M 1-6 Exams

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Pay it forward. MODULE 1 EXAM Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. 1. Convert 845.3 to exponential form and explain your answer. 2. Convert 3.21 x 105 to ordinary form and explain your answer. 1.Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places = 8.453 x 102 2.Convert 3.21 x 10* = negative exponent = smaller than 1, move decimal 5 places = 0.0000321 Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. Using the following information, do the conversions shown below, showing all work: 1 ft =12 inches 1 pound = 16 oz 1 gallon = 4 quarts 1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints kilo (= 1000) milli (= 1/1000) centi (= 1/100) deci (= 1/10) 1. 24.6 grams = ? kg 2. 6.3ft="7inches 1. 24.6 grams x 1 kg /1000 g = 0.0246 kg 2. 6.3ftx12in/1ft=75.6inches please always use the correct units in your final answer Question 3
Click this link to access the Periodic Table. This may be helpful throughout the exam. Do the conversions shown below, showing all work: 1. 28C=7°K 2. 158F =7-C 3. 343K=7°F 1. 28C+ 273 =301°K °C - °K (make larger) +273 2. 158F-32+1.8=70°C °F - °C (make smaller) -32 +1.8 3. 343°K-273=70°Cx1.8 + 32 =158°F oK - °C - °F Question 4 Click this link to access the Periodic Table. This may be helpful throughout the exam. Be sure to show the correct number of significant figures in each calculation. 1. Show the calculation of the mass of a 18.6 ml sample of freon with density of 1.49 g/ml 2. Show the calculation of the density of crude oil if 26.3 g occupies 30.5 ml. 1. M=DxV=149x18.6 =27.7¢ 2. D=M/V=26.3/30.5=0.862 g/ml Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. 1. 3.0600 contains ? significant figures. 2. 0.0151 contains ? significant figures.
3. 3.0600 + 0.0151 = ? (give answer to correct number of significant figures) 1. 3.0600 contains 5 significant figures. 2. 0.0151 contains 3 significant figures. 3. 3.0600 =+ 0.0151 = 202.649 = 203 (to 3 significant figures for 0.0151) Question 6 Click this link to access the Periodic Table. This may be helpful throughout the exam. Classify each of the following as an element, compound, solution or heterogeneous mixture and explain your answer. 1. Coca cola 2. Calcium 3. Chili 1. Coca cola - is not on periodic table (not element) - no element names (not compound) appears to be one substance = Solution 2. Calcium - is on periodic table = Element 3. Chili - is not on periodic table (not element) - no element names (not compound) appears as more than one substance (meat, beans, sauce) = Hetero Mix Question 7 Click this link to access the Periodic Table. This may be helpful throughout the exam. Classify each of the following as a chemical change or a physical change 1. Charcoal burns 2. Mixing cake batter with water
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3. Baking the batter to a cake 1. Charcoal burns - burning always = chemical change 2. Mixing cake batter with water - mixing = physical change 3. Baking the batter to a cake - baking converts batter to new material = chemical change Question 8 Click this link to access the Periodic Table. This may be helpful throughout the exam. 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: *Z,+-» 31 protons, 39 neutrons, 28 electrons 31 protons = Gas;, 39 neutrons = *Ga.,, 28 electrons = (+31 - 28 = +3) — 70Ga31+3 Question 9 Click this link to access the Periodic Table. This may be helpful throughout the exam. Name each of the following chemical compounds. Be sure to name all acids as acids (NOT for instance as binary compounds) 1. PF;s 2. Al (CO,), 3. H.CrO. 1. PF; - binary molecular = phosphorus pentafluoride 2. Al(CO:s); - nonbinary ionic = aluminum carbonate
3. H.CrO, - nonbinary acid = chromic acid incorrect fluoride prefix Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Write the formula for each of the following chemical compounds explaining the answer with appropriate charges and/or prefixes and/or suffixes. 1. Carbon monoxide 2. Manganese (IV) acetate 3. Phosphorous acid 1. Carbon monoxide - ide = binary, mono =10 = CO 2. Manganese (IV) acetate - Mn+, C,H;0,* = Mn(C,H;0,). 3. Phosphorous acid - nonbinary acid of H + phosphite (PO;3) = H;PO; MODULE 2 EXAM Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal. 1. Al(CO;); 2. CHNO.CI 1. 2Al+ 3C + 90 = 233.99
2. 8C+6H+ N+ 40 + Cl = 215.59 Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the number of moles in the given amount of the following substances. Report your answerto 3 significant figures. 1. 13.0 grams of (NH.).CO, 2. 16.0 grams of C;H:NO.Br 1. Moles = grams / molecular weight = 13.0/96.09 = 0.135 mole 2. Moles = grams / molecular weight = 16.0 / 260.04 = 0.0615 mole Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the number of grams in the given amount of the following substances. Report your answer to 1 place after the decimal. 1. 1.20 moles of (NH.).CO; 2. 1.04 moles of C;H:NO,Br 1. Grams = Moles x molecular weight = 1.20 x 96.09 = 115.3 grams 2. Grams = Moles x molecular weight = 1.04 x 260.04 = 270.4 grams Question 4 Click this link to access the Periodic Table. This may be helpful throughout the exam.
