Unit 10: Solutions-Key Regents Chemistry ’14 Mr. Murdoch .
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochUnit 10:SolutionsStudent Name: KeyClass Period: 3, 5, & 10Page 1 of 61Website uploadKey
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Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochUnit 10 Vocabulary:1. Aqueous: A solution in which the solvent is water.2. Colligative Property: A property of a solution that is dependent onconcentration. Examples include boiling point, freezing point, andvapor pressure.3. Molality: The concentration of a solution measured in moles of soluteper kilogram of solvent.4. Molarity: The concentration of a solution measured in moles of soluteper liter of solution.5. Parts per Million: The concentration of a solution measured in massof solute per mass of solution multiplied by one million.6. Percent by Mass: The concentration of a solution measured in mass ofsolute per mass of solution multiplied by one hundred.7. Percent by Volume: The concentration of a solution measured involume of solute per volume of solution multiplied by one hundred.8. Saturated: Any solution that has the maximum concentration of adissolved solute possible in a given quantity of solvent at a giventemperature. A saturated solution is a solution at equilibrium.9. Solubility: The maximum quantity of a solute that may be dissolvedin a given quantity of solvent at a given temperature to make asaturated solution.10. Solute: Any substance that is broken apart by a solvent and keptseparate by the solvent particles.11. Solution: A homogenous mixture formed when a solute dissolvesinto a solvent.12. Solvent: Any substance that attaches to solute particles, breaks thesolute apart, and then keeps the separated particles apart in solution.Page 3 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. Murdoch13. Supersaturated: A solution that has an excess of solute beyond thesolubility point for a given temperature. Excess solute will eitherprecipitate out, or remain in an unstable dissolved state until thesupersaturated solution is disturbed. This causes the excess solute toprecipitate from solution, leaving the solution saturated.14. Unsaturated: A solution in which there are solvent particles thathave no attached solute particles, and therefore has the capacity totake more solute into solution.Page 4 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochUnit 10 Homework Assignments:Assignment:Date:Due:Page 5 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochNotes page:Page 6 of 61Website uploadKey
Unit 10: Solutions-KeyTopic:Regents Chemistry ’14-‘15Mr. MurdochMixturesObjective: What is the result of combining two dissimilar substances?Mixtures: A mixture is made by physically combining two (or more)substances without any type of chemical reaction occurring. Mixing ionic compounds with water forms aqueous solutionscomposed of dissolved ions. The polar water molecules attach to theions, separating them from other ions. The polar water moleculeskeep the ions separate, holding the ions apart. This property is calledmolecule-ion attraction.Page 7 of 61Website uploadKey
Unit 10: Solutions-KeyTopic:Regents Chemistry ’14-‘15Mr. MurdochMolecule-Ion AttractionObjective: How are ions situated within a solution? A solid crystal of NaCl is placed into a beaker of water. Immediatelythe polar water molecules attract the charged ions. The partiallypositive ends of the water molecules (hydrogen atoms) attract thenegatively charged chloride ions. The partially negative ends of thewater molecules (oxygen atoms) attract the positively charged sodiumions. When enough water molecules attach to an ion, the naturalkinetic motion of the liquid water molecules will “tear” the ionsfrom each other. This is the molecule-ion attraction.Page 8 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. Murdoch The hydrated (aqueous) ions are kept separated by the motion of thewater molecules that are attracted to them. In this example, fourwater molecules around each ion are enough to keep the ionsseparated. There are still “free” water molecules in the solutionwhich could “tear” apart, attract, and keep separate additional soluteions. This is an example of an unsaturated solution. If more NaCl(s)were added until all water molecules were attached to ions, thesolution would then become saturated. At the saturation point, anyadditional NaCl(s) added would simply sink to bottom of the beakerwithout going into solution, or drive other Na 1 or Cl-1 ions out ofsolution as a precipitate. The number of water molecules required tokeep ions apart depends on water temperature. The higher thetemperature, the greater the average kinetic energy, and the faster thewater molecules move, so fewer water molecules would be requiredto keep the ions apart.Page 9 of 61Website uploadKey
