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trifluoride A nitroglycerine molecule, C,H, (NO,), , contains 3 C atoms, 5 H atoms, 3 N atoms, x100% equals the total negative charge, moles of OH” from Ca (OH ) : Thus, NH,NO), (reaction 1) produces the (2) mass of aluminum are soluble in water; Al(OH)x(s) is not soluble in water. (39.9624u x 0.99600)+(35.96755ux 0.00337)+K(37.96272u x 0.00063) = 39.948u Thus, the molar mass of X = ——=— “This result assumes that a neutral atom is involved. (b) (a) molecular mass (mass of one molecule). A Thus, Ox(g) is the limiting reactant, and all of the O(g) is consumed. 1 mol C,HBrCIF, 1mol F atoms 39.0983u = (0.932581x 38.963707u)+(0.000117x39.963999 1) +(0.067302x “K) The correct equation ís 2KCIO, (s) ->2KCl(s) +30, (g). Net: 10 1 (aq)+2 MnO, (aq)+16 H'(aq)>5 1,(s)+2 Mn” (aq)+8 H,0(1) S(s)+ 6 0H (aq)+2 OCT (aq)+2H,0() > 50,” (aq)+3 H,0() +2 CI” (aq)+40H' times a ratio of the two volumes. Three of the four remaining atoms mass of *C]= mass of 'Fx1.8406 =18.998ux1.8406= 34.968 u (e) mass KCI=28.3g O, x molO, 2molKCL, 74558KO oe PROBLEMAS RESUELTOS DE QUÍMICA GENERAL CINÉTICA QUÍMICA - 4 de 12 fGrupo A: ASPECTOS TEÓRICOS DE LA CINÉTICA QUÍMICA CINÉTICA - A-01 Cuando se adiciona un catalizador a un sistema reaccionante, decir razonadamente si son ciertas o falsas las siguientes propuestas, corrigiendo las falsas. excess of 100.0 g (it is actually 113 g). 74.6 g. Thus, a 1.00 M KCI solution contains 74.6 g KCI per liter of solution. 0.605g H,Ox A IL x 0.163mol AgNO, e 1 mol Na,S We know the isotopic mass of '*C ¡is 12 u. of —1 in H,O, (aq) to an O.S. 10mm lem? We could use a 100.0-mL flask and a 5.00-mL pipet, a 1000.0-mE flask and A chemical formula is a short-hand representation of a chemical species: atom, ion, or 1 km (a) cobalt-60 Co (b)phosphorus-32 ¿P (c)iodine-131 'I (d) sulfur-35 ¿S 1000mL 1L soln Tmol NaCl (c) amount of Br, =11.3 kg Br, «2008, mol Br =70.7 mol Br, Pouring the milk into the jug is a process that is subject to error; there can be slightly mv 0.485mol_ 32.048 CH,OH__ImL 2 (Two atoms of chromium have a mass 81.398 Zn0 1molZnO 1 mol ions Reduction: (MnO,' (aq)+2 H,0(1)+3 e” > MnO, (s)+4 OH" (aq) 3x2 x 100 % = 79,89 % by mass Cu barium ion (a) Cr” chromium(II) ion 134,00 gNa,C,O, 1 molNa,C,O, $ molC,O,” l1m moles of T” in KI solution = 250.0 mLx = 0.0219 mol” Enter the email address you signed up with and we'll email you a reset link. (e) Sr(CIO 4) strontium perchlorate (bf KHSO4 potassium hydrogen textbook. (20 Ejercicios) by JoeJerez alloy > volume of alloy. (b) 8. (b) no. 4.37%P 59, mass Na=155mL solnx x and thus a bit more than 1 mole of S atoms. . 53. Volume of concentrated AgNO, solution PCI, (1) +4H,0(1) > H,PO, (aq) + 5HC1(aq) We begin with the quantity of S atoms =4.58x10* mol S, x Reaction: P, (s)+6CL, (g) >4PCI, (1) . 331.218 331.21 molecule (1 x 10% nm) (£) No reaction occurs, based on the information in Table 5-3. Only after you have made a determined effort to solve each problem should you turn to the lysine mass = 1.15 mol Nx - Na,CO, (s) —*5 2 Na” (aq)+CO,” (aq) Multiply each of the mole numbers by 4 to obtain an empirical formula of C,¿H,¿O, 1£ lmL ” 62.1368 1mol The problem is most easily solved with amounts in millimoles. Mg is a main-group metal in group 2. drops 28:3: 1.2810" +2=0.640x10""C =6.40x10""C =4e a) La variación de entalpía de la reacción se . Some of the solutions given in the manual differ Step 2: Balance each skeleton half-equation for O (with H,O ) and for H atoms (with H”). 153.33kg POCI, l g Rb sulfurs must be +4. mol of stearic acid x bromine” Br 35 35 45 80 Oxidation-Reduction (Redox) Equations mass of '*F= mass of '*Cx1.5832 =12,00000u x1.5832 =18.998u 45, Asasalt: NaHSO, (aq) > Na” (aq)+HSO, (aq) =3.508 O, (8) “Rb(natural) mol Cl = x + x The nucleus of '¿Ba contains 56 protons and (138 — 56) = 82 neutrons, Thus, the percent 15.9994 g. 1000 g x 100.00 g solution The balanced chemical Rb content (ppm) = 10% = 159 ppm Rb (a) The seven SI base units are those from which all other units are derived. 375 mL. 26.98g Al 2molAl 1molH, 9 daysx 22 -216.000h 3minx P_-0.050h 44sx =0.012h % 22.1747 g H/mol decane lkm 98m 60s Agregar a Mis Libros. Then convert the number of chloride ions to the mass of MgCl». Mm 100 The second 4x12.0g C)+(5x1.0g H 53.0 1. (b)_ 2NO(g)+0, (8) >2N0, (2) 1f lin. (b) Units of Measurement Chapter 5: Introduction to Reactions in Aqueous Solutions Page 5-9 fig.) Of these species, only in 3¿Cr is more 1.000 g P (d) an equation that summarizes the overall result of a process consisting of several multiples of e. of an uncombined, neutral element is O, PP — -1.00088u 25.22 CO(NH,), . 10A First, determine the mass of iron that has reacted as Fe?” with the titrant. Solucionario Cálculo Multivariable - Dennis G. Zill. mol Fe, [Fe (CN), ], mol Fe, [Fe(cn), ], mole OE Fenol Fe,[Fe(CN), Unbalanced reaction: N2Ha(g) + N20x(g) —>H20(g)+ N2(g) «(e 2 | =1.00x10' em 20.768 0x 222. the people of Western Canada, continue to be an inspiration to young and old alike. Page 2-1 =3.176 mmol H* 500.0 mL soln lLsoln 1mLHC,H,O, 60.05g HC,H,O, Enunciados de los problemas resueltos de TERMOQUÍMICA. the point, the difference between the larger number of ions and the smaller number, HCl(ag) reacts with active metals and some anions to produce a gas. (2) Ifan element forms a cation with charge 2+, itis in group 2(2A) Since there are 108 1250mLCH,0H_ 0.7928 1mol CH,OH B is a nonmetal in group 13(3A). Thus, boron-11 ¡is the isotope present in greater abundance. The symbols must be arranged in order of Thus, the O.S. This will help you realize that there is often more than one Thermochemistry Reactores catalíticos heterogéneos Diseño de reactores heterogéneos, Cinética química Velocidad de reacción: Tiempo v/s Concentración Molar, MANUAL DE PRÁCTICAS DE CINÉTICA Y CATÁLISIS, PRÁCTICAS DE LABORATORIO INTEGRAL II (FISICOQUÍMICA II, R E S U M E N F I N A L D E Q U Í M I C A 2016 QUÍMICA MENCIÓN, DESARROLLO DE LA CINÉTICA QUIMICA DE LA REACCIÓN DE TRANSESTERIFICACIÓN DE LA OLEINA DE PALMA, TEXTO DEL ESTUDIANTE MARÍA ISABEL CABELLO BRAVO, UNIVERSIDAD NACIONAL EXPERIMENTAL " FRANCISCO DE MIRANDA " ÁREA DE TECNOLOGÍA DEPARTAMENTO DE QUÍMICA " Compendio teórico de Fisicoquímica " Elaborado por, III Reacciones químicas y sus leyes fundamentales, CUESTIONES Y PROBLEMAS DE LAS OLIMPIADAS DE QUÍMICA III. =1.298mol O +1.298 >1.00mol O O.S. amount of chlorine by the fixed mass of phosphorus with which they are combined. for two Hg is +2 and each Hg has O.S.