Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.
Upcoming SlideShare
×

# Chemistry [QEE-R 2012]

1,056 views

Published on

Published in: Technology, Education
• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• Be the first to comment

• Be the first to like this

### Chemistry [QEE-R 2012]

1. 1. Department of EngineeringLyceum of the Philippines UniversityCavite Campus
2. 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. 3. monoatomic Element polyatomic (molecule) Pure Substance Compound ionicMatter molecule Homogeneous (i.e. solutions) Impure Substance (Mixture) Heterogeneous (i.e. colloids and suspensions)
4. 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. 5. Given: Qty1 Find: Qty2Solution: Given ConversionUnkown = × quantity factor/s QtyQty = (Qty )× 2 2 1 Qty 1Qty 2 = Qty 2
6. 6. Given: Qty1 Find: Qty2Solution: Given ConversionUnkown = × quantity factor/s QtyQty = (Qty )× 1 2 1 Qty 2 (Qty ) 2Qty ≠ 1 2 Qty 2
7. 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. 8. A. Base Quantities and Units Quantity SI Metric Englishtemperatur K C F e amount of mol - -substance electric A - - current luminosity cd - -
9. 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. 10. C. Some Conversion Factors Please refer to QEE Review 2012 – Support Material, re: Table 2
11. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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