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EOC SOS. Chemistry Readiness TEKS and Performance Level Standards. 4A. The student knows the characteristics of matter and can analyze the relationships between chemical and physical changes and properties .
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EOC SOS Chemistry Readiness TEKS and Performance Level Standards
4A The student knows the characteristics of matter and can analyze the relationships between chemical and physical changes and properties. • The student is expected to differentiate between physical and chemical changes and properties
4A possibility The figures below were used to illustrate the arrangement of particles in three different states of matter. 1) Which physical state is illustrated by substance number 1? • A gas B liquid C plasma D solid #3 is a gas #2 is the liquid None of these are plasma Closely and regularly spaced
4A possibility (Level II: Describe chemical reactions qualitatively and quantitatively.) 2) Which of these is an indication that a chemical change occurs in the leaves of a growing plant? • A the green color of the leaves • B difference in temperature of leaves as air temperature changes • C shade in area beneath a tree with leaves • D increase in the amount of oxygen around the leaves Although we know the green color is due to the chemical reaction called photosynthesis, purely observing a color without the color changing is not an indication of a chemical change. The change in temperature in this case is just due to heat transfer between a solid (the leaf) and a gas (the air) and is a physical change. No bonds were broken or formed and no substances were formed. Shade is just the absence of light that was blocked by the leaf and is physical. No bonds were broken or formed and no new substances were formed. A new substance (oxygen) appears to have been formed and would be the best indication that a chemical reaction occurred.
4A possibility (Level II: Describe chemical reactions qualitatively and quantitatively.) 3) When two clear solutions are mixed in a chemistry lab, a yellow precipitate forms and settles to the bottom of the beaker. The total mass of material inside the beaker does not change when the solid precipitates. Which statement is a reasonable conclusion about the formation of the precipitate? • A It is a physical change because precipitates are a change of state. • B It is a physical change because the total mass does not change. • C It is a chemical change because a new material was formed in the beaker. • D It is a chemical change because physical changes do not produce solid products. Although it appears that a liquid turned into a solid (freezing is a physical change), that is not what happened. A solid appeared that never was there before. Two ions met and bonded. Total mass does not differ in either chemical or physical changes. Any time a new substance is formed a chemical change has occurred. It is not the state of the product that determines if it is a physical or chemical change. It is the chemical composition changing that determines if it is a chemical change.
4A posssibility 4) Why does the water in deep lakes stay liquid even when the air above them is very cold? • A There is no heat transfer between water and air. • B Substances dissolved in the water keep it from freezing. • C Heat from inside Earth keeps the water beneath the surface warm. • D Ice is less dense than liquid water so the top of the lake freezes and insulates the deeper water. This is not true. Heat energy will always move from an area of high heat to low heat unless there is an insulator present. Dissolved salts will lower the melting/freezing point but the water will still become colder and freeze if the temperature lowers enough. This does not explain the observed phenomena. In the case of hot springs (water that comes to the surface from deep below the ground), all of the water stays liquid and about the same temperature due to turbulence. This is not true in lakes if it gets cold enough. True for the reason stated.
4D The student knows the characteristics of matter and can analyze the relationships between chemical and physical changes and properties. • The student is expected to classify matter as pure substances or mixtures through investigation of their properties.
4D possibility (Level II: Describe aqueous solutions and distinguish between various reactions that occur in solutions.) 5) In a chemistry lab, a student carefully boiled a clear, colorless liquid in a beaker. After all of the liquid evaporated, the bottom of the beaker was covered by a layer of white crystals. What type of substance was the liquid? • A compound • B element • C heterogeneous mixture • D homogeneous mixture If the substance was one pure compound it would have boiled away to a gas and left nothing behind. If the substance was one pure element it would have boiled away to a gas and left nothing behind. Heterogeneous would not be the same throughout the solution. Since a salt was left after boiling away the liquid, this made the original substance a solution (a salt dissolved in water). All solutions are homogeneous mixtures.
5B The student understands the historical development of the Periodic Table and can apply its predictive power. • The student is expected to use the Periodic Table to identify and explain the properties of chemical families, including alkali metals, alkaline earth metals, halogens, noble gases, and transition metals .