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Show the calculation of the percent of each element present in the following compounds. Report your answer to 2 places after the decimal. 1. (NH.)CrO, 2. CgH:NOI 1. %N =2x14.01/152.08 x 100 = 18.43% %H = 8 x 1.008/152.08 x 100 = 5.30% %Cr =1 x 52.00/152.08 x 100 = 34.20% %0 = 4 X 16.00/152.08 x 100 = 42.08% 2. %C=8x12.01/261.05 x 100 = 36.80% %H = 8 x 1.008/261.05 X 100 = 3.09% %N =1 x 14.01/261.05 x 100 = 5.37% %0 =1 X 16.00/261.05 x 100 = 6.13% %l =1 x 126.9/261.05 x 100 = 48.61% Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the empirical formula for each compound whose elemental composition is shown below. 38.76% Ca, 19.87% P, 41.27% O 38.76% Ca /40.08 = 0.9671/0.6416 =1.5x2 =3 19.87% P/ 30.97 = 0.6416/0.6416 =1x2 =2 41.27% O /16.00 =2.579/0.6416 =4x2=8 - Ca;P.0 Question 6
Click this link to access the Periodic Table. This may be helpful throughout the exam. Balance each of the following equations by placing coefficients in front of each substance. 2. 3. CeHs + Oz - COz + Hzo AS + Oz - ASZOS Al(SO,); + Ca(OH), - AI(OH); + CaSO, 2 C5H5 + 15 Oz - 12 COz + 6 Hzo 4As + 50, - 2As0; Al(SO.); + 3 Ca(OH). - 2 Al(OH); + 3 CaSO. Question 7 Click this link to access the Periodic Table. This may be helpful throughout the exam. Classify each of the following reactions as either: N Combination Decomposition Combustion Double Replacement Single Replacement HzSO4 - SOs + Hzo S + 3F, - SF H,SO. - S0, + H,O = Decomposition, One reactant — Two Products S + 3F, - SF,= Combination. Two reactants- One product
3. H. + NiO - Ni + H,O = Single Replacement, Hydrogen displaces metal ion Question 8 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the oxidation humber (charge) of ONLY the atoms which are changing in the following redox equations. Na,HAsO; + KBrO; + HClI - NaCl + KBr + H;AsO, NazHASO3 + KBr03 + HCI - NaCI + KBI‘ + H3ASO4 Na,HAsO:: Na is metal in group | = +1 (total is +2), H = +1, each O is -2 (total is -6), so As is +3 H;AsO.: H is +1 (total is +3), each O is -2 (total is -8), so As is +5 KBrO;: K is metal in group | = +1, each O is -2 (total is -6), so Br is +5 KBr: K is metal in group | = +1, so Bris -1 Question 9 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the balancing of the following redox equation, including the determination of the oxidation humber (charge) of ONLY the atoms which are changing. KMnO, + KI + HO - KIO; + MnO, + KOH Mn compounds x 2 ; | compounds x 1 = 2KMnO4 + 1Kl + 1 H,O - 1KIO; + 2MnO, + 2KOH KMnO, + KI + HO - KIO; + MnO, + KOH KMnO,: K is metal in group | = +1, each O is -2 (total is -8), so Mn is +7
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MnO,: Each O is -2 (total is -4), so Mn is +4 Kl: Kis metal ingroup | = +1,solis -1 KIO;: K is metal in group | = +1, each O is -2 (total is -6), so l is +5 Since Mn (on left side) is +7 and Mn (on right side) is +4: Mn changes by 3 Since | (on left side) is -1 and | (on right side) is +5: | changes by 6 Multiply Mn compounds by 2 and | compounds by 1 and after balancing other atoms = 2KMnO, + 1KI + 1HO - 1KIO;: + 2MnO. + 2KOH Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the balanced equation and the calculation of the number of moles and grams of CO, formed from 20.6 grams of CsH.. Show your answers to 3 significant figures. CsHe + Oz - COz + Hzo Molar mass of CO2 = 44.01 g/mol mass of CO2 = 1.5822 ml x 44.01 g/mol =69.63 grams mass of CO2 = 69.63 grams moles of CO2 = 1.58 mole 2 CsHe + 15 Oz e 12 COz + 6 Hzo 20.6 g /(6 x12.01 + 6 x 1.008) = 20.6/ 78.108 = 0.2637 mole x 12/2 = 1.58 mole CO.; 1.582 mole CO, x (12.01 + 2 x 16.00) = 69.6 g CO.
