1. Significant Figures Practice worksheet with solutions for converting numbers bewteen the expanded form and the significant figure form.
2. Adding and Subtracting Significant Figures
3. Multiplying and Dividing Significant Figures
4. Density Density practice worksheet and solutions.
5. Unit Conversion Practice worksheet involving metric-metric and english-metric - great practice for chemistry students.
6. Balancing Equations Practice sheet
7. Formula Writing Practice worksheet with solutions for writing formulas/ Nomenclature
SIGNIFICANT FIGURES
The number of significant figures refers to the number of digits reported for the value of a measured or calculated quantity, indicating the precision of the value.
Rules for identifying the number of significant figures in a number.
1. All digits are significant except zeros (zeros may or may not be significant)
2. Placeholder zeros are not significant. Placeholder zeros would disappear if the number is placed in scientific notation. (Ex: 0.00025 has two significant figures 2 and 5)
3. Terminating zeros ending to the left of the decimal point may not be significant. When the decimal point is not shown (Ex: 2000) you must assume that the zeros are NOT significant.
4. Zeros are significant when they trail a number to indicate precision in the number. For example 0.05678000 ( the zeros to the left of 5 are not significant, but the three zeros following the 8 are, thus the digit contains seven significant figures)
5. Zeros trapped between any two significant digits are significant. Ex: 5008 and 500.1
Find the number of significant digits in the following:
PROBLEM ANSWER
1. 0.025 _____________
2. 54000 _____________
3. 0.005670 _____________
4. 2.40 x 104 _____________
5. 5007.00 _____________
6. 0.508 _____________
7. 0.0058363 _____________
8. 0.0600 _____________
9. 34000.00 _____________
10. 10200 _____________
ANSWERS TO SIGNIFICANT FIGURES
1. 2 – the zeros 0.025 are placeholders zeros only
2. 2 – the zeros in 54000 are ambiguous since the decimal is not shown
3. 4 – the 0.00 part is for place holder and the trailing zero is significant
4. 3 – the zero indicates precision
5. 6 – the zeros 5007.00 are trapped and 5007.00 indicate precision
6. 3 – the zero 0.508 is place holding and 0.508 is trapped
7. 5 – the zeros 0.0058363 are place holding
8. 3 – the zeros 0.0600 are place holding and 0.0600 indicate precision
9. 7 – the zeros 34000.00 indicate precision and 34000.00 are trapped
10. 3 – the zeros 10200 are ambiguous and 10200 is trapped
1) 25.01 2) 150.3
+ 9.11 -107.240
3) 24.0045 4) 6.987
- 3.12 45.45
+ 0.2
5) 1543.02 6) 322.44
+ 394.934 0.321
72.0
- 68.9555
7) 553.24 8) 439.00
+ 7 + 6.3
ANSWERS
1) 34.12
2) 43.06 à 43.1
3) 20.8845 à 20.88
4) 52.637 à 52.6
5) 1937.954 à 1937.95
6) 181.1635 à 181.5
7) 560
8) 445.3
3) 1645.54 x 0.1113 x 67.0 4) 22.091 ¸ (1.0 x 103) ¸ 0.999
5) 743.001 ¸ 0.0562 6) 98.777 x 20.033 x 111 x 0.6577
7) 55601 ¸ 348 ¸ 3.0 8) 4789.33 x 6871.00 x (2 x 102)
