×
Log in to StudySoup
Get Full Access to WSU - Chem 101 - Study Guide - Midterm
Join StudySoup for FREE
Get Full Access to WSU - Chem 101 - Study Guide - Midterm

Already have an account? Login here
×
Reset your password

WSU / Chemistry / CHEM 101 / chem 101 exam 3

chem 101 exam 3

chem 101 exam 3

Description

School: Washington State University
Department: Chemistry
Course: CHEM 101
Professor: Finnegan
Term: Fall 2014
Tags: General Chemistry
Cost: 50
Name: Chem 101 Test 3 study guide
Description: These notes over everything that will be on our next exam, on 11/17/2016. They cover the topics from the beginning of the semester to the date of the exam.
Uploaded: 12/02/2016
12 Pages 219 Views 0 Unlocks
Reviews



0 ゜C, what is the gas pressure in the tank?




What is the ∆H for this reaction?




What mass of chromium (III) Chloride will be made?



STUDY GUIDE FOR EXAM 3 (11/17) Limiting Reactants: 1. Consider the equation: 3CuO (s) + 2NH3 (g)  N2 (g) + 3Cu (s) + 3H2O (l)  How many moles of nitrogen gas could be made from 5.00mol CuO and  2.00mol NH3 First determine how many mol each mol of reactant will make of the N2 (g). Do this by multiplying the amount of mol of reactanIf you want to learn more check out gloablization
Don't forget about the age old question of forsman etsu
If you want to learn more check out consumers involved in habitual decision making engage in little conscious decision making.
We also discuss several other topics like gsu music
Don't forget about the age old question of orbital motion is a combination of
We also discuss several other topics like en 355
t by the mol to mol ratio.  5.00 molCuO×1 mol N2 3 molCuO=1.667mol N2 2.00 mol NH3×1mol N2 2mol NH3=1mol N2 Second simply determine which substance produces less mol of N2, that is  your limiting reactant, so the number of mol that that substance produces of N2 is your answer.  The 2.00 mol NH3 produces less mol of N2. So NH3 is the limiting  reactant, so 1.00mol N2 gas can be made from 5.00mol CuO and  2.00mol NH3 2. Given the equation: 4Cr (s) + 3SnCl2 (l)  4CrCl3 (s) + 3Sn (s)  a. If 6.000g of Chromium and 23.000g of tin(IV) chloride react according to this equation, which of these is the limiting reactant? Firstdetermine how many mol the masses of each substance will  produce of either product (literally choose either one, the answer will  be the same, I chose to use chromium(III) chloride). This is simple  stoichiometry so I won’t detail each step.  51.996 g×4 molCr Cl3 6.000 gCr×1 molCr 23.000 gSnCl4×1 mol SnCl 4 4 molCr=0.1154mol CrCl3 118.71 g+( 35.4527 g×4)×4mol CrCl3 3mol SnCl4=0.1177 molCr Cl3 Second as in the question above, determine the limiting reactant. In  this case, because Cr produced less of the product, it is the limiting  reactant.  b. What mass of chromium (III) Chloride will be made?  First conveniently we already found the number of mol that our  limiting reactant, Cr, will produce of CrCl3. So now, simply convert the mol CrCl3 to grams CrCl3 0.1154 molCr Cl3×51.996 g+(35.4527 g×3) 1molCr Cl3=18.27 gCr Cl3 Endothermic and Exothermic Reactions: 1. Classify each of the following reactions as endothermic or exothermic. Then  calculate the energy absorbed or released when 25.00 g of the first  reactant reacts with an excess of the second reactant.  