MODULE 3 EXAM Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. A reaction between HCI and NaOH is being studied in a styrofoam coffee cup with NO lid and the heat given off is measured by means of a thermometer immersed in the reaction mixture. Enter the correct thermochemistry term to describe the item listed. 1. The type of thermochemical process 2. The amount of heat released in the reaction of HCl with NaOH 1. Heat given off = Exothermic process 2. The amount of heat released = Heat of reaction Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. 1. Show the calculation of the final temperature of the mixture when a 22.8 gram sample of water at 74.6°C is added to a 14.3 gram sample of water at 24.3°C in a coffee cup calorimeter. c (water) = 4.184 J/g °C 2. Show the calculation of the energy involved in freezing 54.3 grams of water at 0°C if the Heat of Fusion for water is 0.334 kJ/g 1- - (mwarn H20 X Cwarn H20 X Atwarn HZO) = (mcool H20 X Ccool H20 X Atcool HZO)
-[22.8 g x 4.184 )/g °C X (Tmx - 74.6°C)] = [14.3 g x 4.184 )/g °C X (T - 24.3:C)] - [95.3952 J/°C X (Tmx - 74.6°C)] = [59.8312 J/°C X (T - 24.3°C)] Tmix = 55-2°C 2. Qs =MXAHuw.. = 54.3 g x 0.334 kJ/g = 18.14 k] (since heat is removed) = - 18.14 kJ Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the amount of heat involved if 35.6 g of H.,S is reacted with excess O, to yield sulfur trioxide and water by the following reaction equation. Report your answer to 4 significant figures. 2HS(g) + 30,(g0 - 2S0O.(g) + 2H,0(g) AH = - 1124 k] 1 mol H2s = 34.1 g of H2S = 603.2 kj (35.69/34.1 g) x -603.2 k}] = (1.0439) x (-603.2 k]) = -629. 7 kj 2H.S(g) + 30.,(g) - 2S50,(g) + 2H,0(g) AH = -1124 k] AH, is for 2 mole of H,S reaction uses 35.6 g of H,S = 35.6/34.086 = 1.044 mole of H.,S g = AH. x new moles / original moles q = -1124 kJ x 1.044 mole of H.S / 2 mole H.S = 586.7 given off Question 4 Click this link to access the Periodic Table. This may be helpful throughout the exam.
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Show the calculation of the heat of reaction (AH..) for the reaction: 3 C(graphite) + 4H,(g) - GCH:(Q) by using the following thermochemical data: C (graphite) + 0,(g) - CO,(qg) AH = - 393,51 K] 2H.(s) + 0O,(g) - 2HO() AH = -571.66 K] C3H8 (g) + 5 Oz (g) - 3 COz (9) + 4 HzO(I) AH = - 22200 kJ Your Answer: 3 (C (graphite) + O.(g) - CO.(9) AH = - 393.51 kJ) 2(2H.(s) + 0O:.(g) - 2 H0(l) AH = - 571.66 kJ) 3CO.(g) + 4HO() - GCH.(9) + 50:(9) AH = + 2220.0 kJ 3 C(graphite) + 4H.(g0 - CiH:(g) AH... = - 103.85 k) AH,. = 3(- 393.51) + 2(- 571.66) + 2220.0 = - 103.85 Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the heat of reaction (AH..) for the reaction: 2CH.(g) + 30.(g) - 2CO(g) + 4H.O( by using the following AH/data: AHCCH, (g) = -74.6 kJ/mole, AH°CO (g) = -110.5 kJ/mole, AH°H,O (l) = -285.8 kj/mole 2CH.(g) + 30.(g0 - 2CO(g + 4HO( AHe¢CH., (g) = -74.6 kJ/mole, AH?CO (g) = -110.5 kJ/mole, AH#H,0 (I) = -285.8 kjJ/mole
AH.. = 2(+74.6) + 3(0) + 2(-110.5) + 4(-285.8) = - 1215.0 kJ/mole Question 6 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the new temperature of a gas sample has an original volume of 740 ml when collected at 710 mm and 35°C when the volume becomes 460 ml at 1.20 atm. P1V1/T1 = P2V2/T2 T2 = 1.2 x 460x308 / 0.934 x 740 = 245.98 kelvin = -27.17°C (P xV)/T = (Px V:)/T; 740 mi/1000 = 0.740 liters =V, 710 mm/760 = 0.934 atm = P, 460 mi/1000 = 0.460 liters = V; 1.20 atm = P; 35:C + 273 = 308K =T, (0.934) x (0.740) / 308 = (1.20) x (0.460) / T, Ti= 246 °K Question 7 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the volume occupied by a gas sample containing 0.632 mole collected at 710 mm and 35°C. PxV = nx RXxT
0.632 mole =n R = 0.0821 710 mm/760 = 0.934 atm =P 35°C + 273 = 308°K =T (0.934) xV = (0.632) x(0.0821) x (308) V = 17.1 liters Question 8 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the volume of CO, gas formed by the combustion of 25.5 grams of CsH, at 20°C and 1.25 atm. The combustion of benzene (CsHs) takes place by the following reaction equation. 2 CsH (g) + 15 O, (g) - 12 CO, (g) + 6 H.O (g) (MW = 78) (MW = 32) (MW = 44) (MW = 18) 2CHs(g) + 150,(g) - 12CO.,(g) + 6 H,0O(g) 25.5 grams 37.75 liters l tbyV=nRT/P = (1.9614)(0.0821)(293)/1.25 0.3269 mol o 12/2 x 0.3269 mol Question 9 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the mole fraction of each gas in a 1.00 liter container holding a mixture of 7.60 g of N, and 8.40 g of O, at 25°C. Ny = g/ (MW,.) = 7.60 g/ 28.02 = 0.2712 mol No: = go./ (MW,,) = 8.40 g / 32.00 = 0.2625 mol