Unit 10: Solutions-KeyTopic:Regents Chemistry ’14-‘15Mr. MurdochSolubilityObjective: How may we know the amount that enters into solution?Solubility: The quantity of a solute that may be added to a given quantity ofsolvent at a given temperature and pressure is known as solubility. Saturated: A saturated solution holds as many dissolved particles asit has the capacity to hold. Capacity is the maximum “spaces”available for anything. The chemistry room has 23 desks; this givesRoom 218 the capacity for 23 students unless more desks are broughtin, or if desks are taken out. In a saturated aqueous solution, all thesolvent water molecules (desks) have an ion (student butt) attached tothem, so no more ions (students) could come into the solution (Room218) and find a place to stay (empty desk). Any additional ions(students) would need to pass through the solution (room). Unsaturated: An unsaturated solution holds fewer dissolvedparticles as it has the capacity to hold. Room 218 has 23 desks, andthe largest chemistry class (for 2014-2015) has 21 students, so there iscapacity for two more students. In an unsaturated aqueous solution,there are solvent water molecules (desks) without an ion (studentbutt) attached to them, so more ions (students) could come into thesolution (Room 218) and find a place to stay (empty desk).Page 10 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. Murdoch Supersaturated: A supersaturated solution is very rare solution inwhere the solution holds more dissolved solute than is theoreticallypossible. A supersaturated solution is an unstable solution that willcause the excess solute to leave (precipitate) if disturbed. Thesolution (Room 218) has 23 desks (water molecules) that each canattach to one ion (student butt). If there are more than 23 students(ions) sitting, whereas some desks have more than one student (ion)sitting in them, this is over capacity, or supersaturated. The solutioncan easily be upset if a disturbance (Murdoch) comes in contact withthe system, and forces the extra ions (students) out of the solution(Room 218).Factors affecting solubility: Three factors affect solubility:1. Temperature2. Pressure3. Nature of solute and solventWatch Crash Course Chemistry Solutions - YouTube - 8:20Page 11 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochTopic: Temperature and SolubilityObjective: How does temperature affect the solubility of a solution?Temperature:1. For solid and liquid solutes, solubility in water increases as watertemperature increases.2. For gaseous solutes, solubility in water decreases as watertemperature increases.Effect of temperature on solubility of a solid solute in water: An example of the effect of temperature on aqueous solubility ofsolids is the sugar water needed to form rock candy. A saturatedsolution is formed at high temperature. As the saturated sugarsolution cools, the sugar solute becomes less soluble, causing somewater molecules to “jump” off a sugar molecule and assist other watermolecules in holding other sugar molecules apart. This allows thenow “free” sugar molecules to come out of solution, and as morewater molecules are pulled off sugar molecules, more and more sugarwill precipitate from the solution as sugar rock candy.Page 12 of 61Website uploadKey
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Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochEffect of temperature on solubility of a gaseous solute in water: An example of the effect of temperature on aqueous solubility ofgases is as easy to find as your favorite carbonated beverage. Taketwo identical bottles of the same carbonated beverage, place one inthe refrigerator overnight, and leave the other on the kitchen counter.When you open both bottles side by side you will see that the colderbeverage will release much less carbonation (bubbles) than will thewarmer beverage. The colder solution allowed more gaseous CO2 toremain dissolved than did the warmer solution. If you keep observingboth bottles, as the colder beverage warms, it will continue to losedissolved CO2 (bubbles) during the warming process, while thewarmer bottle will form fewer bubbles during the same elapsed time.Page 14 of 61Website uploadKey
