2. Pages 2 and 3

3.  How close a measured value is to an accepted value

4.  Refers to how close a series of measurements are to one another

5.  A variable that is changed by the scientist

6.  the variable measured in the experiment

7.  Observations based on quantities (numbers)

8.  Observations based on characteristics

9.  An explanation supported by many experiments

10.  Describes a relationship in nature

11.  All of the measurements can be close to the same value (precision) but not be the correct value (accuracy)

12.  Which student has the highest precision? Student A  Which student has the highest accuracy? Student B

13.  A. a balance that is not set to zero  B. not reading a graduated cylinder at eye level  C. altering the procedure during an experiment  D making the same error with each trial

14.  A. a balance that is not set to zero  B. not reading a graduated cylinder at eye level  C. altering the procedure during an experiment  D making the same error with each trial

15.  A. 304,002 --> 6 significant figures  B. 0.00167  3 significant figures  C. 0.0030  2 significant figures  D. 30,020  4 significant figures  E. 6.200x 10 3  4 significant figures  F. 800  1 significant figure

16.  5000  5.0x10 3  123,500  1.235x10 5  Convert into standard notation:  2.40x 10 -3 .00240  2.975 x 10 5  297500

17.  a) 5.000 x 2.0000 x 3.000 = 30.00  b) 5.000 x 2.0 x 3.000 = 30.  c) 0.0003047 – 4 = -4  d) 3.14 + 6.2 + 1.618 = 11.0

18.  D = m v  D = 6.330=.88g/mL 7.2

19.  D = mV=LxWxH v8.0= 2.0 x 2.0 x 2.0  D = 5.2=.65g/cm 3 8.0

20. 6800cm 1m= 68m 100 cm 2.05 moles 6.02x 10 23 = 1.23 x 10 24 atoms 1 mole 3. 75x 10 24 1 mole= 6.23 moles 6.02x 10 23

21. Pages 4 and 5

22.  ATOM- smallest particle of an element  ISOTOPE- same number of protons (same element) but different number or neutrons

23.  Mass number- Add protons + neutrons  Atomic mass- average mass of all the isotopes that occur in nature (periodic table)  Atomic number- number of protons

24. Isotope name Isotopic symbol Atomic number Mass ## of protons # of neutron # of electron Zinc- 66 Zn3066303630 Palladiu m- 106 Pd46106466046 66 30 106 46

25. (.00337 x 35.968) + (.00063x 37.963) + (.9960 x 39.962) = 39.96 amu b. Why is the average atomic mass of argon closet to the value 39.962 amu? Majority of the isotopes weigh 39.962 amu

26. B, C, E The three choices are isotopes because they all have 8 protons.

27.  1. Everything is made of atoms  2. Not all atoms are alike.  3. Atoms combine to form compounds in whole number ratios  4. A chemical reactions is the rearranging of atoms.

28. ScientistModelExperimentResulting discoveries ThompsonPlum puddingCathode ray tubes electrons RutherfordNuclear atomic model Gold FoilNucleus, proton James Chadwick ---------------------------Neutron BohrBohr orbital's--------Energy levels of shells

29.  Neutron- neutral charge, relative size of 1, found in the nucleus  Proton- positive charge, relative size of 1, found in the nucleus  Electron- negative charge, relative size of 1/1840, found in the electron cloud

30. Pages 6 and 7

31.  Lowest energy level (closest to the nucleus)

32.  When an electron gains a quantum of energy it can move up and occupy a higher energy level.  The electron in the higher level is in an excited state.

33.  Minimum amount of energy that can be gained or lost by an atom

34. Visible light is only a small part of the larger ELECTROMAGNETIC SPECTRUM Radio waves Micro waves Infrared. Ultra- violet X- Rays Gamma Rays Low energy High Energy Low Frequency High Frequency Long Wavelength Short Wavelength Visible Light 700 nm600 nm500 nm400 nm

35.  As the frequency of an electromagnetic wave increases, the wavelength DECREASES and the energy of the wave INCREASES.

36.  When an electron moves to a higher energy level it ABSORBS energy. When an electron falls to a lower energy level it EMITS energy which we see as visible light.

37.  The emission of light is produced when an electron falls to a lower energy level.

38.  Each element requires a different quantum of energy to be released.

39.  C= f λ  C= 3.00x10 8 m/s (it is a constant!) λ= (lamda) wavelength (m) F= frequency (Hz)

40.  E=hf  E= energy (J)  h = plancks constant= 6.63x 10 -34 JxS  F = frequency (hz)

41. Pages 8 and 9

42. IonicCovalent Metal + nonmetalNonmetal +nonmetal High Boiling and melting pointLow boiling and melting point Made of SolidsMade of liquids and gases Electrons are transferredElectrons are shared Conducts electricity in aqueous solutions Does not conduct electricity Crystalline structureMolecular structure

43.  Ionic: 1. Name the cation regularly 2. If the anion is a nonmetal change the ending to –ide 3. If the anion is a polyatomic – use the name of the polyatomic 4. If you have a transition metal – the charge goes in parentheses

44.  Covalent: 1. Use prefixes before the names (do not use mono before the first element) 2. The second NM should begin with a prefix and end in –ide