a. 3Fe2O3 (s) + CO (g)  2Fe3O4 (s) + CO2 (g) ∆H= -203kcal First This reaction is EXOTHERMIC because the ∆H is negative Second to calculate the energy absorbed/ released simply multiply  the mass you are given of Fe2O3 by its molar mass to cancel out the  grams. Then multiply by the number of kcal released per mol of Fe2O3STUDY GUIDE FOR EXAM 3 (11/17) in the equation (in this case, 3) the mol Fe2O3 cancels out, leaving you  with kcal. 25.00 gFe 2O3×1mol Fe2O3 (55.867 g×2)+(16.00 g×3 )×−203 kcal 3mol Fe2O3=−10.6 kcal (in  this case, released b. N2 (g) + O2 (g)  2NO (g) ∆H= 378 First this reaction is endothermic because the ∆H is positive Second to calculate the energy absorbed/ released simply multiply  the mass your are given of N2 by its molar mass to cancel out the  grams. Then multiply by the number of kcal released per mol of N2 in  the equation (in this case, 1) the mol N2 cancels out, leaving you with  kcal 25.00 g N2×1 mol N2 (14.01 g×2)×378 kcal 1mol N2=337 kcal(¿ this case ,absorbed)  Heat, ∆H, entropy, etc.: 1. If 100.0 mL of water containing 0.0750mol of HNO3 are mixed with 100.0mL of water containing 0.0750 mol of KOH, the temperature of the mixture rises from 23.4 ℃ to 35.8 ℃ a. Write the chemical equation for the reaction that takes place HNO3 (aq) + KOH (aq)  KNO3 (aq) + H2O (l)  b. Calculate the heat absorbed by the solution (its specific heat is 0.9783 cal/g ℃ . Assume water has a density of 1.000g/mL)  First we have to figure out the mass of the solution. To do this  multiply the sum of the mL of water (so 100.00mL + 100.00mL =  200.00mL) by the density of water, to get mL to cancel out, leaving  you with grams, then figure the grams of each substance in the water,  and add all three masses together to get the total mass of solution 200.00 mLwater×1.000 g 1mL water=200.00 g water 0.0750 mol HN O3×63.018 g mol=4.72635 g , 0.0750 mol KOH ×56.108 g mol=4.2081 g 200g + 4.72635g + 4.2081g = 208.9345g  Second plug your mass, specific heat, and change of temperature  into the equation for heat—which you’ll remember is  heat(cal)=mass(g)xTemp Change(∆ ℃ )xSpecifc heat 208.9345 g×(35.8℃−23.4℃)×0.9783 cal g℃=2534.6 cal c. What is the ∆H for this reaction? (remember that ∆H is given in kcal)STUDY GUIDE FOR EXAM 3 (11/17) First we have to find the kcal per mol of HNO3 (could use either  reactant) by using the mol given to us in the problem. Note that the  cal here are written as negative because this reaction is  EXOTHERMIC. Then, because ∆H is always written for the number of  mols indicated in the equation, we can multiply the kcal/mol by the  number of mol of HNO3 in the equation to cancel out mol, leaving us  with kcal.  - 2534.6 cal×1kcal 1000 cal×1 0.0750 mol HN O3×1mol HN O3=−33.80 kcal=∆ H 2. For each of the following a) predict the sign of ΔS for the reaction, and b)  state the temperature conditions (all, none, high or low) under which the  forward reaction will be spontaneous. - CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) +H2O (l) + CO2 (g) ΔH=  -3.42kcal a. ΔS will most likely be positive because, though there are the  same number of mol of particles on each side, the phases go  from s and aq to aq, l, and g. So it is becoming more chaotic,  and therefore has positive entropy.  b. The reaction is spontaneous at all temperatures (remember  from the sign chart that a –ΔH and a +ΔS means that ΔG is  negative at all temps)  - CH3OH (l)  CO (g) 2H2 (g) ΔH= +39.94kcal a. ΔS will most likely be positive because not only are the  phases going from l to g, but the mol increase from 1 to 4.  b. The reaction will be spontaneous at high temperatures.  (remember back to the sign chart, a +ΔH and a +ΔS makes ΔG  negative only at high temps) 3. a). Draw a reaction energy diagram for a reaction with ΔG=  -100.00kcal/mol and STUDY GUIDE FOR EXAM 3 (11/17) Ea = 50.0kcal/mol at 25 ℃ Y-Value 1 (Close enough right? You get the idea. The reactants start at zero. The  ΔG is where the products are and the Ea is where the top of the hill is) b.) If the ΔH for the reaction is -50.00kcal/mol at 25.0 ℃ , what is the  value of ΔS at 25.0 ℃ ? First remember the equation: ΔG = ΔH – TΔS (where T must be in  kelvin) so plug in all your known values and solve (The three steps  are just using basic algerbra to solve for ΔS)  −100.00 kcal mol=−50.00 kcal mol−( 25.00℃+273.15) (ΔS ) → −100.00 kcal mol+50.00 kcal mol=( 25.00℃+273.15)×ΔS → −50.00 kcal −25.00℃+273.15=ΔS → ΔS=+0.168kcal mol mol ℃at 25.00℃  c.) Is this reaction spontaneous at all temperatures? Yes, this reaction is spontaneous at all temperatures.  (remember back to the sign chart. A –ΔH and a +ΔS, ΔG is  negative at all temperatures) 4. Identify the temperature conditions under which these reactions will be  spontaneous. For all of the following, REMEMBER you sign chart you should  memorize it!!!STUDY GUIDE FOR EXAM 3 (11/17) a. ΔH is + and ΔS is +  The reaction is spontaneous at high  temperatures b. ΔH is – and ΔS is +  The reaction is spontaneous at all temperatures c. ΔH is – and ΔS is -  The reaction is spontaneous at low temperatures Equilibrium shifting: 1. Consider the following system: 2SO2 (g) + O2 (g)  SO3 (g) ΔHrxn =  -99.2kJ/mol Assume the system is in equilibrium before each of the  following changes are made to the system. For each change, determine the  direction in which the equilibrium will shift (left, right, not at all) c. O2 is added RIGHT  adding to left side pushes equilibrium to the  right d. Temperature is increased LEFT in an exothermic reaction, the heat  is part of the products, so by increasing the heat, you are increasing  the products, thus pushing the equilibrium towards the reactant side,  or towards the left.  e. The volume of the container is decreased RIGHT When looking at  volume, you only look at the molecules in gas phase because the  volume of solids and liquids are constant. Anyway, the mol of gas  particles are higher in the left (3, vs 1 on the right), so decreasing the  volume will push the left side towards the right.  f.Sulfur dioxide is removed LEFT When you remove something from a  side, the equilibrium is shifted towards that side, so in this case, you  remove a reactant, so the equilibrium is shifted towards the reaction ,  or to the LEFT. g. Adding a catalystNo shift adding a catalyst does NOT shift the  equilibrium  h. Increasing the concentration of SO3  LEFT Increasing something  on the product side, pushes the equilibrium the opposite direction,  towards the reactants, the left i.Decreasing the concentration of O2  LEFT decreasing something on  one side, in this case the reactant side, will draw the equilibrium  towards that side, in this case, the left 2. Consider the following reaction: H2O (g) + Cl2O (g) ⇌ 2HClO (g) ΔH=  3.00kcal Keq=0.0900 at 25゜C  a. Does the equilibrium favor products or reactants at 25゜C? i. It favors the reactant side. Remember back to lecture #25 if  K>1 the product side is favored, if K<1 the reactant side is  favored. 0.0900 is <1, so the reactant side is favored. HINT that would be a good thing to MEMORIZE for the test. ii. Which direction would the equilibrium shift in the following  cases? 1. The temperature was lowered LEFT the ΔH is positive, so the reaction is endothermic. This means “heat” is part  of the reactant side, so lowering it will draw the reaction STUDY GUIDE FOR EXAM 3 (11/17) (shift the equilibrium) towards the reactants, or towards  the left. 2. H2O (g) was added RIGHT adding more to the reactant  side will push the equilibrium toward the product side,  the right. 