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# Chemistry [QEE-R 2012]

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• 1. Department of EngineeringLyceum of the Philippines UniversityCavite Campus
• 2.  Matter: A Quick Review Dimensional Analysis and Measurement Modern Periodic Table and Electronic Structure of Atoms Mole: Its Concepts and Applications  Chemical Language: Reactions and Equations  Solutions and Concentration Units Gas Laws
• 3. monoatomic Element polyatomic (molecule) Pure Substance Compound ionicMatter molecule Homogeneous (i.e. solutions) Impure Substance (Mixture) Heterogeneous (i.e. colloids and suspensions)
• 4.  approach used in calculations problem solving provides a systematic way of  solving numerical correct use of problems in science conversion factor/s and other disciplines to change one unit as well as checking to another numerical solutions for possible error/s units must be carried through all
• 5. Given: Qty1 Find: Qty2Solution: Given ConversionUnkown = × quantity factor/s QtyQty = (Qty )× 2 2 1 Qty 1Qty 2 = Qty 2
• 6. Given: Qty1 Find: Qty2Solution: Given ConversionUnkown = × quantity factor/s QtyQty = (Qty )× 1 2 1 Qty 2 (Qty ) 2Qty ≠ 1 2 Qty 2
• 7. A. Base Quantities and Units Quantity SI Metric English m, km, cm, length m mm, m, mi, ft, yd, in nm kg, Tg, mg, mass kg cg, g, ng slug, oz s, hr, min, s, ms, hr, time s min day, wk, mo, yr
• 8. A. Base Quantities and Units Quantity SI Metric Englishtemperatur K C F e amount of mol - -substance electric A - - current luminosity cd - -
• 9. B. Some Derived Quantities and Units Quantity SI Metric English m2, km2, mi2, ft2, yd2, area m2 cm2, mm2 in2 km3, cm3, mi3, ft3, yd3, volume m3 mm3; mL in3; gal; qt m/min; ft/s, mi/hr, speed m/s km/hr; cm/s mi/min, in/s Mg/m3; oz/qt; density kg/m3 g/cm3; oz/cm3
• 10. C. Some Conversion Factors Please refer to QEE Review 2012 – Support Material, re: Table 2
• 11. An aluminum foil is found to be 8.0 10-5cm thick. What is it thickness inmicrometers? -2 -5 10 m 1 μm thickness μm = 8.0 × 10 cm × × -6 1 cm 10 m -1 thickness μm = 8.0 × 10 μm or 0.80 μm
• 12. A gas at 25 C exactly fills a container previouslyto have a volume of 1.05 103 cm3. The containerplus the gas are weighed and found to have a massof 837.6 g. The container when emptied of all gas,has a mass of 836.2 g. What is the density of thegas at 25 C? mass gasdensity gas = volume gas (837.6 g - 836.2 g ) gdensity gas = 3 3 density = 1.3 × 10 -3 1.05 × 10 cm gas cm 3
• 13.  most significant  Isotopic Notation tool in ORGANIZING 13Al and SYSTEMATIC 26.98 REMEM-BERING of chemical facts  Atomic No. = # of list the important p = # of e (neutral characteristics of atom) all elements in  Mass No. = p + n increasing atomic no.
• 14. Isotopic Atomic Mass Electronic p n eNotation No. No. structure13Al [10Ne] 3s2, 27 13 27 13 14 13 3p140Ar [10Ne] 3s2, 18 18 40 18 22 18 3p6 [2He] 2s2,13Al+3 13 27 13 14 10 2p6
• 15.  the mass, in grams, per mole of a substance (atom, ion, molecule) how to find: just add the atomic weights of that substance (compound or molecules) thus, in g/mol MM= amu (monoatomic element) MM= FW (ions) MM= MW (molecules)
• 16. Molar Mass atomic mass formula Molecular unit/atomic weight mass/weight mass/weight 1 mol Fe2O3 1 mol O2 weighs1 mol Fe weighs 55.85 weighs 159.70 g 32.00 gg MMFe2O3 = MMO2 = 32.00 MMFe = 55.85 g/mol 159.70 g/mol g/mol 1 mol NaCl weighs 1 mol H2O1 mol O weighs 16.00 g 58.44 g weighs 18.02 g MMO = 16.00 g/mol MMNaCl = 58.44 MMwater = g/mol 18.02 g/mol
