aluminum electrode loses mass and becomes smaller as aluminum atoms dissolve into the solution. 6. Allows + and ions to move from one cell to another (electrolytes) and complete the circuit. Answers to Short Answer Questions Chapter 12: Nuclear Chemistry Am 4 2 He Np 2. Fr-220 has a half-life of 27.5 sec. which is way too short for a smoke alarm. Someone would have to constantly change the sample so that it would not run out. 3. The smoke interrupts the flow of ions, which causes the alarm to sound. 4. Carbon-14 goes through beta decay. 5. 1/8 th the original sample is equal to 3 half-lives (1 ½ ¼ 1/8 ) 3 half-lives x 5730 years = 17,190 years 1 half-life 6. C-13 has 1 more neutron than C-12. Chapter 13: Organic Chemistry 1. Organic Halides or Halocarbons 2. All carbon-carbon bonds are single. There are no double or triple bonds. 3. The boiling point of 2-methylpropane is lower than that for 2-iodo-2-methylpropane because 2- methylpropane is less polar and, therefore, has weaker (less) intermolecular forces (attractions) than for 2-iodo-2-methylpropane. [or visa-versa] 4. Alcohol 5. 1,2-ethanediol is soluble in water because it is a polar molecule.

6. 6690 g. 1,2-ethanediol x 1 mole 1,2-ethanediol = 107.9 moles 1,2-ethanediol 62.0 g. 1,2-ethanediol 7.

9 g. 1,2-ethanediol x 1 mole 1,2-ethanediol = moles 1,2-ethanediol 62.0 g. 1,2-ethanediol Ethanoic acid moles propane x kj = 5548 kj 1 mole propane 12. H H-C-S-H H

Atomic Structure

Atomic Structure Answer Key Matter 1. 3 2. 2 3. 3 4. 1 5. 4 6. 3 7. 1 8. 3 9. 2 10. 3 11. 1 12. 4 13. 3 14. 4 15. density of neon gas = 0.827 grams/liter 16. one physical property is the melting of the wax 17. one indication