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Test One

Test Two

Test Three

Test Four

Test Five

Test Six

Test Seven

Test Eight

Test Nine

TEST ONE

1. CaS
2. Na₂O
3. ZnCl₂
4. NH₄OH
5. AlI₃
6. Ni₃P₂
7. Fe(OH)₂
8. AuBr₃
9. SnS
10. PbO₂
11. potassium sulfide
12. strontium nitride
13. silver phosphide
14. sulfur trioxide
15. magnesium oxide
16. phosphorus (III) chloride
17. ferric selenide
18. chrome (III) oxide
19. cuprous arsenide
20. gold (III) nitride

TEST TWO
Match columns:

1. - D
2. - F
3. - A
4. - E
5. - C
6. - B

Types of Changes

1. - P
2. - C
3. - P
4. - C
5. - C
6. - C
7. - P
8. - C

Class of Matter

1. Compound
2. Mechanical mixture
3. Mechanical mixture
4. Solution
5. Element
6. Solution

Written answers

1. Best bet is to remove the sand first by filtering leaving gasoline, water and sugar. (Sugar of course will be dissolved in the water.
   Using a seperatory funnel separate the gasoline from the aqueous solution.
   The sugar can be removed from the water either by evaporation or applying heat to assist vapourization. We will assume that the water does not need to be kept.
2. Many possible answers ==> in biology I use a plastic model of a frog to show postion of the frog's internal organs and where they are positioned in the body cavity.
3. Mass of zinc = 3.726g, mass of bromine = 9.104g
   Total mass 12.83g
   % zinc 3.726 ÷ 12.83 = 29.04%
   % bromine 9.104 ÷ 12.83 = 70.96%

TEST THREE

Math Problems

1. Assume a 1000 atom sample
   Let x rep the number of ¹¹²Cd atoms , therefore (1000 - x) rep the number of ¹¹⁴Cd atoms
   Mass of ¹¹²Cd atoms is 112x and mass of ¹¹⁴Cd atoms is 114 (1000 - x)
   Total mass = 112 x + 114(1000 - x) = 1000 X 112.46
   Solve for x
   x = 770 atoms ¹¹²Cd and 230 atoms ¹¹⁴Cd
   Therefore the percentaages are 77 % Cd and 23 % Cd
   

2. a) From the Law of Conservation of mass, the mass of chlorine will be 31.3 g - 12.3 g = 19.0 g
   b) Setting up a ratio of chlorine gas to sodium chloride or 6.45 : 19 = ? : 31.3
   and solve for the ?
   ? = 10.6 g
   Therefore 10.6 grams of sodium chloride was produced.

Short Answers

1. The colors are: Red, Green, Indigo, and Purple
   Based on these transitions respectively n = 3 ---> n = 2, n = 4 ---> n = 2, n = 5 ---> n = 2,
   n = 6 ---> n = 2, Note that the transitions are increasing in energy.

2. Excite state :
   Ground state :
   I will not answer this type of question; please check text book for answer.

3. Lyman series ; high energy ultraviolet transitions for all levels to the n = 1 level or ground state
   Pashen series ; Lower in energy infrared transitions form all levels greater than n = 3 to the n = 3 level
   Check handed out diagrams for actual pictures.

TEST FOUR

Part A Multiple Choice

Part B Short Answers

1. Sodium will spin around on the top of the water, Hydrogen gas will be produced. A basic solution will form; phenolphthalien will turn red. May explode in a shower of orange yellow flames.
2. gold and cobalt
3. See text book
4. Extra electron when added to last shell will cause it to expand due to electrostatic forces of repulsion.
   1. ammonium --> triangular pyramid
   2. carbon tetra chloride --> tetrahedral
   3. boron hydride ---> planar triangular
5. Subtract electronegativities; if between 0.2 (approximately) and 1.7 it may be considered to be polar. The closer to 1.7 the greater its polarity.
6. Filled up outer shell of eight electrons; Octet rule based on atomic stability.
7. Unable to answer due to HTML restictions.
8. Same number of electrons in the outer shell. Example Ca²⁺ and Ar
9. Add cyclohexane to a buret and test the fluid flow with an electrostaic charge. If no movement is observed in the fluid stream then the fluid is not polar.