3. The volume is increased NO SHIFT there are the same  number of gas molecules on each side, so an increase in  volume will effect each side equally, so there won’t be a  shift.  Equilibrium Constant Expressions: 1. PCl3 (g) + 3NH3 (g) ⇌ P(NH2)3 (g) + 3HCl (g) a.[ P( N H2)3] [ HCl ]3 [ PCl3][ N H3]3 =K 2. Pb2+ (aq) + 2 I- (aq) ⇌ PbI2 (s)  2+¿ Pb¿ ¿−¿ a. I¿ ¿ ¿2¿¿ 1 ¿ Intermolecular Forces: 1. Identify the strongest intermolecular force present in each of the following  substances: FOR EACH OF THE FOLLOWING ask yourself first, is the bond ionic?  Then, is the a hydrogen bond present, a hydrogen bonded directly to a F, N,  or O? Then, is the bond polar? Lastly, the only bond left is London dispersion forces. a. CO2 Dispersion b. CH3OCH3dipole-dipole c. HCldipole-dipole d. CH3NH3 Hydrogen bonds e. CH2Cl2 dipole-dipole f. BF3London Dispersion  Boiling Points: 1. Which member of each pair has the higher boiling point? a. CH3CH2CH3 or CH3CH3 CH3CH2CH3 because both molecules are held  together with London dispersion forces, so the deciding factor is that  CH3CH2CH3 is a longer chain, so it has a higher boiling point because  it has more surface areaSTUDY GUIDE FOR EXAM 3 (11/17) b. NO or N2 NO because it is a dipole-dipole bond, while N2 is held with a London dispersion force which is a weaker bond, so it has a lower  boiling point  c. H2S or H2O H2O because hydrogen bonds are stronger than dipole dipole bonds (which is the intermolecular force between H2S)  2. Arrange the series of substances in order of increasing boiling point.  a. Propane (CH3CH2CH3) < dimethyl ether (CH3OCH3) < ethanol  (CH3CH2OH) i. Because dispersion bonds< dipole-dipole< hydrogen bonds b. Cl2< Br2< I2 i. Because they each have the same bonds (dispersion), so rank  them in order of increasing mass, because the higher the mass,  the higher the boiling point 3. Rank the following solutions in order of increasing boiling point: a. 0.250m NaCl, 0.400m C2H6O2, and 0.150m K2SO4 First multiply the molality of each substance by the number of  particles the compounds will break into 0.250 m NaCl×2=0.500,0.400 m C2H6O2×1=0.400, 0.150m K2SO4×3=0.450 Second rank them in order of increasing answer: C2H6O2 (0.400) < K2SO4 (0.4500) < NaCl (0.500) Vapor Pressure: 1. Which substance would have the higher vapor pressure at 25゜C:  CH3CH2CH2OH or CH3CH2F? a. CH3CH2F Because it has a weaker bond holding it together (dipole dipole, compared to hydrogen bonds with CH3CH2CH2OH) and for  vapor pressure, the weaker bond has the higher vapor pressure.  THINK the weaker bond has a lower boiling point, so it’s vapor  pressure will increase faster.  Gas Pressure: 1. A Steel tank containing 600.00g of Nitrogen has a gas pressure of 150.0atm at 50.0゜C. If the tank (and the gas) is cooled to -50.0 ゜C, what is the gas  pressure in the tank? a. REMEMBER PV=nRT (note that in this equation, we are using atm as  the units of pressure, so the R constant is 0.0820575) First solve for the volume of the steel tank using the g of nitrogen  you are given (to convert to moles by multiplying by the molar mass of nitrogen—note that nitrogen gas is N2), the gas pressure you are  given, and the first temperature you are given (remembering to  convert to kelvin by adding 273.25):STUDY GUIDE FOR EXAM 3 (11/17) [(600.00 g N2×1mol N2 V= 14.01×2g N2 )×0.0820575×( 50.0℃+273.15)] 150.0 atm→V=3.785L Second Solve for the pressure this time by using the volume you just  found, the same equation for mol of nitrogen and the new  temperature (remembering again to convert to kelvin by adding  273.15) [(600.00 g N2×1mol N2 P= 14.01×2 g N2 )×0.0820575×(−50.0℃+273.15)] 3.785L→ P=103.6 atm 