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In Exercises 53 and 54, an object moving vertically is at

College Algebra | 9th Edition | ISBN: 9781133963028 | Authors: Ron Larson ISBN: 9781133963028 204

Solution for problem 53 Chapter 6

College Algebra | 9th Edition

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College Algebra | 9th Edition | ISBN: 9781133963028 | Authors: Ron Larson

College Algebra | 9th Edition

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19
3
Problem 53

In Exercises 53 and 54, an object moving vertically is at the given heights at the specified times. Find the position equation for the object. At second, feetAt seconds, feetAt seconds, feet

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CHAPTER 4 TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY Aqueous Solutions: Strong and Weak Electrolytes 10. The electrolyte designation refers to how completely the dissolved solute breaks up into ions. Strong electrolytes completely break up into ions when in water, weak electrolytes only partially break up into ions (less than 5% usually), and nonelectrolytes do not break up into ions when they dissolve in water. The conductivity apparatus illustrated in Figure 4.4 of the text is one way to experimentally determine the type of electrolyte. As illustrated, a bright light indicates many charge carriers (ions) are present and the solute is a strong electrolyte. A dim light indicates few ions are present so the solute is a weak electrolyte, and no light indicates no ions are present so the solute is a nonelectrolyte. 11. Solution A: 4 molecules ; solution B: 6 molecules ▯ 1.5molecules 1.0 L 4.0 L 1.0 L 4 molecules 2 molecules 6 molecules 3molecules : C n o i t u l o S ▯ ; solution D: ▯ 2.0 L 1.0 L 2.0 L 1.0 L Solution A has the most molecules per unit volume so solution A is most concentrated. This is followed by solution D, then solution C. Solution B has the fewest molecules per unit volume, so solution B is least concentrated. 12 . Only statement b is true. A concentrated solution can also contain a nonelectrolyte dissolved in water, for example, concentrated sugar water. Acids are either strongeak electrolytes. Some ionic compounds are not soluble in water, so they are not labeled as a specific type of electrolyte. 13. a. Polarity is a term applied to covalent compounds. Polar covalent compounds have an unequal sharing of electrons in bonds that results in unequal charge distribution in the overall molecule. Polar molecules have a partial negative end and a partial positive end. These are not full charges as in ionic compounds but are charges much smaller in magnitude. Water is a polar molecule and dissolves other polar solutes readily. The oxygen end of water (the partial negative end of the polar water molecule) aligns with the partial positive end of the polar solute, whereas the hydrogens of water (the partial positive end of the polar water molecule) align with the partial negative end of the solute. These opposite charge attractions stabilize polar solutes in water. This process is called 56 K g 8 . 8 3 g n i s u d a e t s n i , a t 5 7 . 1 e h t f o L m 4 1 1 g n i s u d a . L 0 0 . 1 f o e m u l o CHAPTER 4 SOLUTION STOICHIOMETRY 19. Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 40.00 g/mol × L 0 0 5 2 . 0 = H O a N s s a M 1molAgNO 20. 10. g AgNO ×3 169.9g 21. Mol solute = volume (L) × molarity 0.30 mol AlCl l C l o M ▯= 0.1000 L × l C g M 2s) ▯ Mg (aq) + 2 Cl (aq) ▯ 0.60 mol MgCl l C l o M = 0.0500 L × + ▯ ) s ( l C a N ▯ Na (aq) + Cl (aq) ▯ 0.40 mol NaCl l C l o M = 0.2000 L × 0 3 . 0 f o L m 0 . 0 0 1 M AlCl 3ontains the most moles of Cl ions. 22. a. M = 0.100 mol Ca(NO ) Ca(N3 2 0.100 L 2+ O N ( a C 3 2s) ▯ Ca (aq) + 2 NO (aq); 2.5mol Na SO . b M Na2S4= 1.25L a N 2O 4s) ▯ 2 Na (aq) + SO H N g 0 0 . 5 . c Cl × 1mol NH Cl 4 53.49g NH Cl M = 0.0935mol NH Cl NH4Cl 0.5000L + H N 4Cl(s) ▯ NH 4aq) + Cl (aq); = n o i t u l o . L m 0 . 0 0 1 f o e m u l o v l a n i f a CHAPTER 4 SOLUTION STOICHIOMETRY 2.00 ▯10 ▯6g steroid 1000 mL ▯ 100.0 mL L 26. Stock solution: 1mol Mn 2▯ n M g 4 8 5 . 1 2+× 54.94 g Mn 2▯ 0 1 × 3 8 8 . 2 = : s n i a t n o c A n o i t u l o S × L m 0 0 . 0 5 1L ▯ 2.883 1000 mL 1.442 ▯ 10▯3 mol = y t i r a l o M 1000.0 mL : s n i a t n o c B n o i t u l o S 1L 1.442 × L m 0 0 . 0 1 1000 mL ▯ ▯5 = y t i r a l o M 1.442 ▯ 10 mol 0.2500 L : s n i a t n o c C n o i t u l o S ▯5 ▯3 5.768▯ 10 0 1 × 0 0 . 0 1 L × L 5.768 ▯ 10▯7 mol = y t i r a l o M 0.5000 L 5.0 ng Hg 27. a. 5.0 ppb Hg in water = mL H O ▯9 5.0 ▯ 10 g Hg ▯ 1mol Hg mL 200.6 g Hg ▯9 . b 1.0 ▯ 10 g CHCl 3 ▯ 1mol CHCl mL 119.4 g CHCl 10.0 μg As = s A m p p 0 . 0 1 . c ▯ mL . L m 0 . 0 0 1 o t k c o t s m p p 0 . 0 0 0 . L m 0 . 0 0 1 o t k c o t s m p p 0 . 0 0 0 . L m 0 . 0 0 1 o t k c o t s m p p 0 . 0 0 0 . L m 0 . 0 0 1 o t k c o t s m p p 0 . 0 0 0 . e l b u l o s e r a s t c u d o r p e l b i s s d n a e t a h p s o h p . e l b u l o s e r a s t c u d o r p e l b i s CHAPTER 4 36. 37. 38. × L 0 0 5 1 . 0 a N L m . 0 5 2 = 39. O N g A l o M l C a C l o M O N g A l o m 0 2 0 . 0 l C d l C d : e r a s n o i r o t a t c e p s e h t f o CHAPTER 4 SOLUTION STOICHIOMETRY b. The spectator ion concentrations will be one-half the original spectator ion concentrations in the individual beakers because the volume was doubled. Or using moles, 6.00 mol K▯ 8.00 mol NO ▯ = 1.50 M and M ▯= 3 = 2.00 M. The concentration of 4.00 L NO3 4.00 L ▯ ▯ OH ions will be zero because OH is the limiting reagent. From the drawing, the number of Cu 2+ionswilldecreasebyafactoroffourastheprecipitateforms. Becausethe 2+ volume of solution doubled, the concentration of Cu ionswilldecreaseby afactorof eight after the two beakers are mixed: ▯ ▯ M Cu▯= 2.00 M ▯ ▯ = 0.250 M ▯ ▯ Alternately, one could certainly use moles to solve for 2▯ Cu ▯ 2▯ 2+ 3.00 mol OH 1mol Cu Mol Cu reacted = 2.00 L × ▯ ▯ = 3.00 mol Cu L 2 mol OH 2.00 mol Cu 2▯ uC l oM 2+ present initially = 2.00 L × = 4.00 mol Cu L 2+ uC s s e c xE present after reaction = 4.00 mol ▯ 3.00 mol = 1.00 mol Cu 2▯ 1.00 mol Cu M Cu2▯ = 0.250 M 2.00 L ▯ 2.00 L 1molCu(OH) 2 97.57 g Cu(OH) Mass of precipitate = 6.00 mol KOH × ▯ 2 molKOH molCu(OH) Mass of precipitate = 293 g Cu(OH) 2 41. 2 AgNO (3q) + Na C2O (a4) ▯ Ag CrO 2s) +42 NaNO (aq) 3 0.100molAgNO 3 1mol Na C2O 4 161.98g Na C2O 0.0750 L × L ▯ 2 molAgNO ▯ mol Na CrO 3 2 4 42. XCl 2aq) + 2 AgNO (3q) ▯ 2 AgCl(s) + X(NO ) (a3 2 × l C g A g 8 3 . 