Part III Bond diagrams

1. Unable to insert or write answer but see text book page page 139
2. Unable to draw and insert

Nomenclature

1. sodium sulphide
2. (NH₄)₂Se
3. Fe₃As₂
4. Nickel (II) oxide

TEST Five

Nomeclature & Balancing

Balancing: Insert coefficients as required

1. _2_ HCl + __ Ca(OH)₂ -------> _2_ H₂O + __ CaCl₂
2. _3_ Zn + _2_ H₃PO₄ -----> __ Zn₃(PO₄)₂ + _3_ H₂
3. __ Fe(NO₃)₃ + _3_ NaOH -----> __ Fe(OH)₃ + _3_ NaNO₃
4. _2_ C₄H₁₀ + _13_ O₂ ----> _8_ CO₂ + _10_ H₂O
5. _3_ Na₂CO₃ + _2_ FeCl₃ + _3_H₂O -----> _2_ Fe(OH)₃ + _6_ NaCl + _3_ CO₂

Word Equation: Write balanced equations for these word equations

1. 3 Ca(OH)₂ + 2 H₃PO₄ ----> Ca₃(PO₄)₂ + 6 H₂O
2. K₂SO₄ + 2 NH₄NO₃ ----> 2 KNO₃ + (NH₄)₂SO₄
3. MgCO₃ + 2 HNO₃ ----> Mg(NO₃)₂ + CO₂ + H₂O
4. Pb(NO₃)₂ + Na₂S -----> PbS + 2 NaNO₃
5. 3 Cu + 8 HNO₃ ------> 3 Cu(NO₃)₂ + 4 H₂O + 2 NO

TEST Six

1. i) Sn = 65.3%, P = 11.2 %, O = 23.3 %
   ii) K = 26.5 %, Cr = 35.4 %, O = 38.1 %
2. # of moles = 8.04 x 10^-2mol
3. # of grams = 1.977 grams
4. be careful: # of molecules = 5.32 x 10²³, # of atoms = 4.79 x 10²⁴
5. Ca₃(PO₃)₂
6. Na₂S₂O₈

TEST Seven

1. The balanced equation is 2Na + 2H₂O ---> 2NaOH + H₂
   Since the mole ratio of sodium to sodium hydroxide is 2:2 or 1:1 the number of moles
   produced are 7.45x10^-2 moles.
2. Mole ratios from balanced equation are 2, 3, 1, 6
   i) grams of aluminum sulfate produced is 1.53 g ii) grams of sodium sulfate needed is 174 g
3. The balanced equation is 2C₄H₁₀ + 13O₂ ----> 8CO₂ + 10H₂O
   grams of water produced are 100.86 g or 1.01x10² g
   Moles of oxygen needed are 7.28 mol
4. Balanced chemical equation is
   3Pb(NO₃)₂ + 2FeCl₃ ----> 3PbCl₂ +2Fe(NO₃)₃
   Since this is a limiting reagent question solve for PbCl₂ produced using both reactants. Remember; this was my way of doing the problem
   On doing so the amount of plumbous chloride precipitated is 13.1 grams
5. The balanced equation is
   2V + 5Zn(ClO₂)₂ ----> 2V(ClO₂)₅ + 5Zn The theoretical amount (using the balanced equation) of zinc produce from 18.94 grams of
   zinc chlorite is 6.181 g. Therefor the actual yield is 67.8% of this or 4.19 grams.
6. THe required equation is CuSO₄5H₂O -----> CuSO₄ + 5H₂O The theoretical yield fron this raction is 2.327 grams of water
   giving a percentage yield of 85.1%
7. complete the reaction
   • ----> Zn(NO₃)₂ + Cu --- single displacement
   • ----> (NH₄)₂SO₄ = 2H₂O --- neutralization
   • ----> Li₂CO₃ + CO₂ + H₂O --- decomposition
   • ----> MfCl₂ + H₂ --- single displacement
   • ----> 2Al₂O₃ --- synthesis
8. reaction type with balancing
   • complete combustion --- 2, 19, 12, 14
   • double decomposition --- 1, 2, 1, 2
   • double decomposition or neutralization --- 3, 2, 1, 6
   • decomposition --- 2, 2, 1

TEST EIGHT

1. Definitions: lokk them up in the text book; learn as exactly and precisely as possible.
2. KMT you must talk about molecules. In I the gas molecules inside the container are increased in numbers and forced together increaseing the number of molecular collisions causing an increase in the average kinetic energy of the system (more molecular collisions per second). In II the molecules lose energy, slow down, less collisions per second on the inner surface so that the outside pressure becomes greater than inside, hence balloon collapses. In III when heated moleculse can spread apart more hence less molecules per unit volume therefore less dense. In IV boiling occurs when the vapour pressure of a fluid is equivalent to the outside air pressure. This means molecules will leave the fluid freely, with nothing to keep them confined (assuming an open container). Remember the bell jar experiment.
3. P₂ = 88.4 kPa
4. 1. volume = 4.86 L
   2. Use PV = nRT. You can use either the first or second conditions
      You get the same answer n = 0.2035 moles changing to grams the mass is 8.954grams.
      Be aware that neither mass nor number of moles can be altered by changing temperature, volume or pressure
   3. Density is mass over volume therefore D = 1.84 g/L at 56.8^oC and 114.8 kPa
5. Use PV = nRT and solve for n. n = 0.3026 moles
   Now change this to grams by multiplying by the formula weight. Mass = 13.92 g
   Calculate density given mass and volume D = 2.05 g/L
6. Total pressure is 500 kPa, take percentage of this amount for each individual gas
   oxygen = 228 kPa argon = 142.5 kPa and carbon dioxide = 129.5 kPa
7. Convert 350 grams into moles (use formula weight) n = 6.034 moles
   Use PV = nRT and solve for P. P = 39.9 kPa

TEST NINE

1. Number of moles is 2.78 x 10^-2 mol.
2. Find n first; n = 1.47 x 10⁴ mol.
   Use PV = nRT where V = 7.21 kL
3. Balance equation first; coefficients are 1, 2, 1, 1
   Since reaction is at STP use 22.4 L/mol in ratio matrix
   Therefore Volume of H_2(g) = 25.7 L at STP.
4. The equation is Na₂CO_3(aq) + 2HCl_(aq) ----> 2NaCl_(aq) + H₂O_(l) + CO_2(g)
   Determine how many moles are produced from 45.6 grams of Na₂CO_3(aq). n = 0.430 mol
   Use this value of n in the equation PV = nRT and solve for V. V = 10.28L or 10.3L
5. Calculate how many moles are in 10.0L of chlorine gas at the given conditions
   using PV =nRT. n = 0.4077 mol
   Using balanced equation and ratio matrix the mass of potassium permanganate required is 25.8 grams

TEST TEN

1. Saturate at a high temperature then carefully cool down.
2. Formula weight is 342 g/mol, therefore the number of moles is 3.74 x 10^-2
   Remember to divide volume in L into the number of moles.
   Concentation is 8.31 x 10^-2 mol/L or M
3. a) potassium chlorate
   b) saturated c) supersturated, mass of salt that will precipitate is 500 g
4. Use an electroconductivity apparatus (light bulb with electrodes) and check for brightness.
5. Note: this is not a complete answer, just a quick breakdown
   Lead out with chloride
   Calcium out with sulfate
   This leaves magnesium behind
6. hydrogen ion concentration 3.98 x 10^-12 mol/L
7. Using number of moles = conc x vol, n = 0.0903 moles
   Formula weight of Fe(NO₃)₂ = 179.8 g/mol, mass is 16.24 grams
8. pH = 3.33
9. Depends on indicator selected. See note for answer.
10. Make sure you balance the equation;set up ratio matrix to get 0.846 mol NaNO₃
    Yielding a concentration of 1.21 mol/L
11. # of moles = 0.15 mol HCl
    Using ratio matrix, grams of MgCl₂ = 28.6 g
12. Tough question folks; probably too tough but
    Change pH to moles of H⁺¹ [H⁺¹] = 6.026 x 10^-4
    moles Change pH to pOH by subtracting from 14 hence pOH = 4.78 with [OH^-1] concentration of 1.66 x 10^-5
    I can avoid this part of the problem by giving you not pH's but actual concentration values. Be aware of this.
    This now becomes a limiting reagent question; you need the balanced equation
    2HCl + Ba(OH)₂ ----> BaCl₂ + 2H₂O Set up mole ratio matrix, select the smallest and you get 1.66 x 10^-5 mol BaCl₂
    Watch you volumes; sum everything total volume is 775 mL.
    This give a final concentration of 2.14 x 10^-5 mol/L
13. Volume need of 18.0 mol/L sulphuric acid is 6.0 mL
14. To quantitatively determine the concentration or molar amount of an unknown. See text book for complete answer.