Day 11: Balancing Equations (balancing Equation, Practice, Application, Stoichiometry)


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Thread: Day 11,12 and 13 i have homework

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Day 11 - Reactions, Slides 60 -80 A. Synthesis (composition): • two or more elements or compounds may combine to form a more complex compound. • Basic form: A + X → AX Examples of synthesis reactions: 1. Metal + oxygen → metal oxide EX. 2Mg(s) + O2(g) → 2MgO(s) 2. Nonmetal + oxygen → nonmetallic oxide EX. C(s) + O2(g) → CO2(g) 3. Metal oxide + water → metallic hydroxide EX. MgO(s) + H2O(l) → Mg(OH)2(s) 4. Nonmetallic oxide + water → acid EX. CO2(g) + H2O(l) → H2CO3(aq) 5. Metal + nonmetal → salt EX. 2 Na(s) + Cl2(g) → 2NaCl(s) 6. A few nonmetals combine with each other. EX. 2P(s) + 3Cl2(g) → 2PCl3(g) B. Decomposition: o A single compound breaks down into its component parts or simpler compounds. o Basic form: AX → A + X Examples of decomposition reactions: 1. Metallic carbonates, when heated, form metallic oxides and CO2(g). EX. CaCO3(s) → CaO(s) + CO2(g) 2. Most metallic hydroxides, when heated, decompose into metallic oxides and water. EX. Ca(OH)2(s) → CaO(s) + H2O(g) 3. Metallic chlorates, when heated, decompose into metallic chlorides and oxygen. EX. 2KClO3(s) → 2KCl(s) + 3O2(g) 4. Some acids, when heated, decompose into nonmetallic oxides and water. EX. H2SO4 → H2O(l) + SO3(g) 5. Some oxides, when heated, decompose. EX. 2HgO(s) → 2Hg(l) + O2(g) 6. Some decomposition reactions are produced by electricity. EX. 2H2O(l) → 2H2(g) + O2(g) EX. 2NaCl(l) → 2Na(s) + Cl2(g) C. Replacement: a more active element takes the place of another element in a compound and sets the less active one free. • Basic form: A + BX → AX + B or AX + Y → AY + X Examples of replacement reactions: 1. Replacement of a metal in a compound by a more active metal. EX. Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s) 2. Replacement of hydrogen in water by an active metal. EX. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) EX. Mg(s) + H2O(g) → MgO(s) + H2(g) 3. Replacement of hydrogen in acids by active metals. EX. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) 4. Replacement of nonmetals by more active nonmetals. EX. Cl2(g) + 2NaBr(aq) → 2NaCl(aq) + Br2(l) D. Ionic • Occurrs between ions in aqueous solution. A reaction will occurr when a pair of ions come together to produce at least one of the following: 1. a precipitate 2. a gas 3. water or some other non-ionized substance. • Basic form: AX + BY → AY + BX Examples of ionic reactions: 1. Formation of precipitate. EX. NaCl (aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s • ) EX. BaCl2(aq) + Na2 SO4(aq) → 2NaCl(aq) + BaS O4(s) 2. Formation of a gas. EX. HCl(aq) + FeS(s) → FeCl2(aq) + H2S(g) 3. Formation of water. (If the reaction is between an acid and a base it is called a neutralization reaction.) EX. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 4. Formation of a product which decomposes. EX. CaCO3(s) + HCl(aq) → H2O(l) CaCl2(aq) + CO2(g) + Combustion of Hydrocarbons: Another important type of reaction, in addition to the four types above, is that of the combustion of a hydrocarbon. When a hydrocarbon is burned with sufficient oxygen supply, the products are always carbon dioxide and water vapor. If the supply of oxygen is low or restricted, then carbon monoxide will be produced. This is why it is so dangerous to have an automobile engine running inside a closed garage or to use a charcoal grill indoors. Hydrocarbon (CxHy) + O2(g) → CO2(g) + H2O(g) EX. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) EX. 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) NOTE: Complete combustion means the higher oxidation number is attained.' Incomplete combustion means the lower oxidation number is attained. The phrase "To burn" means to add oxygen unless told otherwise. RE: Day 11 - Reactions, Slides 60 -80 COLLAPSE Calculating molar mass will be on examination 2 as well as converting grams to moles and moles to grams. Learn: For aA ---> bB + cC 1. Mm(A) = (#gA/#gB)(b/a)(Mm(B). 1. Liter to grams What is the mass in grams of 2.1 liters of water? The density of water is 1.0 g/mL and there are 1000 mL in a liter. Therefore, the mass of 2.1 liters of water is 2.1 kilograms. Calculations: (2.1 L)(1000 mL)(1g/mL)(1 kg/1000 g) = 2.1 kg 2. Grams to moles How many grams of glucose are needed to produce 2.1 L of water in a combustion reaction? Balance the equation: C6H12O6(s) + 6O2(g) ----> 6CO2(g) + 6H2O(l) Set up a table of the atomic weights involved: Aw(C)= 12.0 amu Aw(H)= 1.01 amu Aw(O)= 16.0 amu Find the molecular weight of water. Mw(HOH) = 2Aw(H) + Aw(O) = 2.02 amu + 16.0 amu = 18.02 amu to three significant figures = 18.02 amu. Find the molar mass of water. Mm(HOH) = 18.02 g/mol remember Mm(glucose) = [6(12.0) + 12(1.01) + 6(16.0)] g/mL = 180.12 g/mol Find the stoichiometry between glucose and water. 