Unit Conversions, Important Constants and Relationships

Transcription

1 NOTE: Exact quantities are specified as exact. Consider 1 as exact! mass (M) 1 kg = lb m = oz 1 lb m = 16 oz (exact)= g length (L) 1 m = (exact) angstroms (Å) = 100 cm = in = ft = yd = mi 1 ft = 12 (exact) in = 1/3 (exact) yd = m mmhg volume (L 3 ) 1m 3 = 1000 L = 10 6 ml = ft 3 = gal = qt kwh 1 L = 1000 cm 3 = 1000 ml (memorize this line) Btu 1 ft 3 = 1728 (exact) in 3 = gal = m 3 ev = L = cm 3 force (mass x acceleration) 1 N = 1 kg.m/s 2 = 10 5 dyn = 10 5 g.cm/s 2 = lb f 1 lb f = lb m.ft/s 2 = N = x10 5 dynes pressure (Force/area) 1 atm = x10 5 N/m 2 (Pa) = bar = 760 (exact) mmhg (torr) at 0 C = m H 2O at 4 C = lb f/in 2 (psi) = ft H 2O at 4 C = in Hg at 0 C energy (Force x distance) 1 J = 1 N.m = cal = 2.778x10-7 kw.hr = ft.lb f = 9.486x10-4 Btu 1000 cal = 1 Cal (food) = 4184 J 1 ev = x10-19 J 1 V = 1 J/C 1 L.atm = J power (Energy/time) 1 W = 1 J/s = cal/s = ft.lb f/s = 9.486x10-4 Btu/s = 1.341x10-3 hp Memorize these metric system prefixes (Abbreviations are case sensitive!) Prefix abbreviation value pico p nano n 10-9 micro μ 10-6 milli m 10-3 centi c 10-2 deci d 10-1 deka da 10 1 hecto h 10 2 kilo k 10 3 mega M 10 6 giga G 10 9 tera T Memorize these unit abbreviations (and their quantity of measure) atm atmosphere (pressure) Pa Pascal (pressure) torr torr (pressure) millimeters mercury (pressure) J joule (energy) cal calorie (energy) kilowatt-hour (energy) British thermal units (energy) electron volt (energy) Cal food calorie (energy) L liter (volume) gal gallon (volume) qt quart (volume) fl-oz fluid-ounce (volume) yd yard (length) m meter (length) mi mile (length) in inche (length) ft feet (length) oz ounce (mass) lb m (lb) pound (mass) g gram (mass) hr, min, s hour, min, s (time) K Kelvin (temperature) C celsius (temperature) F farenheit (temperature) mol mole (count) dyn dyne (force) N Newton (force) lb f pound (force) V volt (electric potential) C coulomb (electric charge) hp horse power (power) W watt (power) Useful constants (provided for tests) g = m/s 2 gravitational acceleration c = x10 8 m/s speed of light in vacuum R= L.atm/(mol.K) gas constant R= J/(mol.K) gas constant e = x10-19 C electron charge m e = x kg electron rest mass h = x10-34 J.s Planck s constant F = C/mol charge Faraday s Constant N = 6.022x10 23 count/mol Avogadro s number Page 1 of 6 π = the number Pi 2008, K. Golestaneh (revision 2017)

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3 Memorize the following shapes, definitions and formulas P: Perimeter (L) A: Area (L 2 ) V: volume (L 3 ) Shape Key Perimeter Area (surface) Volume dimensions Circle diameter, d P = π.d or P = A = π.r 2 or A = π.d 2 /4 --- radius, r 2.π.r Rectangle length, l P = 2.l + 2.w A = l. w --- width, w Square side, a P = 4.a A = a Triangle Sphere Rectangular Box (parallelepiped) Cylinder π = base, b height, h diameter, d radius, r length, l width, w height, h See circle & Height, h P = Add 3 sides A = ½ (b. h) A = 4. π. r 2 V = 4/3. π. r A = area of all 6 sides V = l. w. h --- A = 2 x circle areas + π.d.h (unfolded rectangle) V = circle area x height Memorize the Pythagorean Theorem applied to a right triangle (a 2 + b 2 = c 2 ). Memorize the following trigonometric relationships: sin θ = a/c ; cos θ = b/c ; tan θ = a/b Sum of all angles in triangle = 180 degrees example: In a right triangle where; a=3 cm and b=5 cm, find: c=? cm sin θ =? θ =? a b c θ Note: for any shape with a uniform cross sectional area remember the following rule: Volume = Cross sectional area x height (example: cylinder, prism, etc.) Some shapes shown below in 3-dimensional space, with uniform cross-sectional area and a height (or depth). A rectangular box A prism A cylinder (a pipe) Page 3 of , K. Golestaneh (revision 2017)

