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UND 362 Math
| Question | Answer |
|---|---|
| what is a solvent and and solute | solvent is the substance on which something will be dissolved in, the solute is the substance being dissolved. |
| define saturated | max amount of solute in a solution (depending on time, temp, and pressure) |
| define super saturated | more solute than normal (excess lost easily) |
| define concentration | amount of one substance relative to another |
| percent solution definition and example | defined as parts per 100 30% is = to 30 parts in a total of 100 or 30/100 or 30:100 (3 parts solute and 7 parts solvent) |
| give 3 type sof solutions | weight/weight, weight/volume, volume/volume |
| weight/weight define | gram of solute to 100 g of solution EX: make 10% NaCl aqueous soln ANS: 10g NaCl in 90g h20 |
| Weight/volume define | grams solute/100 ml solution EX: calc % W/V of 250MG Nacl in 100ml H20 ANS: .25g NaCl/100 or .25% *convert mg to g and multiply by 100 for %* |
| volume/volume define | Ml of solute per 100ml solution EX: Ml of ethanoh needed to make 75% v/v solution using h20 ANS:100ml - 75ml ethanol = 25% water |
| what is the proper way to measure volume | use graduated cylinder, pipette, flas close to the volume you are using. (read at bottom of meniscus) |
| how do you prepare a (no change in concentration value) | *variation in the total amount solution equation (%needed:100=g/ml:volume desired) multiply outer two and inner two then solve for x EX: 300ml of 10% hcl ANS: 10:100 = X:300 |
| Dilutions formula what are the 2 requirements | V1 x C1 = V2 x C2 concentration must change, and unit of measure must both be the same |
| dilution example give answer how much of 100% alcohol is required to make 500ml of 70% alcohol | v1 x c1 = v2 x c2 100x = 70 x 500 100x = 35000 = 350 |
| what is gravimetric calculations | calculation that compensates for change in dye lots (consistency) Def: ration of 2 dye contents with corrects the weight of stain used |
| what is the formula for gravimetric | conc. of present dye /by conc. of new dye |
| solve this graf. equation dye stains correctly at 1g per 250 ml at 78% but new dye is 68% how much of new dye will stain same as old dye | GF - 78/68 = 1.15 |
| solve this 2 part equation: prepare 160 ml of 10% nacl. The present dye has dye content of 90% and new dye content of 70% | %needed:100 =g/ml:volume desired part 1. 10%:100 = g/ml:160 100x = 1600 therefore X = 16g present dye part 2. 90/70 - 1.29 grav. factor ans 1.29 x 16g = 20.64g of new dye |
| what is a hydrate | addition of water molecule to a molecule of salt |
| hydrate problem give solution make 17% soln of CuSO4*H20 (use periodic tbl) | 1. determine molecular weight cuso4 = 160 and cuso4*h20 is 178 equation is anyhdrous/monohydrate = 17 anhydrous/X 160x = 3026 x = 18.91 therefore diluting 18.91g monohydrate to 100ml = 17% anhydrous CuSO4 |
| give the equation for molarity | 1M = 1g Mol. weight/liter *must be in liter solution* |
| solve this molarity problem 1M solution of NaCl = | molecular weight is 58.5 therefore 1M is 58.5 nacl in 1L solution |
| solve this Molarity problem calculate #grams to produce 1.5L of 1M HNO3 HNO3 = 63g MW | 1 x 1.5 x 63 = 94.5 |
| what is the equation for determining weight of substance needed for a molar solution | weight in grams = Molarity x liter volume x molecular weight |
| solve this molarity problem calc grams for .1L of 1.85M H2s04 (MW =98) | 1.85 x .1 x 98 = 18.13g |
| solve this molarity problem make 300mL of 6M NaCl (58.5 MW) | g = 6 x .3 x 58.5 = |
| how is normality calculated (equation) | grams = normality x volume x MW / valence or dissociable h2 ions |
| solve this normality equation how much solute is need to prepare 1000ml of 1N K2CO3 soln (what about Al(OH)3 | 1N will divide by the K in periodic table (x2) 138.21 / 2 = 69.11 77.98 /3 dissociable H ions = 25.99 |
| when solving for normality the valance can be determined by | the first element of the molecule (using upper most rows of period table (ie Al = 3) |
| 1cc is the same as how many ml | 1 |
| kilogram to gram decigram to centigram to milligram to microgram to | 1000, 1/10, 1/100, 1/1000, 1/1000000 |
| give conversion of F to C solve ex. 150F to C | C = F - 32 x .555 C = 150 - 32 x .5555555555555 ans 65.56 ish |
| give conversion of C to F solve ex. 25C to F | F = 25 x 1.8 + 32 ans 77 |
Created by:
mustangvxd
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