5B possibility: (Level I: Use the periodic table to identify chemical families and determine charges on monoatomic ions.)Look at the periodic table segment below. 6) Which pair of elements has a similar arrangement of outer electrons? • A helium and fluorine • B carbon and nitrogen • C boron and aluminum • D chlorine and oxygen Helium (group 18) has a filled outer shell while fluorine (group 17) needs one more electron to have a filled outer shell. Carbon has 4 valence electrons (group 14) while nitrogen has 5 valence electrons (group15). Boron and aluminum are in the same group and therefore have the same number of outer electrons (3). Chlorine has 7 valence electrons (group 17) while oxygen has 6 valence electrons (group16).
5B possibility: (Level I: Use the periodic table to identify chemical families and determine charges on monoatomic ions.) Based on its position in the periodic table below, what predictions can you makeabout the properties of the element labeled A? • A The element is a poor conductor and has a shiny appearance. • B The element is a good conductor and has a shiny appearance. • C The element is a poor conductor and has a dull-looking appearance. • D The element is a good conductor and has a dull-looking appearance. 7) Element A is a transition metal and all metals are GOOD conductors. Because it is a metal this is what is expected. Element A is a transition metal and all metals are GOOD conductors and shiny as long as the surface is free from dirt and oxidation. Element A is a transition metal and all metals are shiny as long as the surface is free from dirt and oxidation.
5C The student understands the historical development of the Periodic Table and can apply its predictive power. • The student is expected to use the Periodic Table to identify and explain periodic trends, including atomic and ionic radii, electronegativity, and ionization energy.
5C possibility: (Level II: Recognize and explain periodic trends using properties of elements.) 8) Why do the radii of atoms tend to decrease as you move left to right on a period of the periodic table? • A The number of valence electrons increases so the electrons are harder to remove. • B The nuclear charge increases but shielding by inner electrons does not increase. • C The number of electrons in the highest energy level decreases across the table. • D The number of neutrons in the nucleus increases, increasing the effective charge. It is true that the # of valence electrons increase BUT they are still all in the same energy level, with the same # electrons and energy levels between them and the nucleus. More protons are added as you move across a period which means more positive charge that is attracting the same energy level of negative electrons causing them to move in. What ever is between them and the nucleus remains the same. Not true, they increase. The number of neutrons do increase but have nothing to do with charge because they are neutral.
5C possibility: (Level II: Recognize and explain periodic trends using properties of elements.)Based on the information on the graph below, choose the statement on the following slide about ionization energy which is true? 9)
There are definite trends in the graph which means it is NOT independent. If it were independent the data points would be random and scattered all over. • A Ionization energy is independent of period in the periodic table. • B Ionization energies of all elements of a group are about the same. • C Within a group, ionization energy tends to increase with total number of electrons. • D Within a period, ionization energy tends to increase with an increasing number of valence electrons. Not true, follow the line above each group number. The data points are not all in the same place Again looking at each line for a group, the more electrons added the LOWER the ionization energy. Each line represents a period. Notice that, with only a couple of exceptions the lines slant upward (positive slope) meaning it is increasing as group # and valence electrons increase.
5C TEA released (Level II: Recognize and explain periodic trends using properties of elements.) 10) Which of the following elements has the smallest atomic radius? • A Sulfur • B Chlorine • C Aluminum • D Sodium The smallest will be the element that is closest to the top and right of the periodic table. Sulfur is not the closest to the right. Chlorine is in the same period as all of the others so they are all the same distance from the top of the periodic table but chlorine is closest to the right. The smallest will be the element that is closest to the top and right of the periodic table. Aluminum is not the closest to the right. The smallest will be the element that is closest to the top and right of the periodic table. Sodium is not the closest to the right.
6E The student knows and understands the historical development of atomic theory. • The student is expected to express the arrangement of electrons in atoms through electron configurations and Lewis valence electron dot structures.
6E TEA released (Level III: Evaluate periodic trends using electron configurations.) 11) Some properties of scandium are determined by the electron arrangement within scandium atoms. What is the ground state electron configuration of scandium? • A [Ar]4s23p1 • B [Ar]4s23d1 • C [Ar]4s24p1 • D [Ar]4s24d1 Contains the correct # of electrons but 3p1 is a lower energy level that was already used. Scandium is the first element of the d block in the 4th period. Use the last noble gas before the 4th period (which is argon) in brackets, [Ar] then record 4s2 for the s block and the last electron goes into the first spot in the d block. Remember, the energy level number for d block electrons is one less than the period it is in. Correct # of electrons but the last electron in 4p1 skipped the 3d level. Electrons in the ground state cannot skip levels. Correct # of electrons but the last electron in 4d1 skipped the 3d, 4p, and 5s levels. Electrons in the ground state cannot skip levels.