1 glucose molecule is chemical equivalent to six water molecules. Solve the problem, HOH = water and GC = glucose (2100g HOH)(mol HOH/ 18.02 g HOH)(1 mol GC/ 6 mol HOH)(180.12 g GC/mol GC) equals the number of grams of glucose or 3.5 kg glucose 3. Chemical equivalence Nitrogen gas and hydrogen gas yield ammonia gas, how many moles of hydrogen gas is needed to make two moles ammonia gas Balance the equation: N2 (g) + H2 (g) ---> NH3 (g) N2 (g) + 3H2 (g) ---> 2NH3 (g) The answer is three moles of hydrogen gas. 4. Theoretical yield (Actual yield/ Theoretical yield) X 100% = percent yield Suppose three moles of hydrogen is added to one mole of nitrogen gas and only 1.5 moles of ammonia gas is produced. What is the percent yield by mass? The molecular weight of ammonia does not change. Therefore, the actual yield is (1.5 moles)(Mm(ammonia)) and the theoretical yield is (2.0 moles)(Mm(ammonia)), and the percent yield by mass is 75% by mass. Day 11: Balancing Equations (balancing equation, practice, application, stoichiometry); Day 12: Ionic Bonding (Bonds, ions, ionic compounds, naming); Day 13: Reactions of Ionic Bonding (Review naming, solvation, thyroid hormones, metric unit concentration); Day 12 &13 - Reactions - Slides 81 - 105 Rules for Naming Binary Ionic Compounds Containing a Metal Ion With a Fixed Charge A binary ionic compound is composed of ions of two different elements - one of which is a metal, and the other a nonmetal. For example, sodium iodide, NaI, is composed of sodium ions, Na+ (elemental sodium is a metal), and iodide ions, I- (elemental iodine is a nonmetal). Rule 1. The cation is written first in the name; the anion is written second in the name. Rule 2. The name of the cation is the same as the (neutral) element from which it is derived (e.g., Na+ = "sodium"). Rule 3. The anion is named by adding the suffix -ide to the root of the element name (e.g., I- = "iodide"). Note: Greek prefixes are not used to indicate the number of atoms of each element in the formula unit for the compound (e.g., Na2O is named "sodium oxide" not "disodium oxide", or "disodium monoxide"). The key word here is isoelectronic. Ions exist when we change the number of electrons such that the number of protons do not equal the number of electrons. In this case, two things can happen. Case 1: We have more electrons than protons. The will give us an anion. Case 2: We have more protons than electrons. This will give us a cation. Well, what else? Ions have to be stable. When are ions the most stable? Anions are most stable when they are isoelectronic with the closest noble gas to the right of the periodic table, Chloride has 17 electrons. Chloride ion is isoelectronic with argon because it has 18 electrons. So, chloride has gained an electron. When the chlorine atom accepts an electron from a reducing agent it becomes reduced to a chloride anion. Cations are most stable when they are isoelectronic with the closest noble gas to the left of the periodic table. Sodium has 11 electrons. Sodium ion is isoelectronic with neon because it has 10 electrons. So, sodium ion has lost an electron. When the sodium atom losses an electron to an oxidizing agent it becomes oxidized to a sodium cation. The key word is isoelectronic because noble gases have one of several stable electron configurations. Electrolysis of pure water requires excess energy in the form of overpotential to overcome various activation barriers. Without the excess energy the electrolysis of pure water occurs very slowly or not at all. This is in part due to the limited selfionization of water. Hydrogen will appear at the cathode: Reduction at cathode: 2 H+(aq) + 2e− → H2(g) Oxygen will appear at the anode: Anode (oxidation): 2 H2O(l) → O2(g) + 4 H+(aq) + 4e− Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH-(aq) Anode (oxidation): 4 OH- (aq) → O2(g) + 2 H2O(l) + 4 e− Overall reaction: 2 H2O(l) → 2 H2(g) + O2(g) Notice that in the chemical reaction there had to be a charge balance. From the earliest times, people have created myths and legends to explain the natural world. Some stories explain why earthquakes occur or lightning strikes Others tell how rivers, deserts, canyons, and other landforms came to be. This ancient Chinese myth, dating back at least 2,500 years, comes from the area