=+1. d = 2 —=115.76 mi/h Chapter 4: Chemical Reactions Page 4-8 Li,O is an integral multiple of the empirical formula. g ore 0.1002 Mg of N is +2 on the left and -3 on the right side of this equation; N is +(2 mol Ox16.00g O) =146.2 g/mol This is a binary molecular compound: sulfur 6mol Cl, x 70.91 g Cl, 7.16L A chemical formula is a short-hand representation of a chemical species: atom, ion, or molecule. 16.00g O (c) A substance is a pure form of matter; it is either an element or a compound. e 2 9 10 MF andone O? It is calculated as the mass of 14.018 N Chapter 1: Matter - lts Properties and Measurement Page 1-9 Chapter 3: Chernical Compounds Page 3-4 ImolPb——_207.2gPb 0.100g Mg Step 5: Simplify by removing species present on both sides of each half-equation, and 100.20gC,Hs 1 molC,H, 1molC 1 mol CO, Química general 10/e. Consequently there are two oxidation reactions and no reduction reactions, the appropriate units for each. 0.3856 * —_— XK As an acid: HSO,” (aq)+ OH” (aq) > H,0(1)+5S0,* (aq) lkg 159.88 Br 1g KO, x 1 mol KO, x 3 mol O, (UO” (aq)+ HO) > UO,” (aq)+2H" (aq)+2 e"]x3 is 216.59 g/mol; and Pb(NO, ) is 331.2 g/mol. l (i) HCOy hydrogen carbonate ion GQ CON cyanide ¡on The hydrogen ion is the lightest positive ion available. Then determine the number of ions in 1.0 g of ZnO. 1molSO, _ _lmolS has just as many protons as neutrons. Mno, (aq) +4 H" (aq) > Mno, (s) +2 H,0() We combine these two equations and solve the resulting expression. The total for the two chlorines must be +2. 1 kmolPOCL, Chapter 2: Atoms and the Atomic Theory Page 2-12 21720 nes) =2.172 lm The cited reaction is 2 Al(s) + 6HC1(aq) > 2 AICI, (aq) + 3H, (8) The HCl(ag) solution 55, high: "C=3(*F-32)=¿(118'F-32)=47.8'0 =48 %C When copper(ID) sulfate is strongly heated, it decomposes to give SOx(g) and CuO(s). 453.68 Next we need to find the number of moles of anhydrous copper(TI) sulfate and =7.92x10* g solution 1 d 1neck] 107.868u - 55.421 =0.4816'" Ag = 52.45u 19 Ay =108.9 u average atomic mass of argon= 39.803u +0.121u +0.024u = 39,948u C,,H,¿0, : 100.0 gsample 74.093 gBa(OH), 1 molCa(OH), solutions in the manual. element N decreases during this reaction, meaning that NO, (g) is reduced. =24 g acetic acid To see if the Law Oxidation: (Sn” (aq) > Sn” (aq)+2 e” yx3 or 4.7 x 10* m? 0.350 g of rock Thus, the total for 4 oxygens must be 8. name or the symbol of the element involved), the number of electrons (or some way to three Mg” andrwo N? For one conversion factor we need the molar mass of ZnO. imol Na,S x Emol AS 247.80g Ag,S lt would be highly unlikely that all of For every 4 moles of AgNO», 2 moles of l2(s) are produced. (o) volume=65.0 gx > mL =58.6 mL ethylene glycol drop 4: 1.28x107* +8=0,160x10""C =1.60x10"C =le We use the formula expression, so that the resulting equation has the same number and type of abundances converted to fractional abundances by dividing by 100. 1L soin 1 mol CHO, fusionado. This As a molar ratio we have —-_€Q---_—_—_— =4.99 - 5 The empirical formula is CuSO¿*5 H20, L mE soln 2mmol AgNO, - 0.650mmo! lm INTRODUCTION TO REACTIONS Chapter 2: Atoms and the Atomic Theory Page 2-3 1 e (1.0 x 10” nm) mass of *Br= mass of *C1x2.3140=34.968ux 2.3140 =80,917u CrCl, The O.S. Oxidation: (C,H,OH(aq)+5 OH" (aq) > C,H,O, (aq)+4 H,O(1)+4 e. 3x3 (8mol Cx12.0u C)+(5mol Hx1.0u H)+(1mol Nx14.0u N)+(1mol 0x16.0u 0) =131 u. 1 E R-22%T12>5gKCIO, 2molKCIO, ImolO, 24. =18.95u+2.499u+2.861u =24.31u following francium will have atomic number =87+32 =119. U is an inner transition metal, an actinide. 1000. oxygen by difference) and transform these molar amounts to the simplest integral amounts, In this reaction, iron is reduced from Fe** (aq) to Fe?” (aq) and manganese is reduced may be rational numbers whose decimal equivalents are easy to recognize. x Xx = 252, necklaces Chapter 2: Atoms and the Atomic Theory Page 2-6 K,CrO,molarity, dilute solution = =0.0675M ass o AB Comme Sm Y mol K,CrO, — Imol Ag,CrO, CIO” (aq) and oxygen is oxidized from an O.S. denominator by 2. Maximum mass of Pbl (calculated) This compourd is calcium bicarbonate or calcium hydrogen carbonate. IL 0350molC,H,O, 180.168 C,H,O, mass of fuel used = 9000 Ib—82 1b = 8920 lb 0.376 gore (b) [CO(NH,),]= Sample from trona: 6.93 g sample forms 11.89 g AgCl or 1.72 g AgCl per gram sample. 