6E possible(Level III: Evaluate periodic trends using electron configurations.) The chart above shows the relationship between the first ionization energy and the increase in atomic number. Which letter could represent an element with an electron configuration ending in s1? • A W B X C Y D Z 12) Starting from atomic #1, W is always the lowest in the period (requires the least amount of energy from which to remove the e-). The s1 electron is always the easiest to remove. Starting from atomic #1, X is always the highest in the period which means it is the most difficult from which to remove the 1st e- . The s1 electron is always the easiest to remove. Starting from atomic #1, Y is always the 2nd or 3rd lowest in the period which means it is not the easiest from which to remove the 1st e- . The s1 electron is always the easiest to remove. Starting from atomic #1, Z is always second to the highest in the period which means it is second to the hardest from which to remove the 1st e-. The s1 electron is always the easiest to remove.
6E possibilty 13) Which of the following is a possible electron configuration of titanium in the excited state? • A 1s22s22p63s23p64s2 • B 1s22s22p63s23p84s2 • C 1s22s22p63s23p63d24s2 • D 1s22s22p63s23p64s24p2 Add the superscripts. This e- configuration has only 20 e-. Ti has 22. Contains 22 e- but the 3p has 8 e- in it which is impossible. It can only hold 6. Contains 22 e- that are correctly placed but they are in the ground state, not the excited state. Contains 22 e- correctly placed and the last 2 have skipped over the 3D and up to the 4p which means it is in the excited state.
7A The student knowshow atoms form ionic, metallic, and covalent bonds. • The student is expected to nameionic compounds containing main group or transition metals, covalent compounds, acids, and bases, using International Union of Pure and Applied Chemistry (IUPAC) nomenclature rules.
7A possibility 14) What is the correct name for the compound whose formula is Fe2O3? • A iron oxide • B iron(II) oxide • C iron(III) oxide • D diiron trioxide Iron is a transition metal with a possibility of 2 oxidation numbers The correct number must be indicated by a Roman numeral. Iron is a transition metal with a possibility of 2 oxidation numbers The oxidation number in the compound above is 3 and would be indicated by the Roman numeral III not II. Ionic naming rules used correctly indicating the proper oxidation state of iron. The compound is ionic (metal plus a nonmetal). This answer uses the rules for covalent compounds
7A possibility 15) What is the correct name of the molecular compound P2O3? • A potassium oxide • B phosphate trioxide • C diphosphorous trioxide • D triphosphorous dioxide The compound given is covalent (nonmetal + nonmetal). Answer A uses the rules for ionic bonding. The prefix “tri” correctly named the 3 oxygen but the 2 phosphate are not accounted for. Covalent naming rules used correctly indicating the proper number of P and O. Covalent rules used but this name indicates 3 P and 2 O instead of 2 P and 3 O.
7B The student knowshow atoms form ionic, metallic, and covalent bonds. • The student is expected to writethe chemical formulas of common polyatomic ions, ionic compounds containing main group or transition metals, covalent compounds, acids, and bases.
7B possibility 16) What is the chemical formula for potassium sulfide? • A KS • B K2S • C KS2 • D K2S2 To solve this problem first find the oxidation numbers for K and S K1+ S2- Criss Cross to the bottom . DO NOT write + or - and DO NOT write ones. K1+ S2-
7B possibility 17) What is the correct formula for trisulfur pentoxide? • A. SO • B. S3O5 • C. S5O3 • D. SO5 The compound given is covalent (nonmetal + nonmetal). The prefixes are another clue that it is covalent. Answer A uses the rules for ionic bonding. Covalent naming rules used correctly indicating the proper number of S and O. Covalent rules used but this name indicates 5 S and 3 O instead of 3 S and 5 O. The prefix “pent” (short for penta) correctly indicates 5 oxygen but the 3 sulfur are not accounted for.
7C The student knowshow atoms form ionic, metallic, and covalent bonds. • The student is expected to construct electron dot formulas to illustrate ionic and covalent bonds.
7C TEA released (Level III: Predict molecular structure when given only the chemical formulas.) Carbon dioxide is a compound that forms with the combustion of carbon in the presence of oxygen. Electron-dot formulas are used to model the bonds that form. Which of these is the electron-dot formula for carbon dioxide? Each oxygen is missing 2 electrons to complete an octet (8). Carbon is missing 4 electrons to complete an octet (8). Both carbon and oxygen have a complete octet of electrons surrounding them and no more than 8. Each oxygen is missing 2 electrons to complete an octet (8) and the carbon has 4 too many.