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Xw. = 0.2712 / (0.2712 + 0.2625) = 0.5082 Xo. = 0.2625 / (0.2712 + 0.2625) = 0.4918 Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molecular weight of an unknown gas if the rate of effusion of carbon dioxide gas (CO,) is 1.83 times faster than that of an unknown gas. (rNZ /runknown)2 = MWunknown / MWC02 (1.83/1)* = MW.own / 44.01 MW..cown = (1.83)2 x 44.01 = 147.4 calculation error: minus 2.5 MODULE 4 EXAM Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. Write the subshell electron configuration (i.e.1s? 2s?, etc.) for the Fe,; atom. Fe,, = 26 electrons = 1s2 252 2p¢ 352 3p* 452 3d° Question 2 Click this link to access the Periodic Table.This may be helpful throughout the exam. Write the subshell electron configuration (i.e.1s? 2s?, etc.) for the S, atom. S.s = 16 electrons = 1s2 252 2p¢ 3s? 3p* Question 3 Click this link to access the Periodic Table.This may be helpful throughout the exam.
Write the subshell electron configuration (i.e.1s? 2s?, etc.) for the P,; atom and identify which are valence (outer shell) electrons and determine how many valence electrons there are. P.- = 15 electrons = 1s? 2s? 2p¢ 3s2 3p* = 5 valence electrons Question 4 Click this link to access the Periodic Table.This may be helpful throughout the exam. * For the following question, use the "Insert Math Equation" tool (indicated by the X icon on the toolbar and then choose arrows from the window which opens). Using up and down arrows, write the orbital diagram for the Ti,, atom. Ti, = 1s? 252 2p¢ 352 3p¢ 452 3d? LI 2 It I I P I A T IO A Question 5 Click this link to access the Periodic Table.This may be helpful throughout the exam. * For the following question, use the "Insert Math Equation" tool (indicated by the X icon on the toolbar and then choose arrows from the window which opens). Using up and down arrows, write the orbital diagram for the V,; atom and identify which are unpaired electrons and determine how many unpaired electrons there are. V,; = 152 252 2p¢ 352 3p¢ 452 3d® teotl tete e tLote et 1Lttt = 3 unpaired electrons
Question 6 Click this link to access the Periodic Table.This may be helpful throughout the exam. Write the subshell electron configuration (i.e.1ls? 2s?, etc.) for the Fe, atom and then identify the last electron to fill and write the 4 quantum numbers (n, I, m, and m,) for this electron. Fe,s = 152252 2p¢ 352 3pc4s23d¢: n=3,1=2, m =-2, m, = -1/2 Question 7 Click this link to access the Periodic Table.This may be helpful throughout the exam. 1. Arrange the following elements in a vertical list from smallest (top) to largest (bottom) atomic size: CI, F, Br 2. Arrange the following elements in a vertical list from highest (top) to lowest (bottom) electronegativity: S, Si, P 3. Arrange the following elements in a vertical list from lowest (top) to highest (bottom) ionization energy: S, Te, Se Your Answer: 1. F Cl Br Se
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Question 8 Click this link to access the Periodic Table.This may be helpful throughout the exam. 1. List and explain which of the following atoms forms a positive ion with more difficulty. Sn or | 2. List and explain which of the following is the smaller atom. Sn or Te 1. | forms a positive ion less easily than Sn since ionization potential increases as you go to the right in a period which means that | with the higher ionization potential requires more energy to lose an electron and form a positive ion so it does so less easily. 2. Te is smaller than Sn since atomic size decreases as you go to the right in a period which means that Te which is further to the right is smaller. Question 9 Click this link to access the Periodic Table.This may be helpful throughout the exam. On a piece of scratch paper, draw the orbital configuration of the Cs atom and use it to draw the Lewis structure for the C; atom. Then choose the correct Lewis structure for C; from the options listed below. A.) C: B.) :C. C.) :C. D.) :C: E.) C
Question 10 Click this link to access the Periodic Table.This may be helpful throughout the exam. On a piece of scratch paper, draw the orbital configuration of the As:; atom and use it to draw the Lewis structure for the As;; atom. Then choose the correct Lewis structure for the As,; from the options listed below. A) :As: B) :As. C) :As. D) :As: E) :As: Your Answer: As33 = 152 252 2ps¢ 352 3p¢ 3d° 452 4p3 (ORI N 1 A A A A A A A A A A A A O B MODULE 5 EXAM Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the determination of the charge on the ion formed by the Se., atom. Se., (honmetal = gain electrons) 1s2 2s2 2p¢ 3s2 3p¢ 4s2 3d 4p* @gain 2e - Se? Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. L Il N =2