0.0115 moles CuSO, 8.95 x 10% g mL”. Although mass of Mg = 0.500g MgO x (c) Gas evolution: FeS(s)+2 H' (aq)> H,S(g)+ Fe” (aq) There must be one Ca?” and two Cl's: CaCL,. charge 1.602x10"%C Net: 3 C,H,OH (aq)+4 MnO, (aq) > 3 C,H,O,” (aq)+4 MnO, (s)+ OH" (aq)+4 H,0(1) 15mgFr lgF % l mol F 9. total amount OH” = 0.00543 mol from NaOH +0.00048 mol from Ca(OH), =0.00591 mol OH” mass Na,SO, -10H,0 =355 mL soln x (a) mol, ERC A 22 6gKCIO, 2molKCIO, 32.068 S =0.0677g H Of these four nuclides, only ¿Mg? Then the % C and % H are found. ImolCr(NO,), -9H,0 1molH,O FeO The O.S. (b) 5238 Hx 7 =5.18molH +1.298 > 3.99mol H height (¡.e. Xe is a noble gas with atomic The O.S. "“normalized” mass of phosphorus = E ron - 1.000 g of phosphorus 478 (d) Therefore, Rb(natural)_ _ 27.83% 100 yá 36 in 2.54 em lm Step 4: BaCl, (s)+ K,SO, (aq) > BaSO, (s)+2 KCl(aq) (a) (d) 1 mol C¿H,¡¿N,O, k 146.2 g lysine Write the two skeleton half-equations. To gain a truly deep understanding, you must practice using them, both in the **Pd _ 107.90389u Chapter 3: Chemical Compounds Page 3-19 1mol Pb 1000 Pb atoms Its acid is nitrous acid. Chemical Bonding l: Basic Concepts neon Ne?" Tn this reaction, chlorine is oxidized from an O.S. Mass percent copper = express each in terms of e=1.6x10"” C. In other words, for a The reason is that each mass of CH, (OH), POE 5 02210 "molecules 1mol C¿H, (OH), 25,012 mi 138 nucleons 1. The molar mass of KCl is explain a large number of phenomena by leaming and applying a relatively small number of formula is CH,N are those for the proton, a hydrogen ion, H”; and that for the electron. l0mm) lem! e e empirical formula C¿H, has an empirical molar mass o: you should study the Example again and then try another Practice Example, Chemistry is a MnoO," (aq) > Mo, (s) and SO, (aq) ->SO,' (aq) must more than 192 u; that isotope must be '” Ir. 150007 “1x10'4L. (e) 121.9x10*=0.001219 (d) 162x107” =0.162 10 NBx(g) +3 CLO(g) > 6 NH¿CK6) + 2 Nx(g) + 3 H20(0) The total for =221.13g/mol Cu, (OH), CO, moles). should attempt to solve one of the analogous Practice Examples. Thus, the volume for a single stearic acid molecule in nm? 10.00 mL (with correct number of sig. mass of the rock sample, and then multiplying the result by 10% to convert to ppm. will be consumed of the other reactants. = 4318 CH, (OH), Since the three percent abundances total 100%, the percent abundance of “K ¡s found by 87 (d C=0inC,H,O, Hhas O.S.=+l inits non-metal compounds; that of O = -2 This would give an empirical formula of CuO (copper (IT) oxide). 1 mi 1ft lin. 43. DISCLAIMER: Toda la información de la página web www.elsolucionario.org es sólo para uso privado y no comercial. ofisobuty!l propionate is C,H,¿O,. Simplify by removing the species present on both sides. 204”F is below the 212"F boiling mol of stearic acid. (e) An element is a substance that cannot be altered or decomposed chemically, Each of Agis 0 on the left and +1 on the right side of this equation. 9B_ The net ionic equation when solid hydroxides react with a strong acid is OH" + HE" 5 HO. In a synthesis reaction two or more substances combine to form a third. 1mol H 15. mass of carbon-12 is defined as precisely 12 u. 1.97 g alloy x 6.3g Cu total fish = 100 marked fish x —————= 360 fish = 4x10* fish The average atomic mass of boron is 10.811, which is closer to 11.009305 than to (c) Acetic acid mass=1.00 lb vinegar x - Stearic acid mass = 4.03x10* moleculesx appear on both sides of an equation are “cancelled.” The term also is used to describe than NH). First compute the mass of fuel remaining KMnO, The O.S. moles of AgNO; 6 mol K,Cr mol K,C1O, £gNO» Density is necessary to determine the mass of the vinegar, and then the mass of acetic acid. 00 mL. Mass of water present in hydrate = 2.574 g - 1.833 g=0.741 g H20 CHAPTER 3 57, The molarity unit can be interpreted as millimoles of solute per milliliter of solution. 2.2x10 E 1kg q higher temperature, 61. The total of all O.S. 240'F. 3.52x10' mL Enviar por correo electrónico Escribe un blog Compartir con Twitter Compartir con Facebook Compartir en Pinterest acetic acid in the numerator and that of the solution in the denominator, and transform to (b) If, however, you are stumped, =0.0352 kmolPOCI, V = area of base (in nm?) the question, each stearic acid molecule has a cross-sectional area of 0.22 nm, In Reduction: S,(s)+16 e” >8 S” (aq) stoichiometric quantities are two moles of KI (166.00 g/mol) for each mole of If you take this approach, you will never develop the ability to solve This time, however, different Determine mass of PCI, formed by each reactant. Ca(HCO,), HCO),” is the bicarbonate ion or the hydrogen carbonate ion. solution of the desired concentration. Thus, we would expect all other atomic masses to be slightly higher Chapter 4: Chemical Reactions Page 4-13 The volume of a rectangular column is simply its area of the base multiplied by its T” that must be added. The amount of solute (upon filtering, KC (aq) is obtained) The K ImolP, 4molPCI, 137338PCI, 53. pressure = Chapter 5: Introduction to Reactions in Aqueous Solutions Page 5-11 2 - 0.148 mol MgCI, _ 1 mol Mg” Acid-Base Reactions We must convert mass H, > amount of H, > amount of Al > mass of Al > mass of (e) PCls phosphorus pentachloride Thus, the total ImolH For the balanced equation, the order is immaterial; the relative amount of each is same property. 1mol KCIO, 3 mol O, (2) %K=5%K,0x 1.25 L soln Sample derived from manufactured sodium bicarbonate: 6.78 g sample forms 11.77 g E EME O 166 MgO (d) Precision refers to the reproducibility of an experimental measurement; accuracy mass KO, = 100.08 CO, x =323.1g KO, (>) C,H,OH(1)+60, (g) > 4C0O, (g)+5H,0(1) sodium Na 11 11 12 23 35.458 Cl In a 50-year-old chemistry textbook the atomic mass for oxygen would be 16.000: because 0.2358 Nox 1molN, y 2molN 14.0078 N There must be two H*s. This is H,SO,. The > =162.28 H,0/molCr(NO,), -9H,0 Libro “Química General” Petrucci, página 114. la ecuación balanceada, que requiere que usemos las cantidades en moles, de ambas sustancias. 15.9949u numbers, the subscript numbers. 123.908P, ImolP, — 1molPCI, and itis +5 inCIO,”. ImolCl, , 4molPCI, 137.338 PC, .34 . aluminum sulfide AP" andS” two Al” and three S* ALS, SO,” (aq)+H,0(1) > SO,” (aq) +2H' (aq) This is a binary molecular compound: BF, The name of the compound is iron(11) oxide, 20rd +20bl+30gr > 1 necklace (4) TRUE Two-thirds of the S produced does come from the H,S . 1mo!l H,O % 2 mol H AP” (aq)+3 OH” (aq) > AL(OH), (s) =4.58x 10% mol S, N is +5 on the left and +4 on the right side of this equation. 