Portage Chemistry 103 Final Exam Study Guide., Exams of Chemistry

Download Portage Chemistry 103 Final Exam Study Guide. and more Exams Chemistry in PDF only on Docsity! 1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts kilo (= 1000) milli (= 1/1000) centi (= 1/100) deci (= 1/10) Portage Chemistry 103 Final Exam Study Guide. PORTAGE LEARNING CHEMISTRY 103 FINAL EXAM STUDY GUIDE MODULES 1-6 MODULE 1 – EXAM Question 1 – Section 1.1 10 / 10 pts Complete the two problems below: 1. Convert 1005.3 to exponential form and explain your answer. 2. Convert 4.87 x 10-6 to ordinary form and explain your answer. Your Answer: 1.) Convert 1005.3 to exponential form and explain your answer: All number values must be between 1-10. Because this value is greater than 1, the exponent remains positive. In order for this to happen, we have to move the decimal three places to the LEFT thus, the exponential form of 1005.3 is: 1.0053 X 103 2.) Convert 4.87 x 10-6 to ordinary form and explain your answer. All number values must be between 1-10. Because this value has a negative exponent, the final answer will be less than 1. In order for this to happen, we have to move the decimal six places to the LEFT thus: 0.00000487 Question 2 10 / 10 pts Complete the two problems below: Using the following information, do the conversions shown below, showing all work: 1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints 1. 3.6 pounds = ? oz 2. 2680 ml = ? liters Your Answer: 1.) 3.6 pounds = 57.6 oz 1lb = 16oz 3.6(16) / 1 =57.6 oz 2.) 2680 ml = 2.68 liters New/Old Portage Chemistry 103 Final Exam Study Guide. Liters/Milliliters D = M / V = 28.6 / 32.7 = 0.8751. V = M / D = 14.3 / 0.785 = 18.2 ml2. Portage Chemistry 103 Final Exam Study Guide. Be sure to show the correct number of significant figures in each calculation. 1. Show the calculation of the density of benzene if 28.6 g occupies 32.7 ml. 2. Show the calculation of the volume of 14.3 grams of acetone with density of 0.785 g/ml Your Answer: D=M/V M=DXV V=M/D 1.) Show the calculation of the density of benzene if 28.6 g occupies 32.7 ml. ***significant figures are underlined*** D=M/V 28.6/32.7 = 0.875g/ml 3 sig/figs 2.) Show the calculation of the volume of 14.3 grams of acetone with density of 0.785 g/ml V=M/D 14.3/0.785 = 18.2mL 3 sig/figs Question 5 10 / 10 pts 1. 1.35601 contains ? significant figures. 2. 0.151 contains ? significant figures. 3. 1.35601 + 0.151 = ? (give answer to correct number of significant figures) Your Answer: 1.) 1.35601 = SIX significant figures 2.) 0.151 = THREE significant figures. 3.) 1.35601 + 0.151 =1.507 Question 6 10 / 10 pts Classify each of the following as an element, compound, solution or heterogeneous mixture and explain your answer. Portage Chemistry 103 Final Exam Study Guide. 1. Pepperoni Pizza 24Mg12, 10 electrons = (+12 - 10 = +2) = 24Mg12+212 protons = Mg12, 12 neutrons = Portage Chemistry 103 Final Exam Study Guide. 2. Potassium oxide 3. Phosphorus Your Answer: 1.) Pepperoni Pizza = Heterogeneous Mixture because it has MORE than 1 visible layer, color or substance 2.) Potassium oxide = Compound because it is a combination of two elements found on the periodic table of elements. 