2. A sample of gas occupies 453mL at 98.5゜C and 735 torr pressure a. How many moles are present? First solve for n. Use the pressure you are given (note that it is units  of torr, so R will be 62.3637). Use the volume you are given, but don’t  forget to convert it to L. And use the temperature you are given,  remembering to add 273.15 to convert to K 62.3637×(98.5℃+273.15) ¿¿ (735 torr×(453mL×1 L n= 1000mL )) ¿ b. What volume will the sample occupy at 25.0゜C and 1600.0torr? First solve for V. Use the mol you found in the above problem. Use  the temperature you are given, remembering to add 273.15 to convert to K. And use the pressure you are given, note that it is given in units  of torr so R is 62.3637. V=0.0144mol ×62.3637×( 25.0℃+273.15) 1600.0 torr→V=0.16694 L→0.167 L∨167mL 3. 2.400g of a substance are vaporizes to from 625.0 mL of a gas at a  temperature of 87.9゜C and 1.250atm. What is the molar mass of the  substance?STUDY GUIDE FOR EXAM 3 (11/17) First Solve for n. Plug in the values you are given for volume— remember to convert to L. Temperature—remember to convert to K by adding 273.15—and pressure.  n=[1.250 atm ×(625.0mL×1L 1000 mL )] 0.0820575×(87.9? C+273.15)→n=0.02637 mol Second Now that you have mol, you can use the grams given to you to find the molar mass, (mol/g) 0.010987m ol 0.02637mol 2.400 g= g∗¿∗2.400 g 0.02637mol=91.0137g mol 4. Sodium metal reacts with water to produce hydrogen gas. a. Write a balanced equation for this reaction: i. 2Na+ + H2O (l)  Na2O + H2 (g) b. What volume of hydrogen gas at 25゜C and 705.0torr, would be  produced by the reaction of 1.56 g of sodium with excess water? Firststoichiometry. Use the grams of sodium given ¿ you to find  mol of hydrogen gas. (use the molar mass of sodium, multiply  by the mol to mol ratio) 22.99 g Na×1mol H2 1.56 g Na×1mol 2 mol Na=0.033928mol H2  Second Now that you have all the necessary parts of the  equation, plug them in and solve for V. Convert to K by adding  273.15. And remember that since the units of volume are given  in torr, the R constant is 62.3637. V=[0.033928mol×62.3637×(25℃+273.15)] 705.0 torr→V =0.8948L∨894.8mL Partial Pressure: 1. If a mixture of a 10.00g of argon and 10.00g of chlorine exert a total  pressure of 1200.0torr, what is the partial pressure of each gas? a. Remember the equation for partial pressure:  PA= (mol of substance)/(total mol of mixture) x (total pressure)  So First Find the mol of each substance by multiplying the grams of  them by molar mass (Remember that Chlorine always occurs as Cl2) 10.00 g Ar ×1mol 39.95 g Ar=0.2503 mol Ar 10.00 gCl 2×1mol 70.90 gCl2=0.2821 molCl2 Second Add together the mol of each substance to find the total mol  of the mixture 0.2503mol Ar+0.1410mol Cl2=0.3913molTotalSTUDY GUIDE FOR EXAM 3 (11/17) Third find the partial pressure of each substance: PAr=0.2503mol Ar 0.3193mol×1200.0 torr=767.6 torr PCl=0.1410mol Cl2 0.3193mol×1200.0 torr=432.4 torr Solubility: 1. Nitrogen has a solubility of 1.04mM (0.00104M) at a temperature of 0.00゜C  and a partial pressure of 1.00atm. What partial pressure of nitrogen is  required to increase the solubility of nitrogen at 0.00゜C to 0.0562M Remember that the equation for solubility and pressure is (C1)/(P1) =  (C2)/(P2) so in order to get the partial pressure of nitrogen, you have to  solve for P2 P2 = (C2 x P1)/(C1)  so plug in your info for concentration and pressure  and solve P2=0.0526 M ×1.00 atm 0.00104 M=54.04 atm 2. 246.872 g of barium chlorate (Ba(ClO3)2) dehydrate are dissolved in  sufficient water to make 500.0mL of solution. What is the concentration of  this solution in molarity?  