of northern China where Chinese civilization first began. The myth explains how the province of Shaanxi got its mountains. Shaanxi, also known as Shensi, is in northern China. The southern part of the province contains the high and rugged Qinling range, also called the Tsinling Mountains, where the average peak is 8,000 feet high and some are over 12,000 feet high. Every day, Kuafu watched the sun rise in the east and set in the west. When night came, he became greatly saddened. He thought, “I do not like the darkness. All life falls into a silent slumber. If I could catch the sun, then I could keep night as bright as day. The plants could grow forever, and it would always be warm. I would never have to sleep again.” Discharge (dilute sulfuric acid + water): PbO2(s) + Pb(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l) Negative plate reaction: Pb(s) + HSO4- (aq) → PbSO4(s) + 2e- + H+(aq) Positive plate reaction: PbO2(s) + 2e- + HSO4- (aq) + 3H+(aq) → PbSO4(s) + 2H2O(l) Charging (33.5% v/v (4.2 Molar) sulfuric acid (H2SO4)): 2PbSO4(s) + 2H2O(l) → PbO2(s) + Pb(s) + 2H2SO4(aq) Negative plate reaction: PbSO4(s) + 2e- + H+(aq) → Pb(s) + HSO4- (aq) Positive plate reaction: PbSO4(s) + 2H2O(l) → PbO2(s) + 2e- + HSO4- (aq) + 3H+(aq) What is being oxidized? What is being reduced? What is the oxidizing agent? What is the reducing agent? RE: Day 12 &13 - Reactions - Slides 81 - 105 COLLAPSE The average person eats 400 micrograms/day of iodine. This is twice as much as we need. How many ppm of iodine do we need per day based upon this information? hint: A 1/4 teaspoon of iodized table salt provides 95 micrograms (mcg) of iodine. 200 X 10-6 g = 200g/1000000 . . . This is equal to 0.512631579 teaspoons of iodized salt. One teaspoon of table salt converted to gram equals to 5.69 g which would give us about 3 .0 grams of salt (2.99473684 g). so, 200 X 10-6 g/3.0 g = 67 ppm (66.78383128295255 ppm) The actual answer depends upon your age, gender, and health . . . most likely other factors too. The U.S. National Institute of Health recommends: • • • Ages 14 and older: 150 mcg/day Pregnant females of all ages: 220 mcg/day Lactating females of all ages: 290 mcg/day A 1/4 teaspoon of iodized table salt provides 95 micrograms (mcg) of iodine. References: You can calculate those? (I guess it was 2.5 to 3.0 grams of salt? But foods also have it, and people over 50 years old should try to eat 1.5 grams of salt a day) KI dissolves easier in water than NaCl. KI dissolves 144 g/100 mL at 20 oC and 153 g/100 mL at 30 oC; NaCl 36.00 g/100 mL at 20 oC and 36.09 g/100 mL at 30 oC. The units of solubility are given in grams per 100 milliliters of water (g/100 ml) at 1 atmosphere. The ppm values are different too. The density of water is 1 g/mL. Do not forget molecular mass when you think about ions and grams!!! This depends upon how strongly the ions are held together, and how strong the solutesolvent bond is. Well, those calculations are difficult and takes time. So, we usually look at the simple lattice energy first which only depends upon charge and ionic radius. The lattice energy of sodium chloride is higher than that of KI. The Lattice energy, U, is the amount of energy requried to separate a mole of the solid (s) into a gas (g) of its ions. MaLb(s) → a Mb+(g) + b Xa- (g) U kJ/mol This quantity cannot be experimentally determined directly, but it can be estimated using Hess Law in the form of Born-Haber cycle. It can also be calculated from the electrostatic consideration of its crystal structure. As defined, the lattice energy is positive, because energy is always required to separate the ions. For the reverse process, the energy released is called energy of crystallization, Ecryst. a Mb+(g) + b Xa- (g) → MaLb(s) Ecryst kJ/mol Therefore, U = - Ecryst Values of lattice energies for various solids have been given in literature, especially for some common solids. Some are given here. ...
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School: University of Virginia

Here is the slides ;). In regards to the day 11 to 13 reactions ;)

Day 11: Balancing
Equations (balancing equation,
practice, application,

___ HBr + ___ KHCO3 → ___ H2O
+ ___ KBr + ___ CO2

1 HBr + 1 KHCO3 → 1 H2O + 1
KBr + 1 CO2

Day 12: Ionic Bonding (Bonds, ions, ...

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Tutor went the extra mile to help me with this essay. Citations were a bit shaky but I appreciated how well he handled APA styles and how ok he was to change them even though I didnt specify. Got a B+ which is believable and acceptable.

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