SI EXAM #1 REVIEW - David Josh& Jessica CHEM105A SI Exam...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: David, Josh, & Jessica CHEM105A SI Exam Review #1 September 12, 2010 Drs. Parr & Bertolini Welcome to YOUR FIRST CHEM105A SI Exam Review! I know this whole experience may seem overwhelming, but it’s important that you realize this: most of the work you can do will be on your own. Take everything you learn/review in this session and continue to practice it. That’s the key to success! Significant Figures 1. ______________ ______________ always count as significant figures. 2. Zeros. (Three classes) a. ______________ zeros are ( always / never / sometimes ) significant. b. ______________ zeros are ( always / never / sometimes ) significant. c. ______________ zeros are ( always / never / sometimes ) significant. 3. ______________ ______________ never limit the number of significant figures in a calculation. When do you round? ________________ Addition/Subtraction: Result has same number of ________________ as least precise measurement used in calculation. - Example: 52.4 + 291 - 29.78 Multiplication/Division: Result has same number of ________________ as least precise measurement used in calculation. - Example: (2.6 x 103) ÷ 14.5 4. a. [(2.8 x 105) ÷ (4.12 x 104)] + (3.88 – 1.21) = b. (2.86 x 4.559) + 7.99 = c. (1467 – 22.3) x (5.6 x 103) = d. (8.207 x 3.23) ÷ (12.5 – 10.5) = Dimensional Analysis 5. How many µmol are there in 1.5 x 10-3 mol? 6. How many inches are there in 5.67 x 10-4 km? (1 in = 2.54 cm) 7. What is the speed of light, 3.00 x 108 m/s, in mph? (1609 m = 1 mile) 8. A swimming pool holds 205 m3 of water. What is the mass of the water in pounds? (density of water: 1 g/cm3, 1 lb = 454 g.) 9. A particular sample of crude oil has a density of 0.85 g/mL. What is the mass of 1.00 barrel of this crude oil (in kg), where a barrel of oil is exactly 42 gallons? (1 gallon = 3.785 L) 10. A gallon of ice cream contains exactly 32 servings, and each serving has 165 calories, of which 30.0% are derived from fat. How many calories derived from fat would you consume if you ate one half-gallon of ice cream? 1 David, Josh, & Jessica Periodic Table: Group 1: _______________ Group 2: _______________ CHEM105A SI Exam Review #1 September 12, 2010 Group 7: _______________ Group 8: _______________ Drs. Parr & Bertolini Metals are located on the _______________ and tend to _______________ electrons. Nonmetals are located on the _______________ and tend to _______________ electrons. Transition metals are located in the _______________. Atomic Symbols 11. Complete the table. Element Protons 17 Neutrons 9 Electrons 10 Net charge O2 - 8 -2 +1 +2 Ag+ Ca2+ 47 20 63 20 46 18 86 222-Rn 86 136 0 Isotopes and Ions neutrons Isotopes differ in the number of ___________. Ions differ in the number of ___________. electrons ___________ are positively charged ions. Cations ___________ are negatively charged ions. Anion I. Polyatomic Ions Note: Memorize your polyatomic ions! They are important and will come up on exams throughout the semester. Make flashcards. Look for patterns. (e.g. charge, recognize the relationship between –ite and –ate). 12. One way of organizing these ions can be found in your textbook, Table 3.5 on page 91. Let’s try a different format. Charge +1 Ammonium: -1 Dihydrogen phosphate: Hydrogen Carbonate: - -2 H2PO4Hydrogen phosphate: Carbonate: CO3 2Sulfite: S O 42- -3 HPO4 P O 43- NH4+ Phosphate : HCO3- : HSO3 Hydrogen sulfate: SO3 2- HSO4- Sulfate : 2 David, Josh, & Jessica Cyanide: OH - CN CHEM105A SI Exam Review #1 September 12, 2010 Chromate 2CrO4 Dichromate: Peroxide: Drs. Parr & Bertolini : Hydroxide : : Cr2O7 N O 2Nitrate: Nitrite O2 2- NO3 : ClOChlorite: Hypochlorite ClO2: ClO3 Perchlorate: Permanganate: Acetate: Chlorate ClO4- MnO4 C2H3O2 II. Naming Compounds Binary Ionic Binary Covalent What two type of bonding did we discuss? (1)_________________ and (2)____________________. Binary ionic Rules for naming (1)______________ compounds: A. Metal (___________) is named first. Base name of nonmetal (______________) comes next + ____. cation ide anion B. We must indicate the ___________ of the cation if it can vary. oxidation # Quick question: Which three transition metal charges are invariant? 1. Zn 2. Cd 3. Hg Note: It is not necessary to include a roman numeral for these compounds. You can find them listed in table 3.2 in your textbook. 13. Let’s try a few examples. We’ll do the first six. The remaining six are for you to practice. NH4C2H3O2 Zn(OH)2 ammonium acetate zinc hydroxide lithium chloride LiCl CsI cesium iodide 3 David, Josh, & Jessica CHEM105A SI Exam Review #1 September 12, 2010 Drs. Parr & Bertolini AgCl silver chloride iron (III) oxide barium nitrate Fe2 O 3 Ba(NO3)2 MgMnO4 magnesium permanganate strontium flouride SrF2 CaBr2 calcium bromide chromium (II) iodide CrI2 ScF scandium fluoride binary covalent Rules for naming (2)________________ compounds: prefix prefix A. __________ + name of 1st element + __________ + base name of 2nd element + -_________. ide 14. We’ll do three examples. You can use the last two for practice. PCl5 XeF phosphorus pentachloride xenon hexafluoride tetrasilver tetraoxide Ag4 O 4 NCl3 nitrogen trichloride