Chemistry Review Manual 2

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Example 6.7: Using Hess’ Law 74 Equation 6.7: ∆H from enthalpies of formation 74 Figure 6.2: Graphical depiction of Hess' law for formation reactions 75 Example 6.8: ∆H from enthalpies of formation 75 Equation 6.8: ∆H from bond enthalpies 76 Figure 6.3: Graphical depiction of Hess' law for bond dissociations 77 Example 6.9: ∆H from bond enthalpies 77 Problems 77 Solutions 79 7. Math Skills 83 7.1 Scientific Notation 83 Example 7.1: Scientific notation 83 Table 7.1. Units of Distance 84 7.2 Dimensional Analysis 84 Example 7.2: How many Cu atoms are there in a penny? 85 7.3 Solving for a variable 85 Example 7.3: New concentration after dilution 85 Example 7.4: Two equations and two unknowns 86 7.4 Logarithms 86 Periodic Table of the Elements 88 Page 4 of 88
1. The Mole Early chemists discovered that substances combine in certain combinations to form new substances. For example, 32 g (grams) of oxygen combines with 4 g of hydrogen to form 36 g of water. Also, 32 g of oxygen combines with 12 g of carbon to form 44 g of carbon dioxide. Eventually people realized that substances are made of atoms, and that atoms combine in specific ratios. The various combinations of masses that were observed to react is explained when we account for different atoms having different masses. The numbers of atoms encountered in everyday life is very large. Even a tiny quantity of a substance consists of a vast number of atoms. It is inconvenient to constantly talk about such large numbers of atoms when discussing chemical reactions – or substances in general. This approach to dealing with large numbers is actually quite common. For example, when discussing the national debt of the United States, we hear numbers like 14.3 trillion dollars. This number is expressed in units of trillions of dollars. This is much more convenient than writing the number as $14300000000000. It turns out that there are typically a lot more atoms than there are dollars of national debt. Thus, the unit for numbers of particles used in chemistry is much bigger than 1 trillion. It is N A = 6.022×10 23 mol −1 , the Avogadro constant. The unit itself is called the mole (or mol). If we have 1 mol of hydrogen atoms, we have 6.022×10 23 hydrogen atoms. This number arises because chemists wanted 1 mol of hydrogen atoms (the lightest atoms) to have a mass of about 1 g. They achieve this by setting the mass of 1 mol of carbon-12 atoms to exactly 12 g (carbon- 12 is the common isotope of carbon). This convention defines the value of the Avogadro constant. 1.1 How do we use the mole? Counting atoms and molecules The masses of 1 mol of each type of atom 1 are given in the periodic table. With this data, we can determine how many atoms are in a sample by measuring the mass of the sample. Example 1.1: If a sample of Helium has a mass of 6.600 g, then how many helium atoms are in the sample? Approach: Find mol of He. Convert to number of atoms. The molar mass of helium (i.e., the mass of 1 mol of helium atoms) is expressed as 4.003 g mol -1 . You find this in the periodic table. Therefore, 6.500 g of helium consists of 6.600 g 2 = 1.648 1 8 mol of helium 4.003 g mol − To get the number of helium atoms not in units of mol, we simply multiply by N A = 6.022×10 23 atoms mol −1 . The Avogadro constant has units of “particles” per mole. Here, the particle is a helium atom. Thus, there are 1.648 8 mol × 6.022×10 23 atoms mol −1 = 9.929×10 23 atoms 1 Molar masses found in the periodic table are averages of isotope masses, weighted according to natural abundance. 2 The final digit in 1.648 8 is written as a subscript to indicate that we are carrying the extra digit (the second 8) into the next step of the calculation. It is written as a subscript because it is not a significant figure. Carrying the extra figure reduces rounding errors during the steps of a calculation. Page 5 of 88
Page 1 and 2: Chemistry Review Manual 2 nd Editio
Page 3: Example 4.1: Type of bonding 31 4.2
Page 7 and 8: 2 Ca(s) + O 2 (g) → 2 CaO(s) It i
Page 9 and 10: 1.5 Solutions Sometimes we are give
Page 11 and 12: Problems: 1.1 How many grams of cal
Page 13 and 14: 1.3 Mass of water produced = volume
Page 15 and 16: The molar mass is determined from T
Page 17 and 18: pV amount of H 2 consumed = n = RT
Page 19 and 20: Solutions: 2.1 (a) These three gase
Page 21 and 22: The amount of Al = (2/3) × 0.0043
Page 23 and 24: 3.2 Strong and Weak Acids and Bases
Page 25 and 26: Note that weak acids are conjugate
Page 27 and 28: onds to the carbon atom. The result
Page 29 and 30: Solutions: 3.1 (a) 2 Li(s) + Cl 2 (
Page 31 and 32: 4. Bonding 4.1 Types of Bonding The
Page 33 and 34: e.g. δ+ H H δ+ δ+ δ+ H H δ+ O
Page 35 and 36: that pH = 7 tells us that the solut
Page 37 and 38: Solutions: 4.1 (a) Cl 2 (g). The bo
Page 39 and 40: 5. Chemical Equilibrium 5.1 Dynamic
Page 41 and 42: Bases are characterized by the base
Page 43 and 44: Suppose we have a table of equilibr
Page 45 and 46: Whether we add or remove product, t
Page 47 and 48: from which we get 4 4 = = 4 2 ⎛ x
Page 49 and 50: The solution of this quadratic equa
Page 51 and 52: consider dissolving CaCO 3 (s) in a
Page 53 and 54: which gives Using the quadratic for
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(c) H 2 S K a1 = 8.9 × 10 −8 App
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( ) 2 − 1.8× 10 ± 1.8× 10 + 4
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5.2 An equilibrium mixture of N 2 O
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Solutions: 5.1 (a) K = p p p 2 BrCl
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5.5 CH 4 (g) + Br 2 (g) CH 3 Br(g
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(b) HOCl(aq) + H 2 O(l) H 3 O + (
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6. Thermochemistry Chemical reactio
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or which rearranges to q water =
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6.4 Energy Conservation - the First
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The number of moles of H + (aq) is
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Note that an element in the standar
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∆Hgas phase reaction ≅ 2 D[C-C]
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Solutions: 6.1 Heat is related to t
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3 3 × ⎯⎯→ × − 2 2 o −1
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7. Math Skills Chemistry is a quant
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The value of the gas constant used
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If log x < 0, then 0 < x < 1. Note
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