1 mmol OH 0.0962 mmol NaOH =894 g mol"! e=e o 10s Mm ME 0122 M (b) The square brackets, [], surrounding the formula of a species, are the symbol for the of each Clis —1 (rule 7). of Clis —1 in CI” 0.625 molKCI 1 mol CI Je[osas 1 «0.385 mol MgCi, 2 molCI ) Predicting Precipitation Reactions Tn the example, 0.207 g H, is collected from 1.97 g alloy; the alloy is 6.3% Cu by mass. inO, ( 8). 6.94l1u—7.0160lu= 6.01513xu —7.01601xu = -1.00088.xu Tf the seventh period of the periodic table is 32 members long, it will be the same length as Thus, the reactant that produces the smaller amount of ions is the limiting reactant, More to (1) %P=10%P,O, x This factor must be multiplied by the number of degrees Celsius above zero on the M present. This number of moles of acid oceupies 1 cm? 32 Thus the trona sample is purer (i.e., it has the greater mass percent NaHCO; ). lmo!l H,O 67. +1 on the right side of this equation; H is oxidized and thus NO must be an oxidizing describes the agreement between the measurement and the accepted value of the “Rb(natural) 72.17% (c) 2HI(aq)+Na,CO, (aq) >2 Nal (aq) + H,0(1) + CO, (8) 2.72 %(by mass Mg) (8) 740180 molO - 6.162mo1C +1.298> 4.747mol C 0.2612 g cmpd 0.2612 gempd KCl (da) substances HCl and H,; the important conversion factor comes from the balanced chemical M and x=20 Thus, the ratio of the volume of the volumetric flask to that of the pipet =3.0 x 107 mol of stearic acid x - — Thus the symbol is “¿Ag . 2 110. 4 mol P x 30.978 P The formula for stearic acid, obtained from the molecular model, is and then the number of moles of oxygen in that sample, We divide each of these (b) “Mg + "C=25.98259u +12u =2.165216 =0.438 MCI” (4) amountof Br, =2.65L Br PL, 3108B5 ImolBE 61 mol Br, we obtain the maximum amount of product when neither reactant is in excess ( i.e., Gases mass of proton + mass of electron _ 1.0073 u + 0.00055u REVIEW QUESTIONS 0.00236x 4.071x 10 Chemical Kinetics mol S atoms in 0.50 mol S,O . values and are provided (in parentheses) after each element in the following list. 1000 cm 1m ) =2.2x10* g/em? reactant with the smallest molar mass. in precisely 100 grams of the sample it is found in. 53. We see that the mass-to- (a) Libro “Química General” Petrucci, pagina 111. Certain measurements, which are subject to error. 9A — Both the density and the molar mass of Pb serve as conversion factors. 275758 ABC, _ 26 02 Ag,CO, reduced and thus H, must be a reducing agent. 0.0007409 1MnO, 1000 1 47p+6ln=A4=108. Chapter 1: Matter— Its Properties and Measurement Page 1-8 the mass of solute as does 1.00 L of this solution, 373 g. The last description is correct. oxide has less oxygen by mass, hence the empirical formula must have less oxygen or more This is C(OH). (a) product that can be formed from the other reactants, and also limit the quantity that 8 protons (characteristic of the element oxygen) and 3 neutrons, Expression (c) is incorrect because KCIO is potassium hypochlorite, but the stated product 022m (my Chapter 5: Introduction to Reactions in Aqueous Solutions Page 5-5 The boiling point of water is 100 “C, corresponding to ("M) = a 35.12M of O in its compounds is —2 (in most cases). 10 mm lem 9 -KARP biología Molecular, octava edición-, problemas basicos 2 de floyd octava edicion, Solucionario Maquinas Electricas, 5ta edición- Chapman, SOLUCIONARIO Física para Ciencias e Ingenieria - Serway - 7ma edición, Solucionario de Cálculo de Varias Variables 4ta Edición, Banco de preguntas Inmunología ABBAS Capítulos 7-11, Maquinas eléctricas Fitzgerald 6ta Edición - Solucionario, Solucionario Cálculo Multivariable - Dennis G. Zill. amount POCI, =1.00kgPCl, x =0.0121kmolPOCI, This search will, of course, be quite Chapter 2: Atom and the Atomic Theory If the difference is zero, the of 104 u, more than the formula mass of the compound.) Pb(NO,), =(5.000 —x) g. Then we have amount KI =x g KI mass NaNO, =125mL soln xx x reaction. 1 2 3 4 $ 6 7 8 9 10 11 12 13 1 mol C,H,, x 16 mol H yl mol HO 18.015 g HO 1lb -323.4g/mol Pb(C,H,), REVIEW QUESTIONS = 16.8308848 So, the number of stearic The cation is Fe”, iron(IID. 1 +2 1 100% =15.585% H Thus, there are two 1kg 1.118 1000 mL ImolP__3097gP_ Imol(NH,) HPO, SELECTED SOLUTIONS MANUAL Lucio Gelmini . (Cap. One of the primary benefits that you will obtain from your study of chemistry is the ability to =0,206M by first dividing all three by the smallest, Now determine the amount of CI” in 1.00 L of the solution. Does not characterize a specific nuclide; several possibilities exist. 7. 5A The factor 0.00456 has three significant figures. of nueleons that are neutrons is given by 29. PK 4 S,0,” (aq)+12 H,0(1)+16 e” The Transition Metals mass CO,(g)=5.00 mL vinegarx - (d) =4,3x10* mg Mel, — We have expressed each result with an additional significant figure, written as a SIMPLIFY. acid molecule. lmL soln 100.0gsoln 36.46g HCl 6molHCI 1molH, this from the data. number of necklaces = 10.0 kg beads x pr 44.018 CO, 2molCO, 1molKO, from Naci [or]= 0.438 mol NaCl % 1 mol Cl instance, cathode rays, which are beams of “free” electrons, have the same properties no or gas) as the solvent, and the solvent is the component present in the larger amount. 1.000 g P The O.S. PInsuficient data. s of Mg =2.008 MgO of His 0 on the left and (b) The text states that compound B is N¿H>. Vaso, =163mL AgNO, [MnO, (aq)+2 H,0()+3 e” -> Mno, (s)+4 OH (aq))x2 (d) mass Cl, = 0,337 mol PCl, x =35.8g Cl, ions=1.0g ZnOx XK - (e) Fe” (aq)+3 OH (aq) >Fe(OH),(s) (d) Ca” (aq)+C0,” (aq) > CaCo, (s) (OD P=+5inH,PO, The O.S. Advanced Exercises can be found in the Instructors Resource Manual. =1.14mol X 11168 1, ox P01H0_, 2m01H_ 0 1239701 Hx 88H - 012498 H (b) will produce the smallest quantity of product, That reactant will limit the quantity of The O.S. In the next two compounds, the oxidation state of chlorine is —1 (rule 7) and thus the (a) amountof Br, =8.08x10”Br, molecules:———moleBr mass POCI, =0.0121kmolPOCI, x e CH,0H]= x x Reduction: (Cl, (g)+2 e” >2 Cl (aq) yx4 78 Ofall lead atoms, 24.1% are lead-206, or 241 *Pb atoms in every 1000 lead atoms Vicio, = 594mL K,CrO, must be —]. (a) TheO.S. 821x10 EA 211x10* solvent, thus producing a less concentrated (or more dilute) solution. Mass percent oxygen = x 100% =20,11 % by mass O moles of S. The solid sulfur contains 8x0,12 mol = 0.96 mol $ atoms. ImolC,H, 125molO, 320080, The number of stearic acid molecules is: 16.008 O (a) C=-4inCH, H has an oxidation state of +1 in its non-metal compounds 6B This is easier to visualize if the numbers are not in scientific notation. 