ANSWERS
1) 7606.184 à 7,606
2) 16.43187 à 16
3) 12270.956 à 12,300
4) 2.2113 x 10-2 à 2.2 x 10-2 or 0.022
5) 13220.658 à 13,200
6) 144461.67 à 144,000
7) 53.257663 à 53
8) 6.5815 x 109 à 7 x 109 or 7,000,000,000
1. What is the density of benzene in grams per milliliter if 5.5L of benzene weigh 4.8kg?
2. The density of air is 1.3 g/L. How many liters of air weigh 2.0 lb?
3. Given a cylinder with diameter of 39 mm and height of 39 mm, and mass of 1.0 Kg, what is it’s density?
4. How many ounces does a 250ml sample of sulfuric acid weigh? (density = 1.30 g/ml)
5. What is the density of the Earth if 1.00 yd3 weighs 22.0 lb? Express your answer in lb/ft3.
6. The density of brass is 8.0 g/ml. A cube of brass 2.0 inches on an edge will weigh how many grams?
7. An unknown substance has a volume of 0.114 L and has a mass of 467g. What is its density in g/mL?
8. The density of Ethanoic Acetic Acid is 1.05 g/cm3. What is the mass in kilograms of a volume of 5L?
ANSWERS
1) D = mass ?D= 4.8Kg x 1000g x 1L = 0.87 g/mL
vol 5.5L 1Kg 1000mL
2) V= m ?g = 2 lb x 453.6 g = 907.2 g
D 1 lb
?V = 907.2g = 907.2g x 1L = 697.8 L
1.3g 1.3g
1L
3)Volume of a cylinder v= P r2h
r = ½ d, r = ½ (39mm) = 19.5 mm
v= P (19.5 mm)2(39mm) = 46589 mm3
? cm3 = 46589 mm3 x (1m)3 x (100cm)3 = 46.5 cm3
(1000mm)3 (1m)3
D = 1 Kg x 1000g = 21.5 g/cm3
46.5 cm3 1Kg
4) m= vD
?oz = 250ml x 1.3g x 1 oz = 11.5 oz
1 mL 28.35 g
5) ? lb = 22 lb x (1yd)3 = 0.815 lb/ft3
ft3 1 yd3 (3ft)3
6) Volume of a cube v = s3
? v = (2in)3 x (2.54 cm)3 = 131cm3
(1 in)3
? v = 131 cm3 x 1 mL = 131 ml
1 cm3
m= vD = 131 mL x 8g = 1048g
1 mL
7) D = m = 467g x 1 L = 0.467 g/mL
v 0.114L 1000mL
8) ? v = 5 L x 1000mL x 1 cm3 = 5000cm3
1L 1mL
?m = vD= 5000cm3 x 1.05g x 1 Kg = 5.25 Kg
1 cm3 1000g
Dimensional Analysis (Factor Label Method)
1. 1.2 kg = ________dg
2. 2.00 x 10-5m = ________in
3. 25.4 mm = ________cm
4. 1.2 miles = ________km
5. 15.47 m3 = ________km3
6. 17.0 ft/s = ________m/min
7. 342 miles/hr = ________km/s
8. 45.1 yards = ________cm
9. 1.45 L = ________gallons
10. 4.100 g = ________mg
11. 1.2 kg/yard = ________lbs/m
12. 2.00ft3/min = ________L/hour
13. 145 ml = ________cm3
14. 6.51 miles = ________cm
ANSWERS
1. 1.2 x 104 dg
2. 7.87 x 10 –4 in
3. 2.54 cm
4. 1.9 km
5. 1.547 x 10 –8 km3
6. 311 m.min
7. 0.153 km/s
8. 4.12 x 103
9. 0.383 gallon
10. 4.100 x 103 mg
11. 2.9 lbs/m
12. 3.40 x 103 L/hr
13. 145 cm3
14. 1.05 x 106 cmFormatting was lost, but try to interpret the given questions to the best of your abilities :)
1. Al + Fe3O4 ------> Al2O3 + Fe
3. CH3OH + O2 ------> CO2 + H2O
4. P4O10 + H2O ------> H3PO4
5. PCl5 + H2O ------> H3PO4 + HCl
6. SbCl5 + H2O ------> SbOCl3 + HCl
7. MgO + Si ------> Mg + SiO2
8. CaCl2 + Na2CO3 ------> CaCo3 + NaCl
9. C6H6 + O2 ------> CO2 + H2O
10. Al2S3 + H2O ------> Al(OH)3 + H2S
11. C2H6 + O2 ------> CO2 + H2O
12. KClO3 ------> KCl + KClO4
13. KBr + Cl ------> KCl + Br2
14. (NH4)2SO4 + NaOH ------> NH3 + H2O + Na2SO4/
15. Calcium Phosphate reacts with Sulfuric Acid to give Phosphic acid and Calcium Sulfate as a solid. Write the balanced equation for this reaction.
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WRITING FORMULAS
Write the correct chemical formula for the following:
1.Magnesium Nitride
2.Iron (III) Oxide
3.Sodium Sulfate
4.Copper (II) Chloride
5.Barium Nitrate
6.Aluminum Hydroxide
7.Mercury (II) Phosphate
8.Aluminum Silicate
9. Copper (II) Bromide
10.Lead (II) Chlorate
11.Silver Cyanide
12.Ammonium Oxide
13.Aluminum Perchlorate
14.Tin (II) Chloride
15.Nickel (III) Acetate
16.Potassium Sulfide
17.Magnesium Bisulfate
18.Iron (II) Phosphate
19.Cobalt (II) Hydrogen Sulfate
20.Chromium (II) Bicarbonate
21.Sodium Hypochlorite
22.Barium Carbonate
23.Zinc (II) Permanganate
1 mol = 6.022 x 1023 atoms (use with molecular compounds)
1 mol = 6.022 x 1023 formula units (used with ionic compounds)
1 mol = 6.022 x 1023 molecules
Atom- the smallest particle of an element that maintains its chemical identity through all chemical and physical changes.
Molecule- the smallest particle of an element or compound that can have a stable independent existence
Compound- a substance composed of two or more elements in fixed proportions
Element- a substance that cannot be decomposed into simpler substances by chemical means; composed of only one type of atom
Formula Unit- the smallest repeating unit of a substance; for non-ionic substances, the molecule.