7C possibility Using a Lewis diagram how would you show that NaCl is ionic and not covalent? • A. Na:Cl: C. Na+ [:Cl:]- • B. Na Cl: D. Na [:Cl:] Shows transfer of electron from Na to Cl, valence electrons (after transfer) are all in the correct place, the ions that form with their charges are indicated. Answer A indicates that Na and Cl share an electron which is covalent bonding not ionic. : : : : : : : : Correctly shows that one electron has been transferred from Na to Cl (ionic bonding) but does not show the charges that result. Answer B just shows the Lewis diagram for each element with the correct valence electrons – but no bonding.
7C possibility (Level II: relate atomic structure to properties of elements The Lewis diagrams for magnesium and helium are: Mg He: Both show two valence electrons yet each have totally different properties. What best explains these observations? • A. Magnesium needs 8 electrons to complete its outer energy level so the 2 electrons in the Lewis diagram will be very reactive and easily removed . Helium also has only 2 electrons and needs 8 but they are in a lower energy level. This will make helium extremely reactive but will be a gas instead of a solid. • B. The diagrams shown are incorrect. The electrons shown for helium should be separated and on different sides of the symbol to show that they could each bond with another electron. • C. Magnesium has 2 electrons in a 3s orbital which then becomes filled making magnesium very stable and unreactive. Helium also has 2 electrons that complete the 1s orbital making it very stable and unreactive but a gas instead of a solid. • D. Magnesium needs 8 electrons to complete its outer energy level so the 2 electrons in the 3s orbital will be very reactive and easily removed . Helium is in the 1st period (only one energy level and one orbital) in which the 1s is completely filled with 2 electrons and is totally unreactive. Helium does not need 8 electrons to fill its valence shell because it has only 2 electrons which will completely fill the first energy level making He very unreactive. The diagram correctly shows He having 2 paired e- completely filling the first energy level. Although the 3s orbital for Mg is complete the energy level is not. It needs 6 more for the 3p and Mg is very reactive. Correct for the reasons stated.
8B The student can quantify the changes that occur during chemical reactions. • The student is expected to use the mole concept to calculate the number of atoms, ions, or molecules in a sample of material .
8B possibility 18) Calcium has a molar mass of 40.0 g/mol. How many particles are in 18.0 g of calcium? • A 2.71 × 1023 • B 6.02 × 1023 • C 1.33 × 1024 • D 6.34 × 1024 Start with what is given. 18.0g Ca Change grams to moles: (18.0 g Ca ) ( 1 mole Ca) = 0.450 moles Ca ( 40.0 g Ca) Finally change moles to particles using Avogadro’s Number on you formula chart: (0.450 moles Ca) (6.02 x 10 23 particles Ca) = mole Ca
8B TEA released 19) How many oxygen atoms are present in 0.714 moles of SO3? 1)You are told the number of moles so you do not have to calculate that you just need to use Avogadro’s Number from the chemistry formula charts: (0.714 moles SO3 ) (6.02 x 1023 molecules/mole) = 4.29828 x 1023 molecules of SO3(don’t use sig figs until the end) 2) But this time you are asked for the number of OXYGEN atoms. Because of the subscript “3” there are 3 oxygen atoms for each molecule of SO3. This means that once you find the number of molecules in 0.714 moles you will have to multiply by 3: (4.29828 x 1023 molecules of SO3) ( 3 O atoms/molecule of SO3) = Final answer = 1.29 x 024 O atoms
8D The student can quantify the changes that occur during chemical reactions. • The student is expected to use the law of conservation of mass to write and balance chemical equations
8D TEA released (Level I: Determine coefficients to balance chemical equations.) 20) The partial chemical equation above represents the formation of lead(II) chloride. What is the missing reactant? • A PbO(s) • B Pb(s) • C PbO2(s) • D PbCl4(s) Provides the correct number of Pb needed but is short one oxygen. Oxygen is still needed to balance the oxygen on the product side. Exactly the right amount of Pb and O needed on the reactant side to balance the product side. No more chlorine is needed but oxygen is till missing on the reactant side.