63. 0.01968 mol 1lmol H,O Net: 2 8,(s)+24 OH (aq) ->8 S”” (aq)+4 S,0,” (aq)+12 H,O(1), Copyright © 2023 Ladybird Srl - Via Leonardo da Vinci 16, 10126, Torino, Italy - VAT 10816460017 - All rights reserved, Descarga documentos, accede a los Video Cursos y estudia con los Quiz. (c) Rb(natural) = 55,55 1g of Rb convert to grams: 1000 g 1,(s) z 1 mol 1, (s) x 4 mol AgNO), (s) s 169.873 g AgNO, (s) 1L soln (b) — six thousand three hundred seventy eight kilometers=6378 km=6.378x 10* km This means that, based on the relative (0) 30208 Ao =2.515molC +0.6288 > 4,000mol C (a) Am'isacubic meter. contribution from *Ar=35.96755ux0.00337 =0.121u mg, Which is larger than 0.00515 mg. His a main-group nonmetal in group 1. (a) KCON potassium cyanide (b) HCIO hypochlorous acid Molecular mass of oxygen is the mass of one (average) molecule of O,, 31.9988 u. x o Moles of H30 = 0.927 g H20 x = 0.05146 moles of water 3. Ss (a 017 7 mtb) 158mL LL 001581 (a) Sr(NO,),(2q)+ K,SO, (aq): Sr” (aq)+S0,” (aq)> SrSO, (s) C1,O The sum of all oxidation numbers in the compound is 0 (rule 2). The 46% by mass sucrose solution is the more concentrated. will form anions will be on the right-hand side, The number of electrons “lost” when a (d) 0.0047=4.7x107 (e) 938.3=9.383x 10% (f) 275,482=2.75482x 10% 0.4816 Thus, the vapor will be detectable, 4.37%P (b) Chapter 2: Atoms and the Atomic Theory Page 2-13 The cation is iron(ID). 1000 mL 1 Lsoln 1 mol KI OH (aq)+ H'(aq)> H,0() S Chapter 5: Introduction to Reactions in Aqueous Solutions Page 5-14 (e) 2 H,0(g) + CHa(g) > COXg) +8 H(g)+ 80 Determine the Celsius temperature that corresponds to the highest Fahrenheit temperature, ofO is-2, that of Agis+1, and that of Cris +6 on both sides of this 3 Chapter 5: Introduction to Reactions in Aqueous Solutions lead” pp 82 82 126 208 A 0.10 mL sample of this solution contains: carbon atom chain with an acid group on the 1* carbon (terminal carbon atom) excess ion = 3.176 mmol H” - 3.014 mmol OH” =0.162 mmol H*. Balance electric charge by adding electrons. Mg,N, (a) HCIO chlorous acid (b) H2SOz sulfurous acid The volume ef gold is converted to ¡ts mass and then to the amount in moles. Notice that we do not have to obtain the mass of any element in this compound by lmol Cu, (OH), CO, — ImolO 6 NHx(g) +6 H'(aq) > 6 NHs'(a9) Finally, we determine the percent by mass (c) Possible products are calcium nitrate, Ca(NO»3),, which is soluble, and lead(ID) iodide, Cr,O,” The sum of all the oxidation numbers in the ion is —2 (rule 2). 35, Moles of Chloride ion =53,7% PO, immediately before the chemical formula of a species. In each case, we determine the formula with ¡ts accompanying charge of each ¡on in the each element in the sample and transform these molar amounts to the simplest integral podemos ya calcular tanto la cantidad de etanol obtenido como el calor desprendido: x = 2.46. () Ba? pa are known to just three significant figures, our results are only known that well. (a) MgBr, magnesium bromide (b) BaO barium oxide masses have been used for bath of the elements in the second reaction. 1000mm” ler” 32.07g 8molS Measured quantity: the internuclear separation quoted for H) is an estimated value lem” 2078 lmolS 1molS, 3.96 The mass of ANO; required Forthe proton : =1.044x10*g/C 1000g Igrbead 1neckl 2molP____30.97gP_ ImolCa(H,PO,), potassium is +1 (rule 3). ATOMS AND THE ATOMIC THEORY =3.58 -4, molar mass NO, = (2ma an ama Eo) = 92.02 g/mol NO, (e) 1 mol KHP yl mol OH” A mol NaOH speed = =9.83 m/s 0.01508 mol KI _ 2 mol KI 310” (aq)+ Cr,O,” (aq)+8 H' (aq) >3 UO,” (2q)+2 Cr” (aq)+4 H,0(0) amount Pb(NO, ), =(5.000 -x) g Pb(NO,),x 25,012 mi 1000 mm ls ,Imin =2.5min 23 Of course, this calculation can be performed in one step: Representing Molecules molO=0.3378 0x2LO — 00211molO +0.02111>1.00m010 Chemistry of the Living State Net: Fe? Since the hydrate has not been completely dehydrated, there is no problem with The name of each of these ¡onic compounds is the name of the cation followed by that of Expression (a) and (b) are incorrect because O(g) is not 1000mg 186.207 g Re 1 mol Re (e) The speed is used as a conversion factor. Then the expression for the weighted-average atomic mass ¡is used, with the percent reproduce someone else”s solution. containing compounds). 322.21 g Na,SO, -10H,O 2 Felo(ag) +3 Clo(g) —> 2 FeCli(aq) + 2 L(s) (unchanged) find the concepts you need to approach the problem. 2.5038 KI 1.002 g KI 10.012937. Actually, compound A is NH, but we have no way of knowing both chromiums must be +12. Chapter 2: Atoms and the Atomic Theory Page 2-7 22.38(CH,), COx solution on your own. NET: S(s)+2 OH” (aq)+2 OCT (aq) 50,” (2q)+ H,0()+2 CI (aq) oxidizing agent and as a reducing agent in this disproportionation reaction. Again, the total mass is the same before and after the reaction, = 1.00 kg I(s)x (b) An extensive property ís one that depends on the quantity of material present; an equivalent to 4,37% P. (d) CH¿CH(OH)CH, (e) HCO,H 100 yd 36 in. Whereas a chemical formula is rather (c) x height (in nm)). KO, The sum for all the oxidation numbers in the compound is 0 (rule 2). equals 0.) about chemistry. sulfur. from a +7 O.S. 34238C,, H,¿0,, empirical formula for copper(IT) oxide is CuO. massive. nuclide is composed of protons, neutrons, and electrons, none of which have integral Each cation name is the name of the metal, with the oxidation state appended in (b) Possible products are iron(III) sulfate, Fe, (SO, ) , and potassium bromide, KBr, both No reaction occurs. amounts, by first dividing all three by the smallest. Reduction: (BrO, (aq)+6 H'(aq)+6 e” > Br (aq)+3 H,O(1) +2 10 Ib, certainly (“nearly 9000 1b”) not to the nearest pound. 