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=2.14 Br=2.8 Using the electronegativities from the table above, show the determination of the polarity of each different type of bond in the following molecule H—C=0 H H-C bond electronegativity difference = 2.5-2.1 =0.4 <0.5 bond is Nonpolar C-0O bond electronegativity difference = 3.5-2.5=1.0 1.6-0.5 bond is Polar Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. On a piece of scratch paper, draw the Lewis structure for the ClO;t ion. Then choose the correct Lewis structure for ClO;! from the options listed below. -1 1 -1 A (A) :0::Cl: O: (B) :0:Cl: O: (C) :0:Cl::0: (D) :0::Cl::0: U ) ) 0 ) ) :Ei: ) ) :E.'l: (B) Question 4 Click this link to access the Periodic Table. This may be helpful throughout the exam.
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On a piece of scratch paper, draw the Lewis structure for H,SO;. Then choose the correct Lewis structure for H,SO; from the options listed below. (A) HO::8:0:H (B) H:O:S::0:H (C) H:O:S::0:H (D) H:0:S: O:H :0: :0: :0: :0: (D) Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. Determine the electron geometry and explain your answer for the Si atom in SiH.. The Si atom in SiH, has 4 groups of electrons around it in its Lewis structure, therefore, its electron geometry would be tetrahedral. Question 6 Click this link to access the Periodic Table. This may be helpful throughout the exam. Determine the hybridization and explain your answer for the Si atom in SiH.. The Si atom in SiH, has 4 groups of electrons around it in its Lewis structure, therefore, its hybridization would be sps3. Question 7 Click this link to access the Periodic Table. This may be helpful throughout the exam. Determine the shape and explain your answer for HCN.
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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 central C atom, the shape would be linear. Question 8 Click this link to access the Periodic Table. This may be helpful throughout the exam. 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 PH; to determine the molecular polarity of PH; explaining your answer in detail. The shape of PH; is triangular pyramid and since the P-H bonds are all nonpolar, PH; would be nonpolar since all the bonds are nonpolar. Question 9 Click this link to access the Periodic Table. This may be helpful throughout the exam. Is KNO; Polar, lonic or Nonpolar and List and Explain whether it is Soluble or Insoluble in Water? KNO:; is lonic since it has lonic bonds and since it is lonic it is Soluble in water. Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Arrange the following compounds in a vertical list from highest boiling point (top) to lowest boiling point (bottom) and explain your answer on the basis of whether the substance is Polar, Nonpolar, lonic, Metallic or Hydrogen bonding: Ar, NH;, Zn, HBr, NaBr
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Please note in this question you are not being asked to list BPs but the compounds in a list from highest to lowest BP on the basis of the type of compound. NaBr (ionic) = Zn (metallic) NH; (Hydrogen Bonding) HBr (Polar) Ar (Nonpolar) MODULE 6 EXAM Question 1 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. List and explain if each of the following solutions conducts an electric current: sodium chloride (NaCl), hydrochloric acid (HCI) and sugar (CsH.,0s). Sodium chloride (NaCl) is an ionic compound and conducts since it forms ions in solution. Hydrochloric acid (HCI) is a polar compound and conducts since it forms ions in solution. Sugar (C:H.,O¢) is a molecular compound but does not form ions in solution so it does not conduct. Question 2 Not yet graded / 10 pts Explain how and why the presence of a solute affects the boiling point of a solvent. The presence of a solute raises the boiling point of a solvent by lowering the vapor pressure of the solvent. With this lower vapor pressure, more heat (a higher boiling point) is required to raise the vapor pressure to atmospheric pressure. Question 3 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam.