3.). Phosphorus = Element because it is found on the periodic table of elements. Question 7 7 / 10 pts Classify each of the following as a chemical change or a physical change 1. Charcoal burns 2. Mixing cake batter with water 3. Baking the batter to a cake Your Answer: 1.) Charcoal burns =Chemical change 2.) Mixing cake batter with water = Physical change 3.) Baking the batter to a cake = Chemical change Question 8 7 / 10 pts Show the full Nuclear symbol including any + or - charge (n), the atomic number (y), the mass number (x) and the correct element symbol (Z) for each element for which the protons, neutrons and electrons are shown - symbol should appear as follows: xZy+/- n 12 protons, 12 neutrons, 10 electrons Your Answer: Question 9 10 / 10 pts Name each of the following chemical compounds. Be sure to name all acids as acids (NOT for instance as binary compounds) 1. PF5 2. Al2(CO3)3 3. H2CrO4 Your Answer: 1.) PF5 Binary Molecular Compound because there are two non-metals and the formula is written bearing in mind that this chemical compound keeps the first non-metal stem and moves the subscript "5" into the prefix of the second non-metal. Therefore the name of this compound is: phosphorus pentaflouride Portage Chemistry 103 Final Exam Study Guide. Barium has +2 charge Chromium (III) sulfide - ide - binary of Cr+3 and S-2 = Cr2S31. Hydroselenic acid - hydro = binary acid of H + Se-2 = H2Se2. Barium iodide - ide = binary of Ba+2 and I-1 = BaI23. Portage Chemistry 103 Final Exam Study Guide. Iodine has a -1 charge Binary ionic compound Ba+2 I-1 Have to balance out the charges: BaI2 MODLUE 2 – EXAM Question 1 10 / 10 pts Show the calculation of the molecular weight for the following compounds, reporting your answe r to 2 places after the decimal. 1. (NH4)2CO3 2. C8H6NO4Br Your Answer: (NH4)2CO3 N 14.01(2) =28.02 H 1.008(8) =8.064 C 12.01 O 16.00(3) = 48.00 MW=96.09 C8H6NO4Br C 12.01(8) = 96.08 H 1.008(6) = 6.048 N 14.01 O 16.00 (4) = 64.00 Br 79.90 MW=260.04 Question 2 8 / 10 pts Show the calculation of the number of moles in the given amount of the following substances. Report your answer to 3 significant figures. 1. 12.0 grams of (NH4)2CrO4 Portage Chemistry 103 Final Exam Study Guide. 2. 15.0 grams of C8H8NOI Portage Chemistry 103 Final Exam Study Guide. L: N4, O10 2 N2 + 5 O2 → 2 N2O52. Portage Chemistry 103 Final Exam Study Guide. R: N4, O10 3. Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O L: Al1, O7, H5, S1 R: Al2, O12, H2, S3 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O L: Al2, O18, H12, S3 R: Al2, O18, H12, S3 1. 2 C6H14 + 19 O2 → 12 CO2 + 14 H2O 3. 2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O Question 7 10 / 10 pts Classify each of the following reactions as either : Combination Decomposition Combustion Double Replacement Single Replacement 1. Pb+2(NO3-)2 + 2 K+I- → PbI2 ↓ + 2 K+NO3- 2. CaO + SO3 → CaSO4 3. 2 KClO3 → 2 KCl + 3 O2 Your Answer: 1. Pb+2(NO3-)2 + 2 K+I- → PbI2 ↓ + 2 K+NO3- Double Replancement 2. CaO + SO3 → CaSO4 Combination/Synthesis 3. 2 KClO3 → 2 KCl + 3 O2 Decomposition Question 8 10 / 10 pts Show the calculation of the oxidation number (charge ) of ONLY the atoms which are changing in the following redox equations. KMnO4 + H2S → S + MnO2 + H2O + KOH Your Answer: Portage Chemistry 103 Final Exam Study Guide. KMnO4 + H2S → S + MnO2 + H2O + KOH Mn is +7KMnO4: K is metal in group I = +1, each O is -2 (total is -8), so Mn is +4MnO2: Each O is -2 (total is -4), so S is -2H2S: H is +1 (total is +2), so S = 0S: is uncombined so Portage Chemistry 103 Final Exam Study Guide. R: K2, Mn2, O8, H6, S3 Equation