45. state of -2; O has been reduced and thus, O(g) (oxidation state = 0) is the oxidizing agent. ES 187 (d) 53. quimica_general_petrucci.pdf - Google Drive. _52.45u The boiling point of water can serve as our reference. SOLUTIONS MANUAL (a) mass=452mLx e = 502 gethylene glycol (a) (a) 2Clions 6.022x10“%fu. v,O, El primero de ellos procede de la ecuación de velocidad: es el exponente que afecta a la concentración de los reactivos en la ecuación de velocidad, (en este caso y ) mientras que el concepto de α β molecularidad procede de la estequiometría de la reacción: es el número de moles de cada reactivo que aparecen en la reacción ajustada (en este caso a y b). So, the average height of a stearic acid molecule = 9556 nm” _ 2.5 nm is the molecular mass of chlorophyll t(*M) = [t(*C) + 38.9]/3.96. We may alternativel y determine the mass of N by difference: A binary acid consists of hydrogen and one other element. Alternatively, note that the change in temperature in “C corresponding to a change of This is the difference between superscripted and subscripted NH,NO, The cation is NH,*, ammonium ion. 112. combinations that could be used. 8920 lb Solubility and Complex-lon Equilibria Reduction: (MnO, (aq)+2 H,O(1)+3 e” > MnO,(s)+4 OH” (aq) yx4 9% ""pg= 2:02:10 atoms “Re 1000, - 62.5% KI and Pb(NO»)z in the balanced chemical equation. Practice Examples, most of the Review Questions, half of the Exercises questions and selected mass of electron 1 Oxidation: (UO” (aq)+ H,O(1) > UO,” (aq)+2 H' (aq)+2 e 3 250.0 mL 0.423 mmol AgNO, x 1mL conc. (2) %PO,= diluted. IN AQUEOUS SOLUTIONS 1.1468 80, x =0.01789mo!S -+0.01789=1.000molS IL Imol CH,OH 0.7928 0.128 mmol HCl 1 mmolH' 1 mmol OH” of Cl=-1 (rule 7). (a) Anexact number—24 soda cans in a case. 1000 mL solution _,g 1 yy, 5.723 g of Cl 1kg Thus, the total for three oxygens must be -6. value of “one hundred.” CHEMICAL COMPOUNDS Feature Problems that are in the companion textbook, General Chemistry: Principles and Modem (d) volume=23.9 kg x =21.5 Lethylene glycol 331.21+332.00 166.008 KI =| 0.225 Lx We determine how many necklaces can . HBrO Br0O" is hypobromite, this is hypobromous acid. 221.138 100cm of 0 in Cl, (aq) to an O.S, of +1 in The molar mass for oleic acid, C1gH340», is 282.47 g mol”. ofoxygen Chapter 3: Chemical Compounds Page 3-14 moles of OH” from NaOH: element. O.S. The O.S. 9B First we find the number of rhenium atoms in 0.100 mg of the element. l gal lat lat 1E OR 331.21x= 10.00 x 166.00-332.00x First calculate the mass of water that was present in the hydrate prior to heating. L (a) ¿E is the symbol for a nuclide. Cu=1.318H _63gCn Convert this amount to a mass of Mgl, in grams. polyatomic ion equals the charge on that ion.) The ions in each product compound are determined by simply “switching the partners” of the - x =15m Actividad 1. 47. Oxidation: (2 T (aq)> 1,(s)+2 e” xs Mg atomic mass =(23.985042ux 0.7899) +(24.985837ux0.1000)+(25.982593ux0.1101) At the point of stoichiometric balance, amount KI=2 x amount Pb(NO, ), intensive property is like a quality; it does not depend on the quantity of material 1mole x 284.58 The net ionic equation is: 78.058 Na,S ImolNa,S 1molAg,S 10B_ The balanced equation provides stoichiometric coefficients used in the solution. DEDICATION vpo 207287 100% -=64.07% Pb aluminum nitrate Aluminum is Al'*; the nitrate jon is NOy”. =0.624g Na 80.008 (b) =3.515 x 10? of Fe =+2 (rule 2). For one conversion factor we need the molar mass ofMgCl, . 1mol CH, C,H, molecule 0.0693 mol AICI, e 1000 mL the Rb content in the rock sample in ppm by mass by dividing the mass of Rb by the total for the molecule. 31 % *"K =100.0000% -93.2581%-—6.7302% =0,0117% Mult. . mL of carbon disulfide, with a density of 1.26 g/mL, should have a mass somewhat in of There are two sources of OH: NaOH and Ca(OH)z. mass after reaction = magnesium nitride mass + 2.505g nitrogen =3.034g Each chlorophyll molecule contains one Mg atom, which makes up 2.72 % of the total mass Z (a) 100km de) =1.00x10% m? 4730225-Ejercicios-Resueltos-De-Nomenclatura-Organica- 1/2 Downloaded from staging.deliciousbrains.com on by guest Ejercicios Resueltos De Nomenclatura Organica moles of CuSO4 = 1.833 g CuSO4 x _Imol CusO, | The distance between any pair of planetary bodies can only be determined through of space. (a) Solution mass= 1.00 kg sucrose x difference; there is no oxygen present in the compound. 1000g tmb IL = 10.0 y stearic acid x mol stearicacid__ 3.515 x 10? (a) 1 mL HCl(2q) 1 mmol HCl 3 H,0()+ S(s) >S0,” (aq)+6 H" phosphorus trichloride. We need to convert between the Vaso =1.0008 H, x 1mol H, y 2molAl_ 26.98 g y 00. 6.022x10*C,H, molecules , —3atoms by 100. Thus, 2.72 % of the molecular mass is Mg (24.305 g mol”). Answer (c), butanoic acid is the most appropriate name for this molecule. 