45.  a) BaF 2 Barium Fluroide  b) CuSO 4 Copper (II) Sulfate  c) NH 4 HCO 3 Ammonium hydrogen carbonate  d) Sn(OH) 4 Tin (IV) Hydroxide  e) CoCl 2 Cobalt (II) chloride  f) Mn 3 (PO 4 ) 2 Manganese (II) Phosphate  g) Cd(CO 3 ) Cadmium (II) carbonate  h) P 2 F 7 Diphosphorous heptafluoride  i) SO 3 Sulfur trioxide  j) CCl 4 Carbon tetrachloride

46.  Ionic-  1. write the symbol for each element  2. write the charges as superscripts on top  3. criss cross the charges  4. reduce as needed

47.  Covalent-  1. use prefixes in the names as subscripts in the formulas

48.  a) Silver nitrate AgNO 3  b) Iron(III) sulfate Fe 2 (SO 4 ) 2  c) Lead(II) phosphate Pb 3 (PO 4 ) 2  d) Sodium hydroxideNaOH  e) Aluminum fluoride AlF 3  f) Magnesium oxide MgO  g) Copper(II) chlorideCuCl 2  h) Potassium chromate K 2 CrO 4

49.  a) Hexanitrogen octoxide N 6 O 8  b) Sulfur pentoxide SO 5  c) Diphosphorus decachloride P 2 Cl 10  d) Arsenic nonafluoride AsF 9

50. Pages 10 and 11

51.  1 mole = 6.02x10 23 atoms  1 mole = the formula mass

52.  1. Separate out the elements  2. look up and record atomic mass of each element  3. multiply atomic mass by subscript number  4. add them all together

53.  Ca: 40.08 x 1 = 40.08  N: 14.007 x 2 = 28.014  O: 15.999x 6 = 95.994 40.08+28.014+95.994 = 164.088g/mol

54.  Write the formula following ionic rules: NH 4 Cl  N: 14.007 x 1 = 14.007  H: 1.008 x 4 = 4.032  Cl: 35.453 x 1 = 35.453 14.007+4.032+35.453 = 53.492g/mol

55.  19.5 g Ca(NO 3 ) 2 1 mole=.119 moles 164. 088g Molar mass of Ca(NO 3 ) 2

56.  7.45x10 20 atoms 1 mole164.088g 6.02x10 23 1 mole Atoms Answer:.203 grams Molar mass of Ca(NO 3 ) 2 Avogadros number

57.  9.85 grams 1 mole 6.02x10 23 molecules 52.077g 1 mole Formula: (NH 4 ) 2 O Answer: 1.14x10 23 Molecules Molar mass of Ca(NO 3 ) 2 Avogadros number

58.  (NH 4 ) 3 PO 4  N: 14.007 x 3 = 42.021  H: 1.008 x 12 = 12.096  P: 30.974 x 1 = 30. 974  O: 15. 999x 4 = 63. 999 149.087 g/mol N: 42.021 x 100 =28.19% H:12.096 x 100 = 8.11% 149.087149.087 P: 30.974 x 100= 20.78 O: 63.996 x 100 = 42.93% 149.087149.087 Ionic formula Molar mass

59.  a) C 6 H 12 O 6 Molecular – empirical : CH 2 O  b) PbO 2 Empirical  c) Fe 2 O 3 Empirical  d) C 4 H 8 Cl 2 Molecular – empirical C 2 H 4 Cl  Empirical- all the way reduced  Molecular- not reduced

61. Empirical  Percent to grams  Grams to moles  divide by small  Multiply until whole  Molecular Molecular weight = Molar mass of empirical weight Multiply the subscripts in the empirical by this answer

62.  C: 75.7g 1 mol = 6.3 mol= 6.5 x2 = 13 12.011 g.969 mol  H: 8.80g 1 mol = 8.73 mol = 9 x 2= 18 1.008g.969 mol  O: 15.5g 1 mol =.969 mol = 1 x2 =2 15.999g.969 mol Empirical Answer: C 13 H 18 O 2 Divide by smallest mole Not a whole number. Multiply by 2 to Make it a whole number

63.  Empirical Answer: C 13 H 18 O 2 Find the molar mass of the empirical formula: Molar mass = 206g/mol Molecular mass = 412 = 2 Empirical mass 206 Molecular answer C 26 H 36 O 4 Multiply the subscripts in the empirical formula by this number!

65.  C:59g 1 mol = 4.9 mol= 9 12.011 g.54 mol  H: 7.1g 1 mol = 7.04 mol =13 1.008g.54 mol  O: 26.2g 1 mol = 1.64mol = 3 15.999g.54 mol  N: 7.7g1 mol =.54 mol = 1 14.007.54 mol Empirical Answer: C 9 H 13 O 3 N Divide by smallest mole

66.  Empirical Answer: C 9 H 13 O 3 N Find the molar mass of the empirical formula: Molar mass = 180g/mol Molecular mass = 180 = 1 Empirical mass 180 Molecular answer C 9 H 13 O 3 N *The molecular formula and empirical formula can be the same! Multiply the subscripts in the empirical formula by this number!