1 1molAgCl ▯ 1molXCl 2 = 4.81 × 10 ▯3mol XCl 143.4g 2 molAgCl 2 1.00g XCl 2 ▯3 = 208 g/mol; x + 2(35.45) = 208, x = 137 g/mol 4.91 ▯ 10 mol XCl 2 The metal X is barium (Ba). ) H O ( l A g 7 0 1 l A % s s I l T g 4 2 8 1 l T % s s O S a B g 2 3 0 5 = % s s a m e g a r e L 0 2 7 3 0 e t i s a l g u o d g 0 0 9 2 . 0 = = e t i s a l g u o d % s s O S a C g 6 3 M g 2 4 . 1 , m e l b o r p e h t m o = s s a m r a l o u m a 3 2 = M s s a m c i m o t a , ) 0 0 : s e v i g r e b m u n t s a d n o i t a r t i C o s l a s i d i c a c i n i m r a c f o × 0 8 . 1 s n i a t n o c e s a b g n o r t s b a e h t s a h e s a × L m . 0 0 0 1 = s s a m l a t o t : n CHAPTER 4 58. Because KHP is a monoprotic acid, the reaction is: NaOH(aq) + KHP(aq) ▯ H O(l) + ) q a ( P K a N × H O a N L 6 4 0 2 0 . 0 = P H K s s a M 59. HCl and HNO are strong acids; Ca(OH) and RbOH are strong bases. The net ionic equation that occurs is H (aq) + OH (aq) ▯ H O(l). H l o M × L 0 0 0 1 . 0 H O l o M × L 0 0 0 2 . 0 H O f o s s e c x e n a e v a h e W H O l o m 0 0 3 0 . 0 M 60. 39.47 × 10 H N f o y t i r a l o M 61. Ba(OH) (aq) + 2 HCl(aq) ▯ BaCl (aq) + 2 H O(l); H (aq) + OH (aq) ▯ H O(l) 0 1 × 0 . 5 7 0 1 × 8 8 . 1 0 1 × 0 . 5 2 2 0 1 × 8 4 . 2 The net ionic equation requires a 1 : 1 mol ratio between OH mol H ratio is greater than 1 : 1, so OH is in excess. Because 1.88 × 10 neutralized by the H , we have (2.48 ▯ 1.88) × 10 = 0.60 × 10 mol OH in excess. × L 0 C ( s i a l u m r o f r a l u c e l o m e h t t a h t t n e m e l e e h t s i d t c a e h t , s d n u o p m o c c i n o H N ( e g e C ( 4 + e b t s u m m u i r e c f o e t _ _ _ _ _ _ _ _ _ _ _ _ d _ e _ c _ u _ d_ e _ R _ _ _ _ _ _ _ _ _ _ d n a , ) e t a t s n o i t a d i x o 0 e h t r e v l i s f o e t a t s n o i t a d i x o e : d e : d e r r e f s n a r t s n o r t c e l e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ e 6 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ e 0 1 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ . s n o r t c e l e 6 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ O C 6 + _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 0 1 0 1 × 0 8 . 3 0 = × I 4 2 e _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ H 8 H 6 : O I K g 0 1 0 1 CHAPTER 4 I . c S 2 0 1 × 0 0 . 5 2 . d 0 1 × 0 5 . 1 × L 0 0 0 5 . 0 . e O I K g 7 0 . 1 e c a l P Additional Exercises 81. Mol CaCl present = 0.230 L CaCl × l C a C f o e m u l o v e h T 0 1 × 3 3 . 6 H e m u l o V 82. There are other possible correct choices for the following answers. We have listed only three possible reactants in each case. a. b. Na SO , Na CO , and Na PO would form precipitates with the Ca c. O N ( l A f o e l o m 1 h t i w t c a e r . O % 4 . 0 4 d n a , H % 9 0 . 5 , l C % × e F g 9 5 O N ( e F % s s 0 1 × 0 ) 4 4 0 1 × 0 9 2 . 2 ( = l C K % 0 . 0 0 1 = l × C g 3 × B g : r e b m u n t s e l l a m s e h t y b g n i C s i a l u m r o f l a c i r i p m × L 0 7 4 3 . = F R A B f o s s a m CHAPTER 4 SOLUTION STOICHIOMETRY 1molCr(OH) Mol NaOH used = 2.06 g Cr(OH) ×3 103.02g to form precipitate NaOH(aq) + HCl(aq) ▯ NaCl(aq) + H O(l) 2 0.400molHCl Mol NaOH used = 0.1000 L × L to react with HCl totalmolNaOH 6.00▯ 10 ▯2mol M NaOH= = volume 0.0500 L 90. 3 (NH 4 2rO 4aq) + 2 Cr(NO 2 3aq) ▯ 6 NH NO (aq) + Cr (CrO ) (s) 0.307 mol ▯2 × L 3 0 2 . 