8D possibility (Level I: Determine coefficients to balance chemical equations.) 21) How can the following reaction be balanced? Cl2O7 + H2O → HClO4 • A Put a coefficient of 2 in front of Cl2O7. • B Put a coefficient of 2 in front of H2O. • C Put a coefficient of 2 in front of HClO4. • D Put a coefficient of 2 in front of Cl2O7 and in front of HClO4. Too many Cl (4)and too many O(15) on the reactant side. Too many H (4)and not enough Cl on the reactant side. All elements balance – Cl(2), O(8), H(2) Chlorine doesn’t balance (4 in the reactants and 2 in the products).Oxygen doesn’t balance (15 in the reactants and 8 in the products.
9A The student understands the principles of ideal gas behavior, kinetic molecular theory, and the conditions that influence the behavior of gases. • The student is expected to describe and calculate the relations between volume, pressure, number of moles, and temperature for an ideal gas as described by Boyle’s law, Charles’ law, Avogadro’s law, Dalton’s law of partial pressure, and the Ideal Gas law .
9A TEA released (Level II: Apply the gas laws qualitatively and quantitatively to a variety of situations.) Final answer is 1.84 atm When asked to do calculations for gasses, first look at what was given and determine if the problem either gave a number of moles or asked for a number of moles. • 22) A gas was held at a constant temperature in a closed system. The initial pressure of the gas was 1.20 atm, while its initial volume was 2.30 L. The final volume of the gas was 1.50 L. What was the final pressure of the gas to the nearest hundredth of an atmosphere? Record your answer and fill in the bubbles on your answer document. This problem does not mention any moles so use the given numbers in the following equation: P(initial)V(initial) = P(final)V(final) (1.20atm)(2.30 L) = (Xatm)(1.50 L) T(initial) T(final) Now solve for X. Temperature (T) was left out because it did not change. Whenever the questions says that any variable (P, V, or T) doesn’t change then do not include it in the equation. If temperature does change and is used in the equation you must change it to Kelvin degrees if it is not already in Kelvin (add 273 to Celsius). It the problem had given moles or asked for moles then use the Ideal Gas Law: PV = nRT. This equation and the value for the constant R is in the formula chart you are given. “n” is the number of moles. MAKE SURE ALL UNITS MATCH THE UNITS IN THE CONSTANT ”R” THAT YOU USE!!!!
9A possibility (Level II: Apply the gas laws qualitatively and quantitatively to a variety of situations.) 23) Which of the following changes would increase the pressure of the gaseous contents of a container with a fixed volume? • A decreasing the number of moles of gas at a fixed temperature • B decreasing the temperature of the contents of the container • C increasing the number of moles of gas and increasing temperature • D increasing the temperature and opening the container Decreasing the number of moles decreases the number of particles of gas hitting the sides of the container which would decrease the pressure – not increase it. Decreasing the temperature means decreasing the kinetic energy of the gas particles (they slow down) so there would be less collisions with the container and the pressure would decrease not increase. More moles of gas at a higher temperature means more particles moving faster and making more higher energy collisions which is called increased pressure. Answer D would drive all of the gas particles out of the container and decrease the pressure
9A possibility (Level I: Recognize relationships among volume, pressure, and temperature of a gas.) 24) In an ideal gas, how will the volume change if the pressure is doubled and temperature remains constant? • A Volume will double. • B Volume will quadruple. • C Volume will be reduced by half. • D Volume will not change. The mathematical relationship that governs this situation is: P1V1 = P2V2. If P2 AND V2 both double this equation will longer be equal. The mathematical relationship that governs this situation is: P1V1 = P2V2. If P2 doubles (2x) AND V2 quadruples (4x) the right side of this equation will now be 8x larger and no longer equal. The mathematical relationship that governs this situation is: P1V1 = P2V2. If P2 doubles (x2) AND V2 decrease by half (x1/2) they cancel each other out (2 x ½ = 1) and this equation remains equal. In a closed container, if pressure changes either the volume or the temperature MUST change and the problem states that the temperature was constant so therefore volume MUST change.
9A possibility (Level II: Apply the gas laws qualitatively and quantitatively to a variety of situations.) Choice “A” correctly shows volume increasing as the temperature increases (Gay-Lussac’s Law) Graph C shows the balloon getting larger but then abruptly stopping its increase. If this happened the balloon would either have to pop or the pressure would increase and not remain constant. 25) Which graph best shows the relationship between the volume of a gas and its temperature as the gas pressure remains constant? • A. A B. B C. C D. D Imagine a balloon or tire that could expand forever without popping. If heat were applied to it, it would get larger. This graph shows the opposite. D indicates that no change would happen at lower temperatures. Again this would mean the pressure would have to increase and it is not.