1000mL 1L soln 2 mol AgNO, magnesium nitride mass = 3.034 g —2.505g = 0.529g magnesium nitride The symbol “ —25; ” indicates that the mixture is heated to produce the reaction. Nuclear Chemistry The purpose of this manual is to help you master many of the fundamental chemical principles number of F atoms = 12.15mol C,¿HBrCIF, x ———=———x 395: 10% 8 acid 9 8.95 x 10% gofacid. (Mm ? Metals, nonmetals, metalloids, and noble gases are color coded in the periodic moles of K¿CrO4 = C x Y = 0.0855 M x 0.175 L sol = 0.01496 moles K¿CrO, [HC] = 0.00591 mol OH x 1 mol H _ el mol Hal, 1000 mL soln =0.130 M Enter the email address you signed up with and we'll email you a reset link. Step 2: (12 p, 12m), Cr (24p,23 m), $Co*" (27 p, 33 n), and FCT (17 p, 18m). amount € = 73.278 Cx == 6,100 mol € 0.7625 —>8.000molC Multiply by 2 (whole $) 2 NzHa(g) + N20s(g) > 4H20(g)+ 3 Na(g) 32.00g0, 3molO, 1mol KCI 1.14gsol % 28.0g HCl x 1mol HCl y 3 mol H, % 2.016gH, obtain it, such as the charge on the species), and the mass number (or the number of BL. molar mass CuSO, -5H,O = 63.5 g Cu+32.1g S+(9x16.08 0)+(10x1.01g H) It has a four Using a similar procedure as that provided in 8A mass Fe 7 mol Fe 55.85 g Fe (c) A natural law is a summary of experimental results or observations, often expressed In this manual you will find solutions to all of the many moles of bromine are combined with each mole of magnesium in the compound. Table of Contents Spb aroms=8.27x107 mol Poo 222X107Pb atoms 241 "Pb 2toms _, 29,192 Pb ators Only a few hydroxides The oxidation state (O.S.) = 2.25L soln x ol Po(C¿Hs), Imol P(C,H,],. 43. 9 [om-]- 0.132 g Ba(OH), :8H,O_ 1000 mE 1 mol Ba(OH), BHO 2 mol OH” We use the expression for determining the weighted-average atomic mass. 4 (a gx 1kg x10g (b) 000 8 g Oneoxide of copper has about 20% oxygen by mass. (a) 1 Lsoln 1 mol KCI g Al is a main-group metal in group 13. Then determine the mass of fuel used, and finally, the fuel consumption. mass of CuSO, -5H,0 =18.6molx Its There are 0,50x2 Reduction: O, (g)+2 H,0(1)+4 e” >4 OH (aq) Next, for each chapter, you should solve all of the Review best we can state is that we can make at least 163 necklaces, because 164 is uncertain For x= 0.05146 moles H,O its solution, you will have fooled yourself into believing that you would have come up with the M =(2x12.011 g C)+(6x1.008 g H)+(1x32.066 g S)=62.136g/mol C,H,¿S Sign in. (e) BaCl, (aq)+ K,SO, (ag): Ba?” (aq)+50,” (aq) > BaSO, (s) is given first, followed by the explanation for its assignment. Self Check: 6N+8H+40 > 6N+8H+40 mass of proton + mass of electron 1.8x10* (9) Ad” gold(III) ion (1) HSOy hydrogen sulfite ion Fundamental Charges and Mass-to-Charge Ratios Principles of Chemical Equilibrium This is HIO,. 1mol Pb(NO,), _ 5.000-x CHAPTER 5 204.22 gKHP 1 molKHP 1 molOH” Crucigrama DE Elementos Químicos, Defensores DE LOS Derechos Humanos Linea DEL Tiempo, Let 011 Unidad I GUÍA DE Trabajo Sobre LA Comunicación Lingüística 2, Variables, Tipos DE Datos Y Operadores EN Pseint, 431917317 Proporcione 3 ejemplos de disyuntivas que ha enfrentado en su vida docx, 01 lenguaje estimulacion cognitiva ecognitiva, Unidad 7 Trauma Y Politrauma - Alexander Núñez Marzán, Unidad 6 Primeros auxilios (atragantamiento^J hemorragias^J fracturas y ahogado) - Alexander Núñez Marzán, Unidad 3 - Primeros Auxilios^J Triaje Y Cadena DE Supervivencia - Alexander Núñez Marzán, Cultura de la Pobreza y Corona Virus - Análisis - Alexander Núñez Marzán 100555100, Cultura DE LA Pobreza EN Tiempo DE Coronavirus - Alexander Núñez Marzán 100555100, Cuestionario sobre Bioseguridad, SAP-115, Unidad No. ImL IL 1000 mL 1 L soln 1 mol Na,SO, Notice that Each isotopic mass must be divided by the isotopic mass of '?C, 12 u, an exact number. 0.1239mol H 0.0177 >7.00 to make them integral. 79.545 g CuO of the 13 measurements is exceedingly close to a common quantity multiplied by an 2 molI 1 mol Mgl, lg 2 H2S(g) + SOxg) > 3/8 Sg(s) + 2 H20(g) or 55.85 kg Fe z 2 kmol Fe 1kmol Fe,O, S is a main-group nonmetal in group 16(6A). 1 mmol NaOH 1 mL N2a0KH(aq) Combustion Analysis Cinética química ejercicios resueltos velocidad de reacción química Explicación y formulas de velocidad de reacción ,Curso para ser unas máquinas de la cinét. the "spiked" mass spectrum to find the total mass of Rb in the sample. chlorine must have 0.S.=+1. Step 3: Page 4-15 Temperature Scales (b) Since there are 11 H atoms in each C¿H,,NO,S molecule, there are 11 moles of H Oxidation: 5,0,” (aq)+5 H,O() >2 SO,” (aq)+10 H' (aq)+8 e” (b) CaCo,(s)+2 H (aq) Ca” (aq)+ H,O(1)+ CO,(g) 44.588 Clx MOLE 1 258molCI 0.6288 >2.00]mol Cl will produce the greatest mass of CO, per mole on complete combustion. Mg?" Then a net ionic equation is written to summarize this information. 2 I2Y4 356.9 -(-38.9) =0.320M CO(NH,), Bco 12 (SE) > 1/8 Sa(s)+ 2H (g)+2 e pa Then the percents of the two elements in the compound are computed. 12B Since data are supplied and the answer is requested in kilograms (thousands of grams), NO) 1 = HZ 5 0.0820M Notice that we do not have to consider each step separately. (2 NHa(g) > Na(g) +60 +6 H'(aq) jx2 Chemical Bonding II: Additional Aspects Among “1.00 L Write the two skeleton half-equations. =2x 2 series of conversion factors. amounts of hydrogen calculated in part (a), compound A might be N¿Hs and 18.015gH,0 1molH,O This value is slightly higher than the value of 15.9994 in modern A fundamental particle would be expected to be found in all samples of matter. (b) molar mass Fe, [Fe(CN), ], =(7x55.85g Fe)+(18x12.01g C)+(18x14.01g N) fc) TheOsS. of Cr =+3 (rule 2). none that we have encountered in this chapter are precisely integral. 6.022x10* Ca atoms 8B Atomsof He=22.6 g Hex molHe_, 6.022x10" He atoms _ 3 49,10% Ho atoms 4.6x10"*cm* per molecule converted to product. (e) Acetone mass=7.50 L antifreeze x 1000 mL e 0.9867 y antifteeze e 8.50 g acetone Do not sell or share my personal information. typically involves two or more species. Note that the H:N ratio in NH3 and N>Hs are the same, 3H:1N. 1mol C,H,,NO,S 1mol € _ Na is a main-group metal in group 1(1A). amour (29) 1 mL KOH(aq) : 1 mmol KOH mo NH,NO, is 80.04 g/mol; Ag,O is 231.74 g/mol; HgO Fertilizer mass= 775 g nitrogen x NET:2MnO,' (aq)+ 3s0,7 (aq) + H,0() -> 2 Mo, (s)+ 3s07 (aq)+2 OH” (aq) (2 e +2 H'(g) + NO(g) > Y Na(g) + H20(g) )x4 molar mass C¿H, (OH), =(2x12.01g C)+(6x1.018 H)+(2x16.00g O) =62.08g/mol (c) Oxidation: T (aq)+3 H,0(1) > 10, (aq)+6 H (2q)+6 e The balanced 100.208 C,H,, 1molC,H,¿ 2molH 1 mo] HO we produced 165 of them. Thus, it would appear that upon heating to 1000 “C, the sample of CuSO, was (c) O=-linNa,O, Na has O.S.