Unit 10: Solutions-KeyTopic:Regents Chemistry ’14-‘15Mr. MurdochPressure and SolubilityObjective: How does pressure affect solubility of solutes?Pressure: For solids and liquids, pressure has either no effect or a negligibleeffect on solubility. For gaseous solutes, solubility increases as pressure increases.i. In commercial soda machines, as water travels through a tube,flavored syrup is added to the water along with carbon dioxide gas.Gases don’t have an affinity to form aqueous solutions and requirepressure to force the CO2 gas into the water and syrup solution.When the aqueous syrup and CO2 mixture is bottled, the gas is“trapped” within the volume of the container, and equilibriumbetween dissolved CO2 molecules and free CO2 molecules quicklyforms. When the container is opened, the pressure decreases,allowing the CO2 to escape as bubbles. CO2 gas is soluble at highpressures (sealed container) but nearly insoluble at low pressure(open container).Page 15 of 61Website uploadKey
Unit 10: Solutions-KeyTopic:Regents Chemistry ’14-‘15Mr. MurdochPolar Nature of SolventObjective: Does the polarity of a solute or solute apply in solutions?Nature of Solute and Solvent (Like dissolves Like): Polar solutes dissolve in polar solvents.o Water is a polar molecule and therefore has partially charged endsthat can attract other polar or ionic solutes. This is why ionicsolutes such as NaCl and polar molecular solutes such as glucose(sugar) may be dissolved in water.o Water’s polar structure has little attraction for nonpolar molecularsolutes like oil (remember Hank squishing butter?) Oils will notmix (are immiscible) with water, and being less dense, will float ontop of the water. This is why water is NOT used to extinguish oilfires; the water will go under the fire, and float the fire to a newlocation. Nonpolar gases such as CO2 and O2 are even harder toforce into aqueous solution because of this. Nonpolar solutes dissolve in nonpolar solvents.o Oils will not dissolve in polar water solvent, but oils will easilydissolve in the nonpolar organic solvent benzene (C6H6). Whereaswater is a great polar solvent, benzene is a good nonpolar solvent.Acetone is a nonpolar solvent used to remove nonpolar mixturessuch as nail polish.Page 16 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochPage 17 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochSolubility Regents Practice Problems (ungraded):1. At room temperature, the solubility of which solute listed below would the mostaffected by a change in pressure?a) Sugar(s)c) Sodium nitrate(aq)b) Methanol(l)d) Carbon dioxide(g)2. As the pressure on a gas confined above a liquid increases, the solubility of thegas in the liquida) Increasesb) Decreasesc) Is unchanged3. Sublimated carbon dioxide is most soluble in water under the conditions ofa) Low pressure and low temperatureb) High pressure and low temperaturec) Low pressure and high temperatured) High pressure and high temperature4. At which temperature given below could water contain the most dissolved gasat a pressure of 101.3 kPa?a) 183 Cb) 283 Kc) 383 Kd) 483 CPage 18 of 61Website uploadKey
Unit 10: Solutions-KeyTopic:Regents Chemistry ’14-‘15Mr. MurdochSolubility CurvesObjective: How may we use solubility curves to determine solubility?Reference Table G: Solubility Curves at Standard Pressure:1. For Table G, the SLOPE of the curve tells you about the solute!a. IONIC salts solutes have UPWARDS slopes;b. GAS solutes (SO2 and NH3) have DOWNWARDS slopes.2. Interpreting Reference Table G:a. The Table G x-axis is the temperature in C. If you are given atemperature in K, you’ll need to convert (K - 273 C). Note thatthe numbers are significant to TWO places, as all numbers have adecimal (.) at the end. Note also that the interval is to the tensplace (10. C), so precision may be interpolated to the ones placeand will still be significant to TWO digits.b. The Table G y-axis is the solubility in grams of solute per 100.grams of water. Again, the numbers are significant to TWO places,as all numbers have a decimal (.) at the end. Note also that theinterval is to the tens place (10. grams), so precision may beinterpolated to the ones place and will still be significant to TWOdigits.c. Also note the segment of the curve for potassium iodide WAY upin the extreme upper left