4 Work on these problems 1) Show the hidden lines (as shown above for the prism) and hatch the uniform cross-sectional area approximately half-way between the faces. Mark the height (or depth) as h on each diagram. 2) For each shape, draw the top view and name each shape. This is also the top view of the crosssection Scientists often monitor and calculate change of a property over time. Change over time is referred to as rate of change of something. In general terms: Rate = Quantity Changing Time Quantity changing is shown with (delta). For example V = Vf Vi (f: final, i: initial) For liquids and gases that flow (called fluids) rate of flow may be expressed as: Volumetric Flow Rate = Change in Volume Time Example units: ml/min, gal/min Another example is mass flow rate: Mass Flow Rate = Example units: kg/s, lb/min Change in Mass Time In a practical sense it is possible to measure these changes. For example liquid flow rate can be measured this way. Put a container at the end of a pipe and start a timer as the bucket fills with the liquid. Stop the timer and measure the volume of liquid collected in the bucket. Apply the volumetric flow rate equation and the rate measured is the average flow rate between the measured time interval. Chemists, Biologists, Physicians Nurses and Nutritionists often deal with concentration of a pure compound in a homogenous mixture. What is concentration? It is commonly defined for a particular solute: Concentration (solute A) = Solute A Quantity Solution (mixture) Quantity Review the definition and memorize formulas of the following concentration units: Page 4 of , K. Golestaneh (revision 2017)

5 Molarity (M) is very commonly used in chemistry especially when we deal with chemical reactions. (sln is the abbreviation of solution) Solute Molarity =,, or M = Parts per million (ppm) is used when expressing very low (yet important!) level of solute concentrations using volume or mass units (units of mass or volume must be matching!). ppm (by mass) is defined as mass units of the solute in 10 6 mass units of the solution: ppm (solute) = g solute 10 g sln Problem: If a solution contains mg of mercury (Hg) in a total solution mass of 75 kg, calculate ppm of Hg. (don t forget to keep the mass units the same!). Show your work! When we deal with very dilute solute in a water solution, we can safely ignore the mass of solute in the denominator and just use the density of water value of 1.00 g/ml and express a new and convenient (from lab point of view) unit for ppm by converting the mass of solution to the volume (liter) of solution. We measure volumes in lab very commonly in ml and L units. Therefore the new unit that is equivalent to ppm (by mass) is (memorize these relationships): ppm (solute) = g solute mg solute 10 = = g sln L sln μg solute ml sln Problem: Verify the above relationship by converting 25.7 ppm Na (by mass) to mg/l concentration unit. Show your work! The following are other commonly used concentration units which you should have been familiar with from your introductory chemistry. mass % (solute) = volume % (solute) = ( ) x 100 ( ) ( ) x 100 ( ) Don t forget to use matching units of mass or matching units of volume or mass when using the above relationships. The 100 multiplier is exact! Page 5 of , K. Golestaneh (revision 2017)

6 When working with dimensional analysis, you must translate compound units to simpler units and descriptors correctly! For example: If a problem states 12.5 % NaCl solution (by mass): It should be translated as: 12.5 g NaCl 100 g sln 7.2 ppm (by mass) lead (Pb) translates as: (Use whichever that is more useful to solve the problem!): Problems: 7.2 g Pb 7.2 mg Pb 10 = g sln L sln 1) A solution should be prepared as 3.56 M calcium chloride, CaCl2. Calculate mass (g) of the solute that is needed in order to prepare 3500 ml of the solution. 2) A solution is prepared by dissolving 65 g of alcohol (density=0.78 g/ml) in 2.35 L of water. Calculate the solute concentration in percent (by volume) units. 3) If seawater contains gold at 1.75 ppb (parts per billion or 1 part in 10 9 parts) by mass. What volume (gal) of seawater must be processed in a factory to obtain $55,700 value in pure gold assuming pure gold is worth $1320 per oz. Assume that seawater has a density of 1.10 g/ml. Page 6 of , K. Golestaneh (revision 2017)