Tommy Smith Dr. Snyder 10th Grade 2012-2013

The above image was created by Tommy Smith

1. Periodic Table of Elements Handout 2. Conversion Sheet 3. Significant Figures Handout 4. Constructing A Model 5. The Mole Concept Handout 6. The Mole Concept 7. Atomic Models

A. 2.5 cm/1 x 10mm/1 cm B. 3.4dm/1 x 1dkm/1dm C. 6.834hm/1 x 100 000mm/1hm D. 109.87dkm/1 x 1km/100dkm E. 1900.07cm/1 x 1hm/10 000cm F. 1km/1 x 1 000m/1km G. 2km/1 x 1 000 000mm/1km H. 3m/1 x 1km/1 000m I. 1dm/1 x 1hm/1 000dm J. 2cm/1 x 1hm/10 000cm K. 3hm/1 x 1 000dm/1hm L. 1hm/1 x 100m/1hm M. 4m/1 x 10dm/1m N. 3m/1 x 100cm/1m O. 0.5km/1 x 1 000m/1km P. 14m/1 x 1km/1 000m Q. 23.5mm/1 x 1m/1 000mm R. 14.2mm/1 x 1cm/10mm S. 43.9km/1 x 10 000dm/1km T. 5678.901mm/1 x 1m/1 000mm U. 0.234km/1 x 1 000 000mm/1km V. 34.25dkm/1 x 1km/100dkm W. 99.9999hm/1 x 10 000cm/1hm X. 8787.87mm/1 x 1dkm/10 000mm Y. 5555.555cm/1 x 1hm/10 000cm Z. 0.000302km/1 x 10 000mm/1km

1. a. .0120 m has three significant figures. b. 100.5 mL has four signific80ant figures c. 101 g has three significant figures d. 350 cm2 has two significant figures e. 0.97 km has two significant figures f. 1000 kg has one significant figures g. 180. mm has three significant figures h. 0.4936 L has four significant figures i .020700 s has five significant figures 2. a. 5 487 129 m rounded to 3 significant figures is 5490000m b. 0.013 479 265 mL rounded to 6 significant figures is 0.0134793 mL c. 31 947.972 cm2 rounded to 4 significant figures is 31950 cm2 d. 192.673 9 m2 rounded to 5 significant figures is 192.67 m2 e. 786.916 4 cm rounded to 2 significant figures is 790 cm f. 389 277 600 J rounded to 6 significant figures is 389 278 000 J g. 225 834.762 cm3 rounded to 7 significant figures is 225834.8 cm3 8. a. 13.75 mm x 10.1 mm x .91 mm = 126.376 25 mm3 = 130 mm3 b. 89.4 cm2 x 4.8 cm = 429.12 cm3 = 430 cm3 c. 14.9 m3/3.0 m2 = 4.96 m = 5.0 m d. 6.975 m x 30 m x 21.5 m = 4 498.875 m3 = 4 000 m3 9. 752 m x 319 m x 110 m = 26 387 680 m3 = 26 000 000 m3 10. a. 7.38 g + 1.21 g + 4.792 3 g = 13.384 3 g = 13.4 g b. 51.3 mg + 83 mg - 34.2 mg = 100.1 mg = 1.0 x 102 mg c. .007 L â€“ 0.003 7 L + 0.012 L = 0.015 3 L =0.02 L d. 253.05 cm2 + 33.9 cm2 + 28 cm2 = 314.95 cm2 = 320 cm2 e. 14.77 kg + .086 kg - .391 kg = 14.465 kg = 14 kg f. 319 mL + 13.75 mL +20. mL = 352.75 mL = 350 mL

Constructing a Model I. Purpose: To understand how scientists make inferences about atoms II. Materials: closed container, various objects, balance III. Procedure First place your items in a closed container. Then take another student's container and, keeping it closed, shake it and see if you can determine the contents. Measure the mass of the container with the contents inside and estimate the mass of the contents. Then open the container without looking and feel the contents. Make adjustments to your estimation of the mass if needed. lastly remove the contents and observe them. Measure their mass on the balance. V. Data Part A: Closed Container Closed Container

# of Objects

Mass of Objects (g)

Kind of Objects

1

2

71.5

metallic

2

8

28

Hard, wood?