First use grams of barium chlorate to find mol barium chlorate 246.872 g×1mol 137.33 g+(35.45 g×2)+(16 g×6)=0.7256 mol Ba(Cl O3)2 Second divide the mol by the volume given, but don’t forget to convert  it to L (0.7256 mol Ba (Cl O3 )2 ) 0.5000 L=1.451 M 3. What mass of potassium sulfite (K2SO3) is required to make 250.0 mL of a  0.4750M solution of potassium sulfite? First Multiply the volume given (don’t forget to convert to L) by the concentration given, this way the L cancel and you are left with mol.  Then, multiply by the molar mass of Potassium sulfite, this way the  mol cancel and leave you with grams of K2SO3 0.2500 L×0.4750mol 1 L×158.27 g 1mol=18.79g K2 SO3 Electrolytes and Nonelectrolytes: 1. Identify the following solutes as electrolytes or nonelectrolytes: (notice it is  not asking for strong or weak) a. CH3CH2CH2OH nonelectrolyte (molecular compounds that are not  acids are nonelectrolytes) b. FeCl3 electrolyte (ionic compounds will disassociate in water) c. NaCl electrolyte (ionic compounds will disassociate in water) d. KOH electrolyte (acids and bases disassociate in water) e. HNO3 electrolyte (ionic compounds disassociate in water)STUDY GUIDE FOR EXAM 3 (11/17) f. C6H12O6 nonelectrolyte (molecular compounds that are not acids are  nonelectrolytes) Acids, Bases, Conjugates, Disassociation Constants, etc: 1. Write the formula of the conjugate acids of the following compounds: Simply add an H+ to the beginning, if it’s a regular compound, or to the N or O if it  is an organic compound a. C2O42-  HC2O4-  b. HPO42-  H2PO4- c. CHNHCH3  CHNH2+CH3 (remember that when the compound is  written in organic form, the H+ attaches to the N or O, in this case, N) 2. Write the formula for the conjugate bases of the following compounds:  simply take away an H+ from the compound.  a. H2CO3  HCO3- b. CH3COOH CH3COO c. NH3+CH2CH2NH3+  NH3+CH2CH2NH2 3. Which solutions would you expect to be more acidic? Why?  a. 0.0100M Chloric acid (HClO3) or 0.0100M Iodic acid (HIO3) i. Chloric acid has a Ka of 1.0, Iodic acid has a Ka of 0.16 (look at  the acid disassociation chart). The larger number is the  stronger acid (when the concentrations are equal), thus, chloric acid is more acidic.  4. Rank the following acids in order of increasing acid strength: a. Ascorbic acid, hydrocyanic acid, hydrofluoric acid b. Ka: 7.9E-5 4.9E-10 3.5E-4 c. The larger number is the stronger acid, so  i. Hydrocyanic acid< ascorbic acid< hydrofluoric acid 5. Rank the following bases in order of increasing base strength: a. Ammonia, sodium sulfate, sodium acetate b. First find the conjugate acids of each base NH3NH4+ , Na2SO4 HSO4- , NaC2H3O2 HC2H3O3  c. The stronger the acid, the weaker the base. So if you order them by  increasing acid strength, just reverse the order and that is the order  of increasing base strength.  d. NH4+ 5.6E-10, HSO4-  1.2E-2, HC2H3O21.8E-5 e. The order of increasing acid NH4+< HC2H3O2 < HSO4- f. So the order of increasing base Na2SO4< NaC2H3O2< NH3 pH and pOH, etc: 1. calculate the pH and pOH pf the following solutions:  a. 2.756x10-3M HCl  i. pH= -log(2.456x10-3) = 2.6098 pOH= 14- 2.6098=  11.3902 b. 0.00555M KOH i. pOH= -log(0.00555) = 2.2557, pH= 14- 2.2557=  11.7443STUDY GUIDE FOR EXAM 3 (11/17) 2. Calculate the values of [H+] and [OH-] for solutions with the following pH  values: a. pH= 8.00 i. [H3O+]= 10-8.00= 1.0x10-8M [OH-]= 1.00x10-14/1.0x10-8=  1.0x10-6M b. pH=3.80 i. [H3O+} = 10-3.80 = 1.6x10-4M [OH-]= 1.00x10-14/1.6x10-4=  6.3x10-11M

Page Expired
5off
It looks like your free minutes have expired! Lucky for you we have all the content you need, just sign up here