Ion- an atom or group of atoms that carries an electrical charge; unequal # of protons and electrons
Molecular Compound- a compound that does not contain ions
Ionic Compound- a compound that is composed of cations and anions
MASS OF ONE MOLE OF ATOMS OF ELEMENTS
Element | # of Atoms in formula | Sample | Contains |
Carbon: C | 1 atom for each molecule | 12.001g C | 6.022 x 1023 C atoms or 1 mol of C atoms 6.022 x 1023 C molecules |
Hydrogen: H2 | 2 atoms for each molecule | 1.0079g H2 | 6.022 x 1023 H atoms or 1 mol of H atoms 3.011 x 1023 H2 molecules |
Sulfur: S8 | 8 atoms for each molecule | 32.066g S8 | 6.022 x 1023 S atoms or 1 mol of S atoms 0.75 x 1023 S8 molecules |
MASS OF ONE MOLE OF MOLECULAR SUBSTANCES
Substance | Molecular Wt | Sample | # of Atoms in formula | Contains |
Hydrogen | 2.016 | 2.016g H2 | 2 H | 6.022 x 1023 molecules or 1 mol of H2 molecules 2 * 6.022 x 1023 = 1.2044 x 1024 atoms of H or 2 mol of H atoms |
Methane | 16.04 | 16.04g CH4 | 1 C, 4 H | 6.022 x 1023 CH4 molecules or 1 mol of CH4 molecules 6.022 x 1023 C atoms or 1 mol of C atoms 4 * 6.022 x 1023 = 2.4088 x 10 23 H atoms or 4 mol of H atoms |
MASS OF ONE MOLE OF IONIC COMPOUNDS
Compound | Formula Wt | Sample with a Mass of 1 mol | # of Atoms in formula | Contains |
Sodium Chloride | 58.44 | 58.44 g NaCl | 1 Na, 1 Cl | 6.022 x 1023 Na+ ions or 1 mol of Na+ ions 6.022 x 1023 Cl- ions or 1 mol of Cl- ions |
Aluminum Sulfate | 342.1 | 342.1g Al2(SO4)3 | 2 Al, 3 (SO4) | 2 * 6.022 x 1023= 1.2044 x 1024 Al+3 ions or 2 moles of Al+3 ions 3 * 6.022 x 1023= 1.8066 x 1024 SO4-2 ions or 3 moles of SO4–2 ions |
MGCCC, Perk Learning Lab
TLM
CONVERTING ATOMS, MOLECULES & MOLES
1. How many moles of atoms does 355.7 g of gold contain?
2. Calculate the mass of one carbon atom in grams.
3. How many (a) moles of Cl2 (b) Cl2 molecules and (c) Cl atoms are contained in 45 g of Chlorine gas (Cl2)?
4. Calculate the number of (a) hydrogen, (b) oxygen and (c) sulfur atoms in 63g of sulfuric acid (H2SO4)
5. A sample of metal contains 2.516 x 1023 atoms and has a mass of 82.29 grams. (a)How many moles of metal atoms are present in the sample? (b) What is the metal?
6. How many moles of O are in 6 moles of O3?
ANSWERS ON BACK
ANSWERS
1. 355.7g Au * 1 mol Au atoms = 1.805 moles of Au atoms
196.97 g Au
2. 12 g C * 1 mol C atoms = 1.99 x 10-23 g C/ C atoms
1 mol C atoms 6.022 x 10 23 C atoms
3. a. 45g Cl2 * 1 mol Cl2 = 0.635 mol Cl2
70.906 g Cl2
b. 0.635 mol Cl2 * 6.022 x 1023 molecules = 3.82 x 1023 Cl2 molecules
1 mol Cl2
c. 3.82 x 1023 Cl2 molecules * 2 Cl atoms = 7.64 x 1023 Cl atoms
1 Cl2 molecule
4. a. 63 g H2SO4 * 1 mol H2SO4 * 6.022 x 1023 molecules * 2 H atoms = 7.7 x 1023 H atoms
98 g H2SO4 1 mol H2SO4 1 H2SO4 molecule
b. 63 g H2SO4 * 1 mol H2SO4 * 6.022 x 1023 molecules * 1 S atom = 3.87 x 1023 S atoms
98 g H2SO4 1 mol H2SO4 1 H2SO4 molecule
c. 63 g H2SO4 * 1 mol H2SO4 * 6.022 x 1023 molecules * 4 O atoms = 1.54 x 1024 O atoms
98 g H2SO4 1 mol H2SO4 1 H2SO4 molecule
5. a. 2.516 x 1023 atoms * 1 mol = 0.4178 mol of metal
6.022 x 1023 atoms
b. 82.29 g = 196.96 g/mol , Au (gold) is the metal
0.4178 mol
6. 6 mol O3 * 3 mol O = 18 mol O
1 mol O3
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