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Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare, List from lowest freezing point to highest freezing point. GaCls, Al,(SO.);, Nal, MgCl, Your Answer: Al2(S04)3 -> 2Al+, 3504 2 = = 5 ions -> most ions = lowest freezing point GaCl2 -> Ga3*+, CI- =4 ions MgCl2 = 3 ions Nal = 2 ions - > least ions = highest freezing point GacCl, — Ga+* + 3CI Atf = 1.86 x 0.1 x4 = 3rd lowest FP AIz(SO4)3 - 2 AI+3 + 3 504'3 Atf - 1.86 X 0.1 X 5 - lowest FP Nal - Na+ + I Atf = 1.86 x 0.1 x 2 = highest FP MgCl, - Mg+= + 2CI Atf =1.86x 0.1 x3 = 2nd lowest FP FP: AlL(SO.):. < GaCl, < MgCl. < Nal Question 4 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the mass percent solute in a solution of 20.8 grams of Ba(NQO:). in 400 grams of water. Report your answer to 3 significant figures.
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Mass % = (20.8 / 20.8 + 400) x 100 = 4.94% Question 5 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molality of a solution made by dissolving 28.5 grams of CsH.,:0s in 400 grams of water. Report your answer to 3 significant figures. molality = (Gouwe/ MW) / (Geonen / 1000) molality = (28.5 / 240.208) / (400 / 1000) = 0.297 m Question 6 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molarity of a solution made by dissolving 35.9 grams of Mg(NO:;), to make 400 ml of solution. Report your answer to 3 significant figures. Molarity = (Qsoe/ MW) / (Mloopen: / 1000) Molarity = (35.9 / 148.325) / (400 / 1000) = 0.605 M Question 7 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam.
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Show the calculation of the mass of Ba(NO:), needed to make 250 ml of a 0.200 M solution. Report your answer to 3 significant figures. Molarity = (moles) / (Ml.we: / 1000) 0.200 = (moles) / (250 / 1000) Moles = 0.200 x 0.250 = 0.0500 Moles = (G.oue/ MW) 0.0500 = (gsolute/ 261-55) Juwe= 0.0500 x 261.55 = 13.1 g Question 8 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the volume of 0.667 M solution which can be prepared using 37.5 grams of Ba(NOs).. m0|essolute = gsolute/ MW moleS.we = 37.5 9/ 261.55 = 0.1434 mol Molarity = moles / (mL /1000) 0.667 = 0.1434 / (mL / 1000)
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mL /1000 = 0.1434 /0.667 = 0.2150 mL = 0.2150 x 1000 = 215 mL Question 9 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the boiling point of a solution made by dissolving 20.9 grams of the nonelectrolyte C,H:O, in 250 grams of water. K, for water is 0.51, BP of pure water is 100°C. Calculate your answer to 0.01°C. mOIaIity = (gsolute/ MW) / (gsolvent/ 1000) molality = (20.9/120.104) / (250 / 1000) = 0.6961 m At, =Ko xm = 0.51 x 0.6961 = 0.355°C BPsqution = BPsoIvent - Atb = 100°C + 0-355 = 100.35°C Question 10 Not yet graded / 10 pts Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the molar mass (molecular weight) of a solute if a solution of 14.5 grams of the solute in 200 grams of water has a freezing point of -1.35°C. K: for water is 1.86 and the freezing point of pure water is 0°C. Calculate your answer to 0.1 g/mole.