3

3

89

Metallic, hair clips

Mass of Objects (g)

Kind of Objects

Part B: Open Container without Looking Open Container # of Objects 1

3

78.6

Eraser, Duplo Lego, Magnet

2

15

29

Hairbands, Tag, Pieces of Paper

3

4

70

Bracelet, Bow

Mass of Objects (g)

Kind of Objects

Part C: Open Container with looking Open Contianer # of Objects 1

3

50.7

Eraser, Duplo Lego, Magnet

2

3

15.8

Earbuds, Ribbon, Paperclip

3

5

8.25

Bow, Charm Bracelet, Safety Pin, Sequin, Charm

VI. Analyze and Conclude 1. Scientists often use more then one method to gather data. How was this illustrated in the investigation? We used three different methods to gather data, shaking the box, feeling the object and then finally looking at the contents.

2. Of the observations made, which were quantitative and which were qualitative? The measurements of mass and the number of items inside were quantitative data. The types of objects were qualitative data. 3. Using the data you gathered, draw a model of the unknown objects and write a brief summary of your conclusions. In this one, number two, I felt several different textures, giving the illusion of many different objects. Also, the objects, in particular the ear-buds, had several components, which gave the illusion of several objects when shaken.

The above image was made by the author, Tommy Smith.

Problem Type 1 mol to g 1. Analyze: Given: 1.002 mol of chromium Unknown: mass in grams Plan: mol x g/1mol Calculate: 1.002 mol Cr x 52.00 g Cr/1 mol Cr = 52.10 g Cr Evaluate: Units, correct number of significant figures, correct There is a little more than 1 mol of substance present. So there would be a little more than 1 molar mass of chromium 2. Analyze: Given: 4.08 x 10-8 mol of sodium Unknown: mass in grams Plan: mol x g/1mol Calculate: 4.08 x 10-8 mol Ne x 20.18 g Ne/1 mol Ne= 8.23 x 10-7 g Ne Evaluate: Units, correct number of significant figures, correct There is very little substance present. Therefore there would be very little mass. 3.Analyze: Given: 7 mol of titanium Unknown: mass in grams Plan: mol x g/1mol Calculate: 7 mol Ti x 47.88 g Ti/1 mol Ti = 300 g Ti Evaluate: Units, correct number of significant figures, correct There are seven moles present. So seven times the molar mass of titanium would be the answer and 300 is about seven times the molar mass. Problem Type 2 grams to mol 1. Analyze: Given: 11.9 g of aluminum Unknown: number of moles of aluminum Plan: g x 1 mol/g Calculate: 11.9 g Al x 1 mol Al/26.98 g Al = 0.441 mol Al Evaluate: Units, correct number of significant figures, correct

There is less than half the molar mass present. Ergo there would be less than half a mole. 2. Analyze: Given: 3.22 g of copper Unknown: number of moles of copper Plan: g x 1 mol/g Calculate: 3.22 g Cu x 1 mol Cu/63.55 g Cu = 0.0507 mol Cu Evaluate: Units, correct number of significant figures, correct There is roughly one twentieth of the molar mass present, so the answer is about one twentieth. 3. Analyze: Given: 2.72 x 10-7 g of lithium Unknown: number of moles of lithium Plan: g x 1 mol/g Calculate: 2.72 x 10-4 g Li x 1 mol Li/6.94 g Li = 3.92 x 10-5 mol Li Evaluate: Units, correct number of significant figures, correct The already small number is being made smaller, this makes the answer reasonable. Problem Type 3 number of atoms to moles 1. Analyze: Given: 3.01 x 1023 atoms of silver Unknown: number of moles of silver Plan: number of atoms x 1 mol/Avogadro's Number Calculate: 3.01 x 1023 atoms Ag x 1 mol Ag/(6.022 x 1023)atoms Ag = 0.500 mol Ag Evaluate: Units, correct number of significant figures, correct The number is about half of Avogadro's number, so it makes sense that the answer would be about half. 2. Analyze: Given: 2.25 x 1022 atoms of carbon Unknown: number of moles of carbon Plan: number of atoms x 1 mol/Avogadro's Number Calculate: 2.25 x 1022 atoms C x 1 mol C/(6.022 x 1023)atoms C = .0374 mol C Evaluate: Units, correct number of significant figures, correct The number is less than Avogadro's number by about one ten's place. So the answer should reasonably be within the hundred's place.

3. Analyze: Given: 2 x 106 atoms of oxygen Unknown: number of moles of oxygen Plan: number of atoms x 1 mol/Avogadro's Number Calculate: 2 x 106 atoms O x 1 mol O/(6.022 x 1023)atoms O = 3 x 10-18 mol O Evaluate: Units, correct number of significant figures, correct The number of atoms is so few that it would be so little of a mole. Problem Type 4 number of atoms to grams 1. Analyze: Given: 5.0 x 109 atoms of neon Unknown: number of grams of neon Plan: number of atoms x molar mass/Avogadro's Number Calculate: 5.0 x 109 atoms Ne x 20.18g Ne/(6.022 x 1023)atoms Ne = 1.7 x 10-13 g Ne Evaluate: Units, correct number of significant figures, correct The amount of atoms is so much less than Avogadro's number, so the mass would be much smaller than the molar mass. 2. Analyze: Given: 20 x 108 atoms of copper Unknown: number of grams of copper Plan: number of atoms x molar mass/Avogadro's Number Calculate: 20 x 108 atoms Cu x 63.55g Cu/(6.022 x 1023)atoms Cu = 2 x 10-13 g Cu Evaluate: Units, correct number of significant figures, correct The amount of atoms is so much less than Avogadro's number, so the mass would be much smaller than the molar mass. 3. Analyze: Given: 2.0 x 1023 atoms of gold Unknown: number of grams of gold Plan: number of atoms x molar mass/Avogadro's Number Calculate: 2.0 x 1023 atoms Au x 196.97g Au/(6.022 x 1023)atoms Au = 66 g Au Evaluate: Units, correct number of significant figures, correct The number of atoms is about one-third the mass of Avogadro's number, 66 is about one third of 196.97.

The Mole Concept I. Purpose: to identify each sample based on observation and numerical data II. Materials: sample set, balance, periodic table III. Procedure Make observations about sample A. Measure the mass on a balance. Use a periodic table of elements to identify the substance. Repeat for samples B, C, and D. IV. Data A. Observations Sample

Observations

A

Metallic, silver color; streaking of the metal

B

Silver color; silver color

C

Tarnished; dark color

D

Metallic, coppery color

B. Numerical Data Sample Mass (g) Number of moles A B

65 26.9

Number of atoms

Identity of the element

1

6.022 x 1023

Zinc

1

23

Aluminum

23

6.022 x 10

C

55.5

1

6.022 x 10

Iron

D

63.2

1

6.022 x 1023

Copper

V. Analyze and Conclude 1. What led you to the determination of each element? The molar mass of each substance is unique, so our masses would be similar to the average atomic mass. 2. If each element was exactly one mole, why were the masses different? The mass of each individual atom is different. 3. Determine the actual mass of a single atom of each sample. Sample A: 1.1 x 10-22 Sample B: 4.67 x 10-23 Sample C: 9.22 x 10-23 Sample D: 1.05 x 10-22 4. Write a one-two paragraph summary of the lab. Draw some general conclusions. This lab was helpful in going beyond book knowledge. I knew the basic concepts; this went and applied them. In this lab, the ones with the highest measured mass had the atoms with the highest mass. Each object was one mole, so they were all different shapes and sizes. This illustrated the differences in molar masses. Overall it was a really fun lab and I lokk forward to others like it.

Atomic Models The Billiard Ball Model Early 1800s Proposed by Dalton ~Explains many chemical properties and confirms the laws of conservation of mass and definite proportions ~Lacks a nucleus and all subatomic particles

The Raisin Bun Model Mid 1800s Proposed by Thomson ~Adds subatomic particle to the model and introduces the nucleus ~Does not explain the existence of electrons outside of the nucleus ~Does not include neutrons

The Planetary Model Late 1800s Proposed by Rutherford ~First real modern view of the atom ~Explains why the electron spins around the nucleus. Proposes that an atom is made mostly of empty space. ~Does not place electrons in definite levels around the nucleus. Does not relate the valence electrons atomic charge. ~Does not include neutrons in the nucleus

All images in this lab were created by Tommy Smith

Quarter one chemistry portfolio

Published on Oct 10, 2012

a tenth grade chemistry portfolio

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