is balanced! KMnO4 + H2S → S + MnO2 + H2O + KOH Question 9 10 / 10 pts Show the balancing of the following redox equation, including the determination of the oxidation number (charge ) of ONLY the atoms which are changing. Ca(OH)2 + NaOH + ClO2 + C → NaClO2 + CaCO3 + H2O Your Answer: Ca(OH)2 + NaOH + ClO2 + C → NaClO2 + CaCO3 + H2O Redox Table Ca(OH)2 NaClO2 Ca +2 Na +1 O -2, O(2) = -4 Cl +3 H +1, H2= +2 O -2, O2=-4 Sum=0 Sum=0 NaOH Na +1 CaCO3 Ca +2 O -2 H +1 C +4 O -2, O3= -6 Sum=0 ClO2 Sum=0 H2O Cl +4 O -2, O2= -4 H +1, H2=+2 O -2 Sum=0 C Sum=0 Sum = 0 +2 -2 +1 +1 -2 +1 +4 -2 0 +1 +3 -2 +2 +4 -2 +1 -2 Ca(OH)2 + NaOH + ClO2 + C → NaClO2 + CaCO3 + H2O Portage Chemistry 103 Final Exam Study Guide. +2. -4 +2. +1. -2 +1. +4 -4. 0 +1 +3 -4 +2 +4 -6 +2 -2 Since Cl (on left side) is +4 and Cl (on right side) is +3: Cl changes by 1 Since C (on left side) is 0 and C (on right side) is +4: C changes by 4 Multiply Cl compounds by 4 and C compounds by 1 and after balancing other atoms Cl is +4ClO2: Each O is -2 (total is -4), so Cl is +3NaClO2: Na is metal in group I = +1, each O is -2 (total is -4), so C = 0C: is uncombined so C is +4CaCO3: Ca is metal in group II = +2, each O is -2 (total is -6), so + 4 NaOH + 4 ClO2 + 1 C → 4 NaClO2 + 1 CaCO3 + 3 H2O1 Ca(OH)2 Portage Chemistry 103 Final Exam Study Guide. C has change of 4 (reduction) ; coefficient is 1 4x1=4 Cl has change of 1 (oxidation) ; coefficient is 4 1x4=4 LCM is 4 Ca(OH)2 + NaOH + 4ClO2 + 1C → 4NaClO2 + 1CaCO3 + H2O L: Ca1, O11, H3, Na1, Cl4, C1 R: Ca1, O12, H2, Na4, Cl4, C1 Equation is unbalnaced! Ca(OH)2 + 4NaOH + 4ClO2 + 1C → 4NaClO2 + 1CaCO3 + 3H2O L: Ca1, O14, H6, Na4, Cl4, C1 R: Ca1, O14, H6, Na4, Cl4, C1 Equation is balanced! Ca(OH)2 + NaOH + ClO2 + C → NaClO2 + CaCO3 + H2O Question 10 7 / 10 pts Show the balanced equation and the calculation of the number of moles and grams of CO2 formed from 11.9 grams of O2. Show your answers to 3 significant figures. C6H14 + O2 → CO2 + H2O Your Answer: C6H14 + O2 → CO2 + H2O L: C6, H14, O2 R: C1, H2, O2 Tmix = 55.2oC reaction uses 20.7 g S = 20.7/32.07 = 0.6455 mole S ΔHrxn = -1270.18 + 2(-393.51) + 2(-178.32) = - 2413.84 ql↔s = m x ∆Hfusion = 54.3 g x 0.334 kJ/g = 18.14 kJ (since heat is removed) = - 18.14 kJ2. is for 2 mole of SΔHrx x new moles / original moles255.6 = ΔHrx x 0.6455 mole S / 2 mole S = -791.9 kJ-255.6 = ΔHrx Portage Chemistry 103 Final Exam Study Guide. - [95.3952 J/oC x (Tmix - 74.6oC)] = [59.8312 J/oC x (Tmix - 24.3oC)] Question 3 7 / 10 pts Show the calculation of the ΔH of the reaction if 20.7 g of S is reacted with excess O2 to yield 255.6 kJ and sulfur trioxide by the following reaction equation. Report your answer to 4 significant figures. 2 S (s) + 3 O2 (g) → 2 SO3 (g) Your Answer: q = ΔHrx x new moles / original moles 2 moles of S reaction uses 20.7g / (16x2) = 0.646875 moles of S q = 255.6kj x 0.646875 / 2 = = 82.67kJ 2 S (s) + 3 O2 (g) → 2 SO3 (g) Question 4 Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 2 Ca (s) + 2 C (graphite) + 3 O2 (g) → 2 CaCO3 (s) by using the following thermochemical data: 2 Ca (s) + O2 (g) → 2 CaO (s) ΔH = - 1270.18 kJ C (graphite) + O2 (g) → CO2 (g) ΔH = - 393.51 kJ CaO (s) + CO2 (g) → CaCO3(s) ΔH = - 178.32 kJ 2 Ca (s) + O2 (g) → 2 CaO (s) ΔH = - 1270.18 kJ 2 (C (graphite) + O2 (g) → CO2 (g) ΔH = - 393.51 kJ) 2 (CaO (s) + CO2 (g) → CaCO3(s) ΔH = - 178.32 kJ) 2 Ca (s) + 2 C (graphite) + 3 O2 (g) → 2 CaCO3 (s) ΔHrxn = - 2413.84 kJ Question 5 3 / 10 pts Portage Chemistry 103 Final Exam Study Guide. Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 740 ml/1000 = 0.740 liters = 710 mm/760 = 0.934 atm = Pi 460 ml/1000 = 0.460 liters = Vf 1.20 atm = Pf 35oC + 273 = 308oK = Ti 740 mm/760 = 0.974 atm = P R = 0.0821 V = 230 ml/1000 = 0.230 liters 29oC + 273 = 302oK = T 0.456 = g (0.974) x (0.230) = (0.456/MW) x (0.0821) x (302) MW = 50.5 C2H6 (g) = -84.0 kJ/mole, ΔHf0 CO (g) = -110.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/moleΔHf0 = 2(+84.0) + 5(0) + 4(-110.5) + 6(-285.8) = - 1988.8 kJ/moleΔHrxn x Vi ) / Ti = (Pf x Vf ) / Tf(Pi = (1.20) x (0.460) / Tf(0.934) x (0.740) / 308 = 246 oKTf Px V = (g/MW) x R x T Portage Chemistry 103 Final Exam Study Guide. 2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l) by using the following thermochemical data: ΔHf0 C2H6 (g) = -84.0 kJ/mole, ΔHf0 CO (g) = -110.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole Your Answer: ΔHrxn = Σ n ΔHf0 (products) - Σ m ΔHf0 (reactants) -110.5 + (-285.8) - (-84.0) ΔHrxn= -312.3kJ 2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l) Question 6 0 / 10 pts Show the calculation of the new temperature of a gas sample has an original volume of 740 ml when collected at 710 mm and 35oC when the volume becomes 460 ml at 1.20 atm. Question 7 0 / 10 pts Show the calculation of the molecular weigh t of a gas sample with a mass of 0.456 grams which has a volume of 230 ml when collected at 29oC and 740 mm. Question 8 7 / 10 pts Show the calculation of the volume of H2O gas formed by the combustion of 18.6 grams of C6H6 at 30oC and 1.10 atm? The combustion of benzene (C6H6) takes place by the following reaction equation. 2 C6H6 (g) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) Portage Chemistry 103 Final Exam Study Guide. MWunknown= 69.81 The orbital diagram should have spaces that clearly illustrate the levels and the orbitals. The orbital diagram should have spaces so that levels and orbitals are clearly = 1s2 2s2 2p6 3s2 3p6 4s2 3d7Co27 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ ↑↑↓ = 1s2 2s2 2p6 3s2 3p6 4s2 3d7Co27 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ ↑ = 3 unpaired electrons↑↓ Portage Chemistry 103 Final Exam Study Guide. MODULE 4 – EXAM Question 1 10 / 10 pts Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the Fe26 atom. Your Answer: Fe26= 1s2 2s2 2p6 3s2 3p6 4s2 3d6 Question 2 10 / 10 pts Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the Si14 atom. Your Answer: Si14= 1s2 2s2 2p6 3s2 3p2 Question 3 10 / 10 pts Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the Cl17 atom and identify which are valence (outer shell) electrons and determine how many valence electrons there are. Your Answer: Cl17= 1s2 2s2 2p6 3s2 3p5 Valence electrons are 3s2 3p5 and there are seven (7) total valence electrons Question 4 5 / 10 pts * For the following question, use the "Insert Math Equation" tool (indicated by the icon on the toolbar and then choose arrows from the window which opens). Using up and down arrows, write the orbital diagram for the Co27 atom. Question 5 6 / 10 pts * For the following question, use the "Insert Math Equation" tool (indicated by the icon on the toolbar and then choose arrows from the window which opens). Using up and down arrows, write the orbital diagram for the Co27 atom and identify which are unpaired electrons and determine how many unpaired electrons there are. Question 6 10 / 10 pts = 1s2 2s2 2p6 3s2 3p6 4s2 3d3 : n=3, l=2, ml = 0, ms = +1/2V23 Portage Chemistry 103 Final Exam Study Guide. Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the V23 atom and then identify the last electron to fill and write the 4 quantum numbers (n, l, ml and ms) for this electron. Your Answer: V23= 1s2 2s2 2p6 3s2 3p6 4s2 3d3 Last electron to fill is 3d3 Four Quantum Numbers for 3d3 N= 3 I= 2 MI= 0 Ms= +1/2 Question 7 10 / 10 pts 1. Arrange the following elements in a vertical list from smallest (top) to largest (bottom) atomic size: Br, Cl, I 2. Arrange the following elements in a vertical list from highest (top) to lowest (bottom) electronegativity: S, P, Cl 3. Arrange the following elements in a vertical list from lowest (top) to highest (bottom) ionization energy: S, O, Se Your Answer: 1. Arrange the following elements in a vertical list from smallest (top) to largest (bottom) atomic size: Br, Cl, I Cl (Smallest) Br I (Largest) 2. Arrange the following elements in a vertical list from highest (top) to lowest (bottom) electronegativity: S, P, Cl Cl (Highest) S P (Lowest) 3. Arrange the following elements in a vertical list from lowest (top) to highest (bottom) ionization energy: S, O, Se Se (Lowest) S O (Highest) Question 8 8 / 10 pts 1. List and explain which of the following atoms holds its valence electrons more tightly. Br or As 2. List and explain which of the following atoms more easily forms a positive ion. Si or S Your Answer: 1. As < Br The S atom in H2S has 4 groups of electrons around it in its Lewis structure, therefore, its electron geometry would be electronegativity difference = 3.5 - 2.1 = 1.41.6 - 0.5bond is PolarH-O bond electronegativity difference = 3.5 - 2.1 = 1.41.6 - 0.5bond is PolarP-O bond Portage Chemistry 103 Final Exam Study Guide. Your Answer: H-O 3.5-2.1 = 1.4 (0.5-1.6) = polar H-O bond is POLAR O-P 3.5-2.1 = 1.4 (0.5-1.6) = polar O-P bond is POLAR Question 3 10 / 10 pts On a piece of scratch paper, draw the Lewis structure for the ClO4-1 ion. Then choose the correct Lewis structure for ClO4-1 from the options listed below. Your Answer: B Question 4 10 / 10 pts On a piece of scratch paper, draw the Lewis structure for H2SO3. Then choose the correct Lewis structure for H2SO3 from the options listed below. Your Answer: D Question 5 0 / 10 pts Determine the electron geometry and explain your answer for the S atom in H2S. The shape of HCCH is linear and since all of its C-H and C-C bonds are nonpolar, it would be nonpolar since it has all nonpolar Electronegativity differences determine the bond polarity, not the molecularity has 3 groups of electrons around it in its Lewis structure, therefore, its hybridization would be sp2.The C atoms in H2CCH2 central C atom, the shape would be linear. The C atom in HCN has 2 groups of electrons around it in its Lewis structure, therefore, its electron geometry would be linear and since there are 2 atoms around the be considered when determining the shape of the molecule, and both should be noted in your answer. The shape of a molecule is determined by the number of atoms around the central atom, and by the number of electron groups around the central atom. Both need to Portage Chemistry 103 Final Exam Study Guide. Question 6 0 / 10 pts Determine the hybridization and explain your answer for one of the C atoms in H2CCH2. Your Answer: Question 7 5 / 10 pts Determine the shape and explain your answer for HCN. Your Answer: Question 8 5 / 10 pts H = 2.1 Li = 1.