0 L = 6.23 ×10 mol (NH ) CrO 0.269mol ▯2 × L 7 3 1 . 0 = 3.69 ×10 mol Cr(NO ) L 0.0623 mol/0.0369 mol = 1.69 (actual); the balanced reaction requires a 3/2 = 1.5 to 1 mole ratio between (NH4 2CrO4and Cr(NO 2 3 Actual > required, so Cr(NO ) (the denominator) is limiting. ▯2 1molCr (2rO ) 4 3 × 9 6 . 3 10 mol Cr(NO 2 3 2 molCr(NO )2 3 actual yield = 0 8 8 . 0 , actual yield = (8.34 g)(0.880) = 7.34 g Cr (CrO ) isolated 8.34g 1mol 91. Mol of KHP used = 0.4016 g × = 1.967 × 10 204.22g Because 1 ▯3le of NaOH reacts completely with 1 mole of KHP, the NaOH solution contains 1.967 × 10 mol NaOH. ▯3 1.967 ▯10 mol 7.849 = H O a N f o y t i r a l o M ▯3 ▯ 25.06▯ 10 L ▯3 = y t i r a l o m m u m i x a M 1.967▯ 10 mol 7.865 25.01▯ 10▯3 L ▯ ▯3 = y t i r a l o m m u m i n i M 1.967 ▯ 10 mol ▯ 7.834 25.11 ▯ 10▯3L H O a N L 7 1 5 3 : o s , d e t c a e r s a w A H g 0 1 2 . 3 = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 92 98. Let x = mass of NaCl, and let y = mass K SO . So b P : r u c c o s n o i t c a e r o w T b P Molar mass of NaCl = 58.44 g/mol; molar mass of K SO = 174.27 g/mol; molar mass of PbCl2= 278.1 g/mol; molar mass of PbSO = 303.3 g/mol x = moles NaCl; 58.44 l C b P f o s s a m x (1/2)(278.1) + 58.44 : s n o i t a u q e o w t e v a h e W ) 9 7 3 . 2 ( x + (1.740)y = 21.75 and x + y = 10.00. Solving: x = 6.81 g NaCl; 99. Zn(s) + 2 AgNO Let x = mass of Ag and y = mass of Zn after the reaction has stopped. Then x + y = 29.0 g. Because the moles of Ag produced will equal two times the moles of Zn reacted: 0 . 9 1 ( ▯ y) g Zn : g n i y f i l p m i S 0 1 × 9 5 0 . 3 g n i t u t i t s b u S x = 29.0 ▯ y into the equation gives: 0 1 × 9 5 0 . 3 : g n i v l o S 1 8 5 . 0 ▯ (3.059 × 10 )y = 0.269 ▯ (9.268 × 10 )y, (2.132 × 10 )y = 0.312, y = 14.6 g Zn 14.6 g Zn are present, and 29.0 ▯ 14.6 = 14.4 g Ag are also present after the reaction is stopped. 100. 0.2750 L × 0.300 mol/L = 0.0825 mol H ; let y = volume (L) delivered by Y and z = volume (L) delivered by Z. l o m / g 8 . 7 8 1 = 0 9 . 9 7 + 9 . 7 0 1 : n o i t a m r o f n 0 1 ( + : : e c u d o r p n a c e w d n a , g n i t i m i u c c o I 0 1 × 0 8 . 1 ( , d e n r u _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ e F ( _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ : d e n i a t n o c 0 1 0 1 × 3 6 . 8 = g n i t = g n i t a l p u C o w t o t s u C o t d e t r e v n o c s i u C l l A 0 1 0 a + ) 5 5 . 3 6 ( 3 + ) 3 . 7 3 1 ( 2 u C h t o B : a t a d I _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ c u d n o c r e p u + ) 3 + ( 0 5 . 1 + ) 2 + ( 0 5 , t l u s e r e m a s e h t e v i g s t n : u C f o e t a t s n a N f o 2 e h t e c n a l a b o t s n o i t a c + 1 K % 1 6 s i e r u t ) 7 5 5 6 . 0 ( + y ) 3 7 2 5 . 0 ( ) 0 5 3 6 . 0 ( = s l a t e m g 4 8 6 4 . 0 s n o i ) 0 5 ) 3 7 2 5 . 0 ( = 4 0 100 CHAPTER 4 SOLUTION STOICHIOMETRY 2+ 2+ O S i T 4(aq) + Ba (aq) ▯ BaSO (s) + 4i (aq) Because there is a 1 : 1 mole ratio between mol BaSO and m4l TiSO , you ne4d 9.970 × 10 ▯4mol of TiSO . Because 0.1472 g of salt was used, the compound would have a molar 4 mass of (assuming the TiSO for4ula): × 079 . 9 / g 2741 . 