9A possibility (Level II: Apply the gas laws qualitatively and quantitatively to a variety of situations.) Choice “A” correctly shows pressure increasing as volume decreases and also as you get closer to the limits of the container the pressure will rise faster. Graph C shows no change which is impossible 26) Which graph best shows the relationship between the pressure and volume of gas at constant temperature? • A. A B. B C. C D. D A. C. Imagine a cylinder that you could push one of the ends in or pull out (a piston). As you push in (volume decreases) it becomes more difficult because the gas molecules are being squeezed and the pressure increases. This known as Boyle’s Law. B shows the pressure increasing when the volume increases – wrong choice. D indicates pressure increasing as volume decreases but it doesn’t account for a finite space in the closed container B. D.
10B The student understands and can apply the factors that influence the behavior of solutions. • The student is expected to develop and use general rules regarding solubility through investigations with aqueous solutions
10B possibility 27) For molecular compounds, which property has the greatest effect on the solubility for the compound in water? • A melting point • B molecular mass • C density • D polarity Compounds that are attracted to water will dissolve easiest no matter what their melting point. Most molecular compounds (not ionic) have low melting points so it wouldn’t make much difference. Compounds that are attracted to water will dissolve easiest no matter what their molecular mass. Density determines if something will float or not. It has little to do with solubility. When considering solubility in general always remember “like dissolves like.” So polar molecular compounds like alcohol or acetone will dissolve in polar water but oil will not because it is not polar.
10B possibility 28) A crime lab is investigating a fresh looking stain that was found on a victim’s clothing. The substance needs to be lifted intact (the chemical composition must remain the same) so it can be analyzed. Which of the following would be the best method to use? • A. Since water is the universal solvent dissolve in a few milliliter of distilled water then evaporate the water. • B. Dissolve in a few milliliters of distilled water with detergent, then evaporate the water. • C. Dissolve in a few milliliter of distilled water and evaporate the water. Then use a few milliliters of dry cleaning fluid and evaporate. • D. Use only dry cleaning fluid to dissolve then evaporate the fluid.
10E The student understands and can apply the factors that influence the behavior of solutions. • The student is expected to distinguishbetween types of solutions such as electrolytes and nonelectrolytes and unsaturated, saturated, and supersaturated solutions.
10E possibility 29) Which of the following materials could be used as a solute in the experiment pictured above? • A dextrose • B carbon dioxide • C ethyl alcohol • D lithium iodide Solute in water In order to light the bulb the solute must be an electrolyte in an aqueous solution which means it must dissolve and form ions in water. Dextrose is a type of sugar and does not form ions. Carbon dioxide does not create ions to any great extent. It remains a gas. Ethyl alcohol (drinking alcohol) is miscible with water but does not form ions. Lithium Iodide is an ionic compound (metal + a nonmetal) and will dissolve and form ions in water.
10E possibility 30) A student forgot to label 3 test tubes which contained a saturated, an unsaturated, and a supersaturated solution. The student drops a crystal into each of the test tubes and gets the following results shown in the table above. Given these results the student should conclude--- • A A=saturated, B=supersaturated, C=unsaturated • B A=unsaturated, B=saturated, C=supersaturated • C A=supersaturated, B=saturated, C=unsaturated • D A=unsaturated, B=supersaturated, C=saturated If A was saturated the crystal added would not dissolve. The table indicates that it did. If C was supersaturated the crystal added would cause many more crystals to form. That did not happen. If A is supersaturated the added crystal would not dissolve. The table indicates that it did. A is unsaturated because more crystals can still be dissolved, B is supersaturated because when even one crystal was added the extra solute came out of solution. C is saturated because the one crystal added did not dissolve but did not cause any extra to fall out.
10E possibility KNO3 31) Based on the graph shown above, which saturated solution would have the largest mass of solute at 10oC? • A KNO3 • B NaCl • C Ce2(SO4)3 • D They are all the same NaCl Ce2(SO4)3 At 10 degrees, KNO3 has the least mass of solute because its saturation line is the lowest. At 10o, NaCl has not the least nor the greatest mass of solute because its line is in between. The Ce2(SO4)3 line is the highest at 10 degrees. If they were all the same at 10o the lines would all intersect at 10o on the graph. That happens at about 24 degrees,
10F The student understands and can apply the factors that influence the behavior of solutions. • The student is expected to investigatefactors that influence solubilities and rates of dissolution such as temperature, agitation, and surface area.