disulfur decaflouride S2F10 III. Naming Acids no oxygen oxygen present What two types of acids did we discuss? (1) ___________________ and (2) _____________________. Rules for naming acids: ic hydro (1). __________ + base name of nonmental + _______ +acid (2). Base name of the ________________ + (______or ______)+ acid. anion ous ic 15. We’ll do four. You do four. HCl HI hydrochloric acid hydroiodic acid dichromic acid H2Cr2O7 HMnO4 permanganic acid phosphoric acid 4 H3PO4 David, Josh, & Jessica CHEM105A SI Exam Review #1 September 12, 2010 Drs. Parr & Bertolini HF hydrofluoric acid hydrocyanic acid HCN HC2H3O2 acetic acid Average Atomic Mass. General Formula: AAM = sum(atomic mass x relative abundance) 16. The atomic weight of potassium 39.10 amu. There are 3 naturally occurring isotopes of potassium, and their atomic mass and relative abundance are presented in the following table. Calculate the missing isotopic mass for 41K. Isotope % Isotopic Mass 39.10 amu = .933(38.96371) + .00012(39.964) + .06688(x) x= 40.9997 amu The Mole. 39 40 K K 41 K Abundance 93.3% 0.012% 6.688% (amu) 38.96371 39.96400 ? Avogadro’s Number: 6.02 x 10^23 particles/mole 17. Adenosine (C10H13N5O4, MM = 267.2 g/mol) is administered intravenously to diagnose arrhythmias in patients with problematic tachycardia (high heart rate). Calculate the number of atoms of adenosine used in a typical initial dose, 6.00 mg. (6.00mg)(1g/1000mg)(1 mol/267.2g)(6.02x10^23 atoms/mol) = 1.35x10^19 atoms of adenosine Balancing Equations. 18. Please balance the equations kthanxbye! a. Cu2O(s) + C(s) ! 2 Cu(s) + b. c. CO(g) Co2S3(s) + 2 Co(NO3)3(aq) + 3 (NH4)2S(aq) ! 3 N2H4(l) ! 4 NH3(g) + 6 NH4NO3(aq) N2(g) Molar Mass and % Composition. General Formulas: MM = mass/ # mol % Comp. = mass / total mass x 100% 5 David, Josh, & Jessica CHEM105A SI Exam Review #1 Drs. Parr & Bertolini September 12, 2010 19. Plavix (clopidogrel), an oral antiplatelet agent used to inhibit blood clots in cardiovascular disease, has the molecular formula C16H16ClNO2S. Determine (a) the molar mass of Plavix, and (b) the mass percent of each element present in Plavix. Determining Formulas. Empirical Formula: Molecular Formula: 20. Testosterone, a steroid hormone, is the principle male sex hormone as well as an anabolic steroid. It is composed of 79.12% C, 9.79% H, and the remainder oxygen. Determine the molecular formula of testosterone given its molar mass MM = 288.42 g/mol. 21. Menthol, an organic compound obtained from peppermint and other mint oils, is made up of carbon, hydrogen, and oxygen. Combustion of 7.0 g of menthol yields 19.7 g of CO2 and 8.07 g of H2O. Determine the empirical and molecular formula of menthol. (The molecular weight of menthol is 156.27 g/mol). Stoichiometry, Limiting Reactant, and Percent Yield 1. The numerical relationships between chemical amounts in a balanced chemical equation are called stoichiometry reaction _______________________. Stoichiometry can be used to predict the amounts of reactant required to form a certain amount of product in a chemical reaction and vice-versa. Conversions often involve measurements in amount of substance (moles); mass (grams, kilograms, etc.); or both. Let’s try a simple example. The following balanced reaction represents the process of photosynthesis. 6CO2 + 6H2O + light energy ! C6H12O6 + 6O2 How many moles of carbon dioxide would be required to produce 94.1g of glucose? 2. The limiting reactant (or limiting reagent) limits the amount of product in a chemical equation. If one of the reactants is limiting, then there must also be another reactant in _____________. excess Here’s an example: The Haber process is used to synthesize ammonia on an industrial scale. The final step proceeds by the following balanced equation: If you have 48.00g nitrogen gas and 5.20 moles of hydrogen gas, what is the limiting reactant? 3. Percent Yield Percent yield is calculated via the following equation: 6 David, Josh, & Jessica Percent yield = CHEM105A SI Exam Review #1 September 12, 2010 Drs. Parr & Bertolini actual yield / theoretical yield x 100% Comprehensive Problem 22. You will see how this applies in the following A. Write the unbalanced chemical equation for the combustion of the hydrocarbon octane, which has the molecular formula C8H18. 2 C8H18 + 25 O2 -> 18 H2O + 16 CO2 B. Balance the above reaction. C. If you have 10.5 grams of octane, how many moles of oxygen gas do you theoretically need to react completely with the octane? (10.5g)(mol/42g)(16 mol O2/ 2 mol octane) = 2 mol O2 D. If you have 4.2 moles of C8H18 and 51 moles of gaseous oxygen, what is the LR? How many grams of the excess reactant remain? (4.2 mol octane)(25 mol O2/ 2 mol octane) = 52.5 mol O2 required = LR (4.2 mol)(42g/mol) - (51 mol O2)(2 mol octane/ 25 mol O2)(42g/mol) = 5.04g octane left E. Assume that the reaction in (D) goes to completion. If the percent yield is 63.7%, what mass of water vapor is (51 mol O2)(18 mol H2O/ 25 mol O2)(18g/mol)(.637) = 420g of H2O formed, in micrograms? = 4.2 x 10^8 micrograms of H2O 7 ...
View Full Document

This note was uploaded on 12/12/2010 for the course CHEM 105A taught by Professor Bau during the Fall '04 term at USC.

Ask a homework question - tutors are online