• 17. use unitgrams use MM moles 6.022 x 1023 particles interconversion of mass of a substance, in grams, and the no. of particles of that substance (atoms, ions, or molecules) the no. of moles of substance is central to stoichiometry
• 18. Calculate the no. of moles of glucose,C6H12O6 (MM = 180.0 g/mol), in 5.380 g ofthis substance. 1 mol C 6 H 12 O 6 mol glucose = 5 .380 g C 6 H 12 O 6 × 180.0 g C 6 H 12 O 6 mol glucose = 0 .02989 mol C 6 H 12 O 6
• 19. How many copper atoms, Cu (MM = 63.5g/mol), are there in a traditional copper pennyweighing 3 g. Assume the penny to be 100%copper. 23 1 mol Cu 6.022 × 10 Cu atomsCu atoms = 3 g Cu × × 63.5 g Cu 1 mol Cu 22Cu atoms = 3 × 10 Cu atoms
• 20. Consider the combustion of butane, C4H10, thefuel in disposable cigarette lighters. C 4 H 10(l) + O 2(g) → CO 2(g) + H 2 O (l)2C 4 H 10(l) + 13O 2(g) → 8CO 2(g) + 10H 2 O (l) 2 mol 13 mol 8 mol 10 mol C4H10 O2 CO2 H2O
• 21. Consider the combustion of butane, C4H10, the fuelin disposable cigarette lighters. 2C 4 H 10(l) + 13O 2(g) → 8CO 2(g) + 10H 2 O (l)Determine the mass in grams of CO2 (MM = 44.01g/mol) formed when 1.34 mol of C4H10 (MM = 58.14g/mol) reacts. 8 CO 2 44 . 01 g CO 2 g CO 2 = 1 .34 mol C 4 H 10 × × 2 mol C 4 H 10 1 mol CO 2 g CO 2 = 2 34 g CO 2
• 22. Consider the combustion of butane, C4H10, the fuelin disposable cigarette lighters. 2C 4 H 10(l) + 13O 2(g) → 8CO 2(g) + 10H 2 O (l)Determine the mass in grams of C4H10 (MM = 58.14g/mol) required to react with 6.00 g of O2 (MM = 32.00g/mol). 1 mol O 2 2 mol C 4 H 10 5 8.14 g C 4 H 10 g C 4 H 10 = 6 .00 g O 2 × × × 32.00 g CO 2 13 mol O 2 1 mol C 4 H 10 g C 4 H 10 = 1 .68 g C 4 H 10
• 23. SOLUTE >> the substance to be dissolvedSOLVENT >> the dissolving mediumConcentration units: mol1. Molarity, M = solute V in L of solution mol solute2. Molality, m = mass in kg of solvent mass3. mass percent = x 100 solute solute total mass of solution
• 24. A solution used for intravenous feeding contains 4.80 gof glucose, C6H12O6 (MM = 180.16 g/mol) in 600.0 mLof solution. What is the molar concentration of glucose? mol glucose M glucose = V in L of solution 1 mol C 6 H 12 O 6 4.80 g C 6 H 12 O 6 × 180.16 g C 6 H 12 O 6 M glucose = 1 L sol n 6 00 . 0 mL sol n × 1000 mL sol n M glucose = 0 .0444 M or 0.0444 mol/L
• 25. A solution used for intravenous feeding contains 4.80 gof glucose, C6H12O6 (MM = 180.16 g/mol) in 90.0 g ofwater. What is the molal concentration of glucose? mol glucose m glucose = m in kg of water 1 mol C 6 H 12 O 6 4.80 g C 6 H 12 O 6 × 180.16 g C 6 H 12 O 6 m glucose = 1 kg water 90 . 0 g water × 1000 g water m glucose = 0 .296 m or 0.296 mol/kg
• 26. A solution used for intravenous feeding contains 4.80 gof glucose, C6H12O6 (MM = 180.16 g/mol) in 90.0 g ofwater. What is the molal concentration of glucose? mol glucose m glucose = m in kg of water 1 mol C 6 H 12 O 6 4.80 g C 6 H 12 O 6 × 180.16 g C 6 H 12 O 6 m glucose = 1 kg water 90 . 0 g water × 1000 g water m glucose = 0 .296 m or 0.296 mol/kg
• 27. In a solution prepared by dissolving 24 g of NaCl in152 g of water, determine the mass percent NaCl. g mass %( / g ) NaCl = x 100 NaCl mass of solution g 24 g %( / g ) NaCl = x 100 (24 g + 152 g) % (by mass) NaCl = 1 4%
• 28. A 1.13 molar solution of aqueous KOH (MM = 56.11 g/mol) has a density of 1.05 g/mL. calculate its molality or molal concentration. mol KOH m KOH = m in kg of water 1 . 13 mol KOHm KOH = 1000 mL sol n 1.05 g sol n 56.11 g KOH [1 L sol n × × ] - [1 . 13 mol KOH × ] 1 L sol n 1 mL sol n 1 mol KOH m KOH = 1 .14 m or 1.14 mol/kg