=-+1 in its compounds. The compourd is silver perchlorate. 284.48 g stearic acid 159994 g O case, have combined to give two different compounds. 6:022x107 molecules - ¿ 35,19%0, molecules y= 0.0411 moles H,O (g) NCl nitrogen trichloride (h) BrFs bromine pentafluoride This question is similar to question 10 in that two elements, phosphorus and chlorine in this The O.S. (e) 1OS periodate ion (D cio, chloriteion (b) = 4.64x10'g CuSO, -5H,O (b) We need to convert yards to meters. the thermometer, this thermometer cannot be used in this candy making assignment. masses of oxygen that are in the ratio of small positive integers for a fixed amount of 0.1370molC +0.02111>6.49molC 2 AglKís) + Fe(s) > Felr(aq) + 2 Agís) (multiply by 2) For the electron : = 5.686x10* g/C 1 Lsoln 1 mol MgCl, inefficient because you will not be familiar with the material in the chapter. Au(s) (oxidization state = 0), is the reducing agent. We let x be the fractional abundance of lithium-6. HC) A Robert K. Wismer p GENERAL CHEMISTRY Principles and Modern Applications Eighth Edition Petrucci . 41 Mixture Net jonic equation > Vago, = E = 20229 mol ABNO: 0,1995 Lor 2.00 x 10? =0.235 gsamplex 2 2(OH), Imolca(oM), 2 mol =0.00048 mol OH” Chapter 5: Introduction to Reactions in Aqueous Solutions Page 5-1 6.022x10* molecules IL“ ImLBr, 159.8gBr, This gives the simple whole number ratio 3/2. 400.2 g/molCr(NO, ), -9H,0 Ogalloy_ em'alloy The ratios thus obtained may either be integers or they R A AH of hydrogen in the three compounds end up in a ratio of small whole numbers when 1mol(CH, ), CO 249.7 g CuSO, -5H,O Determine the amount in moles of acetone and the volume in liters of the solution. (a) KBr potassium bromide (b) SrCl strontium chloride Thus, each oxygen must have OS. graph (see next page) 4Nal(aq) + 4AgNOx(aq) + 2Fe(s) + 3CL(g)> 4NaNOx(aq) + 4Ag(s) + 2FeClz(aq) + 2Lx(s) (d) '“O is the symbol for the isotope of oxygen that has 16 nucleons in its nucleus: Thus, for this sample 1.12 ( 1 Esoln 1 mol KCI 1 Lsoln 1 mol MgCl, A theory is a hypothesis 4.000 5.005 6.026 7.013 8.012 9.038 10.05 10.97 12.03 12.96 13.97 15.94 16.93 Es molecule. mass of a proton plus that of an electron for the mass of a hydrogen atom. lithium nitride Li" and N” three Li* and one N” Li,N The equation for the combustion reaction is: C¿H,, (1) + Zo, (8) >8C0, (8) +9H,0(1) 98.3 mg solid_97.9mg CO(NH,), _ ImmolCO(NH), ¿Cuándo se produce el equilibrio químico? OCT (aq) +2H* > CT (aq)+H,0(1) 2.131x10' (e) 438x107 (d) TheO.S. 1mL dilute soln 0.650 mmol AgNO, The desired oxidation state is given first, followed by the method used to assign the 2 The information obtained in the course of calculating the molar mass is used to determine K,CrO, = 0,600 = 6:10 or 3:5 [mer] the mass of nitrogen in all three compounds is normalized to a simple value (1.000 g = o ? boron Both elements are nonmetals. 1 Lsoln — 0.895 g acid With this information, we volume =2.43x 10% km? Each mole of CO, is produced from a mole of C. Therefore, the compound with the largest In each balanced reaction, one mole of O,(g) is produced from two moles of solid reactant. If the answer comes easily to solution is neutral. 2Bratoms 6.022x10”Br, molecules the material (in grams) divided by its volume (in mL orem?). =— 4.0026 g He 1 mol He (b) S=-2 in BaS The O.S. arsenic* As 33 33 49 75 Chapter 5: Introduction to Reactions in Aqueous Solutions Page 5-6 If we assume a 100 gram sample, 1kg 198g 20rd beads cation forms is the periodic group number; the number of electrons added when an anion This information provides the conversion factors we need. 61. (c) (NHx)2SO, ammonium sulfate (d) KIO; potassium iodate The atomic number 47 ¡is that of the element silver. == 21,3 Chapter 3: Chemical Compounds Page 3-15 side of this equation. 1kg 1.824 — 30gr beads labeled “Int.” we give the integer closest to each of these multipliers. ? mass _ 9.109x10*g 39.0983 u — (36.3368u + 0.00468u) We.can calculate the charge on each drop, express each in terms of 107” C, and finally A species with greater than 50% more neutrons than protons will have a mass integer. We recall that “M” stands for “mol /L soln.” 100 %í(total mass) 391.0gFe (b) 2.44x10* (b) 1.5x10* te) 40.0 Teorã A Y Problemas Resueltos De Quã Mica Orgã Nica By Rafael Gã Mez Aspe sirva mucho 13 escribe y nombra todos los hidrocarburos de cinco átomos de carbono que tengan un doble enlace qué les ocurrirá 108, (a) number of moles of CuSOa (x = ratio of moles of water to moles of CuSO4) Thus, 102*Cis the Since the oxidation state of H is O in Hz (g) and is +1 in both NHx(g) and H20(g), hydrogen (b) 1.00 m? McGraw Hill. 1.152 g cmpd - 0.7440 g C-0.1249g H)=0.2838 0x2 =0,0177mol O (b) — mass=18.6 Lx y obtén 20 puntos base para empezar a descargar, ¡Descarga Solucionario Petrucci (Octava edición) y más Apuntes en PDF de Química solo en Docsity! in the formula unit must be oxygen. contribution from “Ar= 39.9624u x 0.99600 = 39.803u In each case we use the solubility rules to determine whether either product is insoluble. mass NO, =7.34mol N,O, x Reni, = 6758 N,0, oxygen isotopes. number of moles of CuSO, (x = ratio of moles of water to moles of CuSO4) 5 100.0 g sample 39.997 g NaOH 1 mol NaOH Thus, the empirical formula E, EA EE — 183, necklaces fundamental principles. —[_1mol Mg % 24.305 g Mg + 2 mol Cl x 35.453g Cl] _ 95.211g MgCl, Roman numerals in parentheses if there is more than one type of cation for that metal. divide all of them by the smallest. 57. aqueous solution, that is, it indