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Chemistry Portfolio

Louis Raiger

Dr. Snyder 2011-2012


Quarter 1


Answer 1. Red 2. Scandium 3. Iodine 4. 14 5. Alkaline earth metals. They are called this because their oxides give simple alkaline solutions and they remain solid in fire. 6. 17 7. 0 and 0 8. 18 + hydrogen 9. Because it is in the alkaline earth metal section of the periodic table of elements. 10. 1 and 2. 11. To indicate that the lanthanide and actinide series belong to the elements lanthanum and Actium. 12. That they are also excellent conductors of electricity.


Additional Problems 1. A. 12.75 Mm x 1000 km/1 Mm = 12750 km B. 6.5 m2 x 10cm/1 m x 1 cm/1m = 6500000 cm2 C. 0.035 m3 100 cm/1 m x 100 cm/1 m x 100 cm/1 m = 35000000 cm3 D. 0.49 cm2 x 10 mm/1cm x 10 mm/1 cm = 49 mm2 E. 300000 km x 1 Mm/1000 km = 300 Mm

2. A. 0.62 km x 1000 m/1 km = 620 m B. 3857 g x 1000 mg/1 g = 3857000 mg C. 0.0036 mL x 1000 µL/1 mL = 3.6 µL D. 0.342 t x 1000kg/1 t = 342 kg E. 68.71 kL x 1000 L/1 kL = 68710 L

3. A. 856 mg x 1 kg/1000000 mg = .000856 kg B. 1210000 µg x 1 kg/1000000000 µg = .00121 kg C. 6598 µL x 1 cm3/1000 µL = 6.598 cm3 D. 80600 nm x 1 mm/1000000 nm = 0.0806 mm E. 10.74 cm3 x 10 L/1 cm3 = 107.4 L

4. A. 7.93 L x 1 cm3/10 L = 7930 cm3 B. 0.0059 km x 100000 cm/1 km = 590 cm C. 4.19 L x 1 dm3/100 L = 0.0419 dm3 D. 7.48 m2 x 100 cm/1 m x 100 cm/1 m = 74800 cm2 E. 0.197 m3 x 1000 L/1 m3 = 197 L

5. 20000 km x 1 mL/20 km = 1000 mL


Additional Problems 1. A. 3 B. 4 C. 3 D. 2 E. 2 F. 1 G. 3 H. 4 I. 5

2. A. 5490000 m B. 0.0134793 mL C. 31950 cm2 D. 192.67 m2 E. 790 cm F. 389278000 J G. 225834.8 cm3 3. A. 49000 cm2 B 3.1 kg/L C. 12.3 L/S D. 170000 cm2 E. 41 m3 F. 3.129 g/cm3 4. A. 90.2 J B. 0.0006 m C. 900 g D. 31.1 kPa E. 278 dl F. 1790 kg


5. 87.59 cm x 3.51 cm = 307.4409 cm2 A. 307.4409 cm2 B. 3074.409 mm2 C. 3.074409 m2


I. II.

Title: Constructing A Model Purpose: To understand how scientists make inferences about atoms without touching or seeing them. III. Materials: 1. Closed container; 2. Various objects; 3. Balances. IV. Procedure: Write out steps (methods). V. Data: Part A: Closed container. Group No. No. of objects Mass of objects (g) Kind of objects 1 12 80 g metal 2 15 80 g Plastic 3 35 22 g Plastic/metal 4 22 292 g Plastic/metal 5 5 98 g Plastic/rubber 6 20 50 g Metal Part B: Open container without looking. Group No. No. of objects Mass of objects (g) Kind of objects 1 20 70 g Metal 2 9 50 g Plastic 3 30 30 g Plastic/metal 4 8 150 g Plastic/metal 5 6 70 g Rubber/metal 6 7 40 g Plastic/metal Part C: Open container while looking. Group No. No. of objects Mass of objects (g) Kind of objects 1 7 24 g Metal/cloth 2 9 47.5 g Plastic/wood 3 83 32 g Plastic/metal 4 10 234.5 g Plastic/metal/stone 5 6 43 g Rubber/metal/plastic 6 4 50 g plastic


Additional Problems 1. B. Analyze Given: 8200 g of iron Unknown: 8200 g of iron in moles Plan g iron ďƒ  mol iron

g iron x 1 mol iron/ g iron = mol iron

Compute 8200 g iron x 1 mol iron/55.847 g iron = 150 mol iron Evaluate a) Units are correct. b) Sig figs are correct. c) This is correct because a mol of iron is significantly smaller than a g of iron. 2. D Analyze Given: 7 mol of titanium Unknown: 7 mol of titanium in (g) Plan Mol titanium ďƒ  g titanium

Mol Ti x g Ti/ mol Ti = g Ti

Compute 7 mol Ti x 47.88 g Ti/1 mol Ti = 3 x 102 g Ti Evaluate a) Units are correct. b) Sig figs are correct. c) This is correct because the answer is approximately 50 x 7.

4. B Analyze Given: 1.06 x 1023 atoms W Unknown: 1.06 x 1023 atoms W in moles Plan


# atoms W  mol W

Atoms W x 1 mol W/ Avogadro’s # = mol W

Compute 1.06 x 1023 atoms W x 1 mol W/Avogadro’s # = 1.76 x 10-1 mol W Evaluate a) Units are correct. b) Sig figs are correct. c) This is correct because the exponents cancel out. 5. F Analyze Given: 37 µg of U Unknown: 37 µg of U in # of atoms. Plan µg U  # of atoms U

µg iron x 1 mol U/ µg U x atoms/1 mol U = atoms U

Compute 37 µg iron x 1 mol U/ 0.000280289 µg U x 6.022 x 1023 atoms/1 mol U = 9.4 x1016atoms U Evaluate a) Units are correct. b) Sig figs are correct. c) This is correct because the other decimal would be about eight so they cancel out.


I. II. III. IV. V.

VI.

Title: The Mole Concept Purpose: To identify the identity of each sample based on observation and numerical data. Materials: 1. Mole sample set; 2. Balance. Procedure: Write out steps in list or paragraph form. Data: Part A: Observation Sample Observations A Somewhat heavy, shiny, silver B Light, more duller color, gray C Small, dull, heavy, tarnished, dark gray D Small, heavy, brownish Part B: Numerical Data Sample Mass (g) # of moles # of atoms ID of element 23 A 65.39 1 6.022 x 10 Zinc B 26.98 1 6.022 x 1023 Aluminum 23 C 55.84 1 6.022 x 10 Iron D 60.64 1 6.022 x 1023 Copper Analysis and Conclusion 1. What lead you to the determination of each element? We found the approximate mass then used then used the periodic table to match the masses and identify the correct element. 2. If each element was only one mol, why were the masses different? Because although they have the same number of atoms the atoms of different elements weigh different amounts. 3. Determine the actual mass of a single atom of each sample (mass of sample in g / Avogadro’s constant). Zinc – 1.096 x 10-22 Aluminum – 4.480 x 10-23 Iron – 9.276 x 10-23 Copper – 1.006 x 10-22.



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