0 10 ▯4 mol = 147.6 g/mol From atomic masses in the periodic table, the molar mass of TiSO is 143495 g/mol. From just these data, TiSO 4eems reasonable. Chris thinks the salt is sodium sulfate, which would have the formula Na SO . 2he 4quation is: 2+ + aN 2SO (4q) + Ba (aq) ▯ BaSO (s) + 24Na (aq) As with TiSO , 4here is a 1:1 mole ratio between mol BaSO and molN4 2SO .4For sodium sulfate to be a reasonable choice, it must have amolarmassofabout147.6g/mol.Using atomic masses, the molar mass of Na SO i2 144.05 g/mol. Thus Na SO is als2 re4sonable. Randy, who chose gallium, deduces that gallium should have a 3+ charge (because it is in column 3A), and the formula of the sulfate would be Ga (SO )2. Th4 3quation would be: aG 2(SO 4 3aq) + 3 Ba (aq) ▯ 3 BaSO (s) + 24Ga (aq) 3+ The calculated molar mass of Ga (SO 2 wou4 3be: 0.1472g Ga (S2 ) 4 3 ▯ 3molBaSO 4 = 442.9 g/mol ▯4 molGa (SO ) 9.970▯10 molBaSO 4 2 4 3 Using atomic masses, the molar mass of Ga 2(SO 4 3s 427.65 g/mol. Thus Ga 2(SO 4 3s also reasonable. Looking in references, sodium sulfate (Na SO ) 2xis4s as a white solid with orthorhombic crystals, whereas gallium sulfate Ga (SO2) is4 3itepowder.Titaniumsulfateexistsasa green powder, but its formula is Ti (SO2) . B4 3use this has the same formula as gallium sulfate, the calculated molar mass should be around 443 g/mol. However, the molar mass of Ti2(SO 4 3s 383.97 g/mol. It is unlikely, then, that the salt is titanium sulfate. To distinguish between3+a SO 2ndGa4(SO ) , o2e co4 3 dissolve the sulfate salt in water and add NaOH. Ga would form a precipitate with the hydroxide, whereas Na SO woul2 4 not. References confirm that gallium hydroxide is insoluble in water. CHAPTER 4 SOLUTION STOICHIOMETRY Marathon Problems 111. M(CHO )2 2q) + Na 2O (4q) ▯ MSO (s4 + 2 NaCHO (aq) From the balanced molecular equation, the moles of M(CHO ) present initially must equal the moles of MSO (s) formed. Because moles = mass/molar mass and letting A = the 4 atomic mass of M: massMSO 4 O S M l o m 4= ▯ molarmassof MSO 4 massM(CHO ) 2 2 O H C ( M l o m 2 2= molarmassof M(CHO ) O S M l o m e s u a c e B 4= mol M(CHO ) 2 2 9.9392 9.7416 ▯ , (9.9392)A + 894.9 = (9.7416)A + 935.9 A M 96.07 A M 90.04 41.0 A M = = 207; from the periodic table, the unknown element M is Pb. 0.1976 From the information in the second paragraph, we can determine the concentration of the KMnO s4lution. Using the half-reaction method, the balanced reaction between MnO and 2- C2O 4 is: C 5 O 2(aq) + 2 MnO (aq) + 16 H (aq) ▯ 10 CO (g) + 2 Mn (aq) + 8 H O(l) 2 4 4 ▯ 1mol Na C O O n M l o M 4 = 0.9234 g Na2C2O4▯ 134.00 g 0 1 × 6 5 7 . 2 = ▯ mol MnO 4 2.756 M KMnO4▯ M MnO4▯▯ ▯ volume From the third paragraph, the standard KMnO solution reacts with formate ions from the filtrate. We must determine the moles of CHO ▯ KMnO 4 solution. The moles of CHO2 ions present initially are: mol Pb(CHO ) O H C ( b P g 6 1 4 7 . 9 2 2▯ 2 2 297.2 g O H C f o s e l o m e h T 2▯present in 10.00 mL of diluted solution are: f i ( g 0 5 . 6 3 = g 0 l o l o m 0 0 0 . 1 l

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Chapter 6, Problem 53 is Solved
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Textbook: College Algebra
Edition: 9
Author: Ron Larson
ISBN: 9781133963028

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In Exercises 53 and 54, an object moving vertically is at