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Atf= fom molality = At: /K= 1.35/1.86 = 0.726 m mOIaIity = (gsolute/ MW) / (gsolvent/ 1000) 0.726 = (moles) / (200 / 1000) Moles = 0.726 x 0.200 = 0.1452 0.1452 = (14.5 / MW) MW = 14.5/0.1452 = 99.9 FINAL EXAM Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. 1. Convert 845.3 to exponential form and explain your answer. 2. Convert 3.21 x 10+ to ordinary form and explain your answer. 1.Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places = 8.453 x 102 2.Convert 3.21 x 10° = negative exponent = smaller than 1, move decimal 5 places = 0.0000321 Question 2
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Click this link to access the Periodic Table. This may be helpful throughout the exam. Do the conversions shown below, showing all work: 1. 246°K=7°C 2. 45:C=7°F 3. 18F =7K 1. 246°K-273 =-27 °C °K - C (make smaller) -273 2. 45Cx1.8+ 32=113°F °C - °F (make larger) x 1.8 + 32 3. 18F-32+18=-7.8+ 273 = 265.2°K °F - oC - °K Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the number of moles in the given amount of the following substances. Report your answer to 3 significant figures. 1. Moles = grams / molecular weight = 12.0/132.15 = 0.0908 mole 2. Moles = grams / molecular weight = 15.0/179.17 = 0.0837 mole Question 4 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the percent of each element present in the following compounds. Report your answer to 2 places after the decimal.
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1. Al(SO.)s 2. GCH;NOBr 1. %Al =2x26.98/342.17 x 100 = 15.77% %S = 3 x 32.07/342.17 X 100 = 28.12% %0 =12 x 16/342.17 = 56.11% 2. %C =7 x12.01/199.02 x 100 = 42.24% %H =5 x1.008/199.02 x 100 = 2.53% %N =1 x 14.01/199.02 = 7.04% %0 =1 x16.00/199.02 x 100 = 8.03% %Br = 79.90/199.02 x 100 = 40.15% Question 5 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the heat of reaction (AH...) for the reaction: 2CHs(g) + 50,(g) - 4CO(g) + ©6HO( by using the following thermochemical data: AH°C,H: (g) = -84.0 kJ/mole, AH°,CO (g) = -110.5 kJ/mole, AH H,O (I) = -285.8 kJ/mole 2CHs(g) + 50,(g) - 4CO(g + 6HOI(I AH°C,H: (g) = -84.0 kJ/mole, AH°CO (g) = -110.5 kJ/mole, AH*H,O (I) = -285.8 kj/mole AH.. = 2(+84.0) + 5(0) + 4(-110.5) + 6(-285.8) = - 1988.8 kJ/mole Question 6
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Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the calculation of the number of moles of a 1.25 liter gas sample collected at 740 mm and 28-C. PxV = nxRXxT 740 mm/760 = 0.974 atm =P R = 0.0821 1.25 liters =V 28°C + 273 =301 K=T (0.974) x (1.25) = n x(0.0821) x (301) n = 0.0493 mole Question 7 Click this link to access the Periodic Table.This may be helpful throughout the exam. Write the subshell electron configuration (i.e.1s? 2s?, etc.) for the Fe, atom and then identify the last electron to fill and write the 4 quantum numbers (n, I, m, and m,) for this electron. Fe,, = 1s2 252 2p¢ 352 3pc4s2 3de: n=3,1=2, m =-2, m, = -1/2 Question 8 Click this link to access the Periodic Table.This may be helpful throughout the exam. 1. List and explain which of the following is the smaller atom. CorN 2. List and explain which of the following atoms holds its valence electrons more tightly. Br or |
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1. N is smaller than C since atomic size decreases as you go to the right in a period which means that N which is further to the right is smaller. 2. Br holds its valence electrons more tightly than | since electronegativity decreases as you go down a group which means that Br which is further up the group has the higher electronegativity and therefore the higher attraction for its valence electrons. Question 9 Click this link to access the Periodic Table. This may be helpful throughout the exam. Is CH. Polar, lonic or Nonpolar and List and Explain whether it is Soluble or Insoluble in Water? CH, has all nonpolar bonds which makes it Nonpolar and since it is Nonpolar it is Insoluble in water. Question 10 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the determination of the charge on the ion formed by the Br;; atom. Bris (nonmetal = gain electrons) 1s2 2s2 2ps 352 3p¢ 4s2 3d* 4ps gain le - Br: Question 11 Explain how and why the presence of a solute affects the boiling point of a solvent. The presence of a solute raises the boiling point of a solvent by lowering the vapor pressure of the solvent. With this lower vapor pressure, more heat (a higher boiling point) is required to raise the vapor pressure to atmospheric pressure. Question 12 Click this link to access the Periodic Table. This may be helpful throughout the exam.