corner. Students have missed the KIcurve; be aware of its presence BEFORE you need it!Page 19 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. Murdoch3. Table G is only useable at standard pressure (1 atm or 101.3 kPa); ifyou have other than standard pressure, you’ll need to infer.Solubility of a solute in 100. grams of water at various temperatures:1. Each line on Table G represents a saturated aqueous solution of thegiven solute for a given temperature. The farther up the y-axis thesaturation line is at a given temperature, the more soluble the soluteis. To find the saturation point at a given temperature, simply find thegiven temperature and then go straight vertically up until youintersect the solubility line you need, and then traverse level left andread the solubility on the y-axis. The table is set for grams ofsolute/100. grams of H2O; if you have other than 100. g of water youwill need to find the correct proportion.Page 20 of 61Website uploadKey
Unit 10: Solutions-KeyTopic:Regents Chemistry ’14-‘15Mr. MurdochDegree of SaturationObjective: Howdo we usesolubility curves to determine saturation?DeterminingMolecularPolarity:Degree of saturation according to Table G:1. Unsaturated:2. Saturated:Page 21 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. Murdoch3. Supersaturated:Page 22 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochTopic: Solubility in other than 100. gObjective: Will Table G determine solubility not in 100. g of water?Solubility for other than 100. g of water solvent: Reference Table G may only be directly used in the mass of thesolute is dissolved in 100. grams of water as a solvent. With simplemath, by finding the proportionality of the given amount of solvent to100. g of water as listed on Table G, we can easily calculate thesolubility at the given mass of water. If given a question with OTHER than 100. grams of water:1. For the stated temperature in the question, find the saturationsolubility in grams of solute per 100. grams of water.2. Place the question’s stated amount of water as a numerator and100. g of water as the denominator. This will give you amultiplier.i. If you had MORE than 100. grams of water in your question,the multiplier factor will be greater than 1.ii. If you had LESS than 100. grams of water in your question, themultiplier factor will be less than 1.3. Multiply the saturation solubility from Table G by the multiplierfactor from Step #2, and you will have your saturation solubilityfor other than 100. grams of water.Page 23 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochI. How many grams of KClO3 are required to make a saturatedaqueous solution in 100. grams of water at 30. C?a. Find 30. C on the Table G x-axis, then move straight up to theKClO3 curve. Move straight over to the y-axis, and read the scale(12 g). This means that 12 grams of KClO3 are soluble in 100.grams of water at 30. C.Page 24 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochII. How many grams of KClO3 are required to make a saturatedaqueous solution in 50. grams of water at 30. C?a. We are now asked for the solubility of KClO3 in 50. grams ofwater. Place 50. g of water in the numerator, and 100. g of waterin the denominator, and solve.50. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟100. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 0.50 multiplierb. We used Table G to determine that 12 g of KClO3 are soluble in100. g of water. We can use the multiplier factor with the 12 g ofKClO3 and determine the mass of KClO3 soluble in 50. g ofwater.c. 12 g of KClO3 x 0.50 6.0 g of KClO3 soluble in 50. g of water at30. CIII. How many grams of KClO3 are required to make a saturatedaqueous solution in 200. grams of water at 30. C?a. We are now asked for the solubility of KClO3 in 200. grams ofwater. Place 200. g of water in the numerator, and 100. g of waterin the denominator, and solve.200. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟100. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 2.00 multiplierb. We used Table G to determine that 12 g of KClO3 are soluble in100. g of water. We can use the multiplier factor with the 12 g ofKClO3 and determine the mass of KClO3 soluble in 200. g of water.c. 12 g of KClO3 x 2.00 24 g of KClO3 soluble in 200. g of water at30. CPage 25 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochIV. Let’s say you have to make a saturated NaCl solution using tapwater (10 C) but you only have an empty (355 mL) soda can. Ofcourse you know that water has a density of very close to 1.0gram/mL near 10 C, so you can use the 355 mL volume of yoursoda can to measure out 355 grams of water. How much NaCl willbe needed to form the saturated solution with the 10 C water?a. First, start at 10 C on Table G and go up until you find thesolubility curve for NaCl (blue dashed line), and then move leftuntil you reach the y-axis scale (green dotted line). I’ll call it 37.However, that 37 g of NaCl is for 100 grams of water! Since youhave 355 grams of water, 355 g needs to go in the numerator, withthe Table G standard of 100. grams of water as the denominator.355 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟100. 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 3.55 multiplierb. We can now take the interpolated 37 grams of NaCl per 100.grams of water and multiply that by 3.55 to give us the saturationpoint of NaCl in 355 grams of water of 131.55 or 130 g NaCl.37gPage 26 of 61Website uploadKey
Unit 10: Solutions-KeyRegents Chemistry ’14-‘15Mr. MurdochTable G example questions:1. What is the solubility of HCl in 100. g of water at 70 C?a. Find 70 C on the x-axis, move straight up to the HCl curve, andthen transit level left to the y-axis.b. 52 g of HCl are soluble in 100. g of water at 70 C2. What type of solution saturation do you have if 20. g of KClO3 aredissolved in 100. g of water at 40. C?a. Find 40. C on the x-axis, move straight up to the KClO3 curve, andthen transit level left to the y-axis.b. 16 g of KClO3 are soluble in 100. g of water at 40. C. As youwere asked what 20. g of KClO3 in solution would be, since 20 isgreater than 16, the solution would be supersaturated.3. For the question above, how can we make the existing solutionsaturated?a. As calculated using Table G, at 40 C only 16 g of KClO3 could bedissolved to make th
ions. This is an example of an unsaturated solution. If more NaCl (s) were added until all water molecules were attached to ions, the solution would then become saturated. At the saturation point, any additional NaCl (s) added would simply sink to bottom of the beaker without going into solution, or drive other Na 1 or Cl-1 ions out of
Mathematics 36 Science 50 World Languages 67 Art 76 Music & Theatre Arts 86 Business Education 96 . Course Selection Sheet 146 . 3 REQUIREMENTS FOR HIGH SCHOOL GRADUATION Regents Diploma Advanced Regents Diploma Test Requirements Test Requirements English Regents English Regents Algebra Regents 3 - Math Regents Global History Regents Global .
Pass nine required Regents exams with scores of 65 or higher: English Regents exam Global History Regents exam U.S. History Regents exam Three math Regents exams: Integrated Algebra, and Geometry, and Algebra 2/Trigonometry Two science Regents exams: Living Environment, and any one of the following: Earth Science
Pass nine required Regents exams with scores of 65 or higher: English Regents exam Global History Regents exam U.S. History Regents exam Three math Regents exams: Integrated Algebra, and Geometry, and Algebra 2/Trigonometry Two science Regents exams: Living Environment, and any one of the following: Earth Science
Mathematics Regents Exams. Each student taking the Regents Examinations in Integrated Algebra, Geometry and Algebra 2/Trigonometry must have a graphing calculator without symbol manipulation. Science Regents Exams. For the Regents Examination in Living Environment, all students who wish to use a four-function or scientific calculator must have one.
2/Trigonometry. These new Regents examinations in mathematics will replace the Regents Examinations in Mathematics A and Mathematics B. To fulfill the mathematics Regents examination requirement for graduation, students must pass any one of these new commencement-level Regents examinations. The
REGENTS EXAM IN GLOBAL HISTORY AND GEOGRAPHY II (GRADE 10) REGENTS EXAM IN GLOBAL HISTORY AND GEOGRAPHY II (GRADE 10) The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION REGENTS EXAM IN GLOBAL HISTORY AND GEOGRAPHY II (GRADE 10) Tuesday, August 13, 2019 — 12:30 to 3:30 p.m., only
The general procedures to be followed in administering Regents examinations are provided in the publications Directions for Administering Regents Examinations (DET 541), and Regents Examinations, Regents Competency Tests, and Proficiency Examinations: School Administrator’s Manual, 2001 Edition.
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