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O= 3.5 F = 4.0 Na = 1.0 Mg = 1.2 Al = 1.5 Si = 1.8 P = 2.1 S = 2.5 Cl = 3.0 K = 0.8 Ca = 1.0 Ga = 1.6 Ge = 1.8 As = 2.0 Se = 2.4 Br = 2.8 Use the electronegativities above and your knowledge of the shape of HCCH to determine the molecular polarity of HCCH explaining your answer in detail. Your Answer: C = 2.5 H = 2.1 (between 1.6-0.5 = polar) Thus, H-C 2.5-2.1 = 0.4 NONPOLAR C-H 2.5-2.1 = 0.4 NONPOLAR Question 9 10 / 10 pts Is O2 Polar, Ionic or Nonpolar and List and Explain whether it is Soluble or Insoluble in Water? Your Answer: O2 is nonpolar and is INSOLUBLE Portage Chemistry 103 Final Exam Study Guide. O=3.5 and there are two oxygen atoms so 3.5-3.5 = 0 (nonpolar) and Oxygen is insoluble because it is a gas. FP: Mg3(PO4)2 < AlBr3 < BeBr2 < KBr / gsolute + gsolvent) x 100%Mass % = (gsolute 4.51%Mass % = (18.9 / 18.9 + 400) x 100 = Portage Chemistry 103 Final Exam Study Guide. 0oC - 0.74 = -0.74oC BeBr2 = Be + Br2 = 3 0.100 x 1.86 x 3 = 0.56 0oC - 0.56 = -0.56oC KBr = K + Br = 2 0.100 x 1.86 x 2 = 0.37 0oC - 0.37 = -0.37oC HIGHEST FREEZING POINT BeBr2 → Be+2 + 2 Br- ∆tf = 1.86 x 0.1 x 3 = 2nd lowest FP AlBr3 → Al+3 + 3 Br- ∆tf = 1.86 x 0.1 x 4 = 3rd lowest FP Mg3(PO4)2 → 3 Mg+2 + 2 PO4-3 ∆tf = 1.86 x 0.1 x 5 = lowest FP KBr → K+ + Br- ∆tf = 1.86 x 0.1 x 2 = highest FP Question 4 10 / 10 pts Show the calculation of the mass percent solute in a solution of 18.9 grams of Ba(MnO4)2 in 400 grams of water. Report your answer to 3 significant figures. Your Answer: Mass Percent = grams of solute / (grams of solute + grams of solvent) ×100% 18.9grams / (18.9 +400) (100) = 18.9 / 418.9 (100) = 4.51% Question 5 10 / 10 pts Show the calculation of the molality of a solution made by dissolving 26.5 grams of C6H12O6 in 300 grams of water. Report your answer to 3 significant figures. Your Answer: Molality (m) = (grams of solute ÷ molecular weight of solute) / ( gram solvent ÷1000) MW C6H12O6 = 180.156 C6 = 72.06 H12 = 12.096 Portage Chemistry 103 Final Exam Study Guide. O6 = 96.00 0.250 = (moles) / (200 / 1000) / MW) / (gsolvent / 1000)molality = (gsolute 0.490 mmolality = (26.5 / 180.156) / (300 / 1000) = / MW) / (mlsolvent / 1000)Molarity = (gsolute 0.253 MMolarity = (27.5 / 310.18) / (350 / 1000) = / 1000)Molarity = (moles) / (mlsolvent Portage Chemistry 103 Final Exam Study Guide. (26.5 / 180.156) / (300/1000) 0.1470947 / 0.300 = 0.4903156 or 0.490m Question 6 10 / 10 pts Show the calculation of the molarity of a solution made by dissolving 27.5 grams of Ca3(PO4)2 to make 350 ml of solution. Report your answer to 3 significant figures. Your Answer: Molarity (M) = (grams of solute ÷ molecular weight of solute) / (ml of solution ÷1000) MW Ca3(PO4)2= 310.18 Cax3 = 40.08 x 3 = 120.24 Px2 = 30.97 x 2 = 61.94 Ox8 = 16.00 x 8 = 128.00 (27.5 / 310.18) / (350/1000) = 0.0886581 / 0.350 = 0.2533088 or 0.253M Question 7 7 / 10 pts Show the calculation of the mass of Mg(NO3)2 needed to make 200 ml of a 0.250 M solution. Report your answer to 3 significant figures. Your Answer: molarity x mw solute x (mL solution / 1000) = grams solute MW = 148.325 Mg = 24.305 N x2 = 14.01 x 2 = 28.02 O x6 = 16.00 x 6 = 96.00 0.250 x 148.325 x (200/1000) = 37.08125 x 0.200 = 185.406 grams