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Show the calculation of the molar mass (molecular weight) of a solute if a solution of 13.5 grams of the solute in 200 grams of water has a freezing point of -1.20°C. K: for water is 1.86 and the freezing point of pure water is 0°C. Calculate your answer to 0.1 g/mole. Atf= Kf XmMm Atf= fom molality = At/ K, = 1.20 / 1.86 = 0.645 m mOIaIity - (gsolute/ MW) / (gsolvent/ 1000) 0.645 = (moles) / (200 / 1000) Moles = 0.645 x 0.200 = 0.129 0.129 = (13.5/ MW) MW =13.5/0.129 = 104.7
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Do the indicated arithmetic and give the answer to the correct number of significant figures. I.   II. 0.71 + 92.2 III. 934 × 0.00435 + 107 IV. (847.89 - 847.73) + 14673
between significant digits are always significant. figures. Example: 5,007 has 4 significant figures. 3) Trailing zeros in a number are significant only If the number contalns a decimal polnt. Example: 100.0 has 4 significant figures. 100 has 1 significant figure. 4) Zeros in the beginning of a number whose only function Is to place the decimal polnt are not significant. Example: 0.0025 has 2 significant figures. 5) Zeros following a decimal significant figure are significant. Example: 0.000470 has 3 significant figures. 0.47000 has 5 signlficant figures. Determine the number of significant figures in the following numbers. 1. 0.02 6. 5,000. 4. 2. 0.020 2. 7. 6,051.006 3. 501 3 8. 0.0005 1. 4. 501.0 9. 0.1020 5. 5,000 10. 10,001 5 Determine the location of the last significant place value by placing a bar over the digit. Example: 1.700) 9,010.0 4.7x10-8 10,8 00,000 3.0Tx 1021 0.00410 1. 8040 6. 90,100 0.0300 699.5 2,000X162 0.90100 2. 0.0300 7. 4.7 x 108 3. 699.5 8. 10,800,000. 3.…
55. Round the number on the left to the number of significant figures indicated by the example in the first row. (Use scientific notation as needed to avoid ambiguity.) Rounded to Rounded to Rounded to 4 Significant 2 Significant 1 Significant Figures Number Figures Figure 1.45815 1.458 1.5 8.32466 84.57225 132.5512
g and a glass of water has a mass of 0.050 kg. An oxygen atom has a mass of 2.66 × 10 Use this information to answer the questions below. Be sure your answers have the correct number of significant digits. What is the mass of 1 mole of oxygen atoms? Round your answer to 3 significant digits. How many moles of oxygen atoms have a mass equal to the mass of a glass of water? Round your answer to 2 significant digits. 0 g x10 X S
because calculating the area is your final step. If you want to use that information in further calculations you must use the unrounded value. Lne Base. It is correct to round to the requested number of significant figures in Part A Part C Using the correct answer from Part B, calculate the volume of a rectangular prism with a length of 5.6 cm, a width of 2.1 cm, and a height of 6.6 cm. H=6.6 cm W=2.1 cm L = 5.6cm Enter your answer to two significant figures and include the appropriate units. P Pearson Copyright © 2021 Pearson Education Inc. All rights reserved. Terms of Use | Privacy Policy Permissions Contact Us| 9:07 AM 82% 69°F 8/30/2021 Insert Delete PrtSc
General Chemistry Laboratory I Lab 1 Significant Figures and Dimensional Analysis Student name PRE-LABORATORY QUESTIONS 1. How many significant figures are in each measurement? (1) 0.0044500 m (2) 5.003 km (4) 1.00 x 10³ s (5) 0.002 mm 2. Define accuracy and precision. 3. Convert 1.567 kilometers to centimeters.ccurate d 4. Convert 3.70 milligrams to kilograms. O Date: Page 1 IN-LABORATORY WORKSHEET 1. Give the number of significant figures in each of the following numbers. (1) 1,278.50 (2) 120,000 (4) 0.0053567 (5) 670 (7) 8.002 (8) 823.012 (10) 2.60 (Prepared by W. Song) (3) 10 dm (6) 10,000 m (3) 90,027.00 (6) 0.00730 (9) 0005789
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Leaming Goal: To learn how to round an answer to the correct number of significant figures. Multiplication and division When multiplying or dividing, the final answer is rounded to the same number of significant figures as the measured number with the fewest significant figures. When we report a measurement in science, we are careful to report only digits known with certainty, plus a final digit that is recognized to be uncertain. We call this set of digits significant figures. Often, these numbers are used in calculations. When we use a calculator, the calculator does not automatically account for significant figures, so we have to do so ourselves. There are some basic rules for handling significant figures in calculations as shown in the table and outlined to the right. Addition and subtraction When adding or subtracting, the final answer is rounded to the same number of digits to the right of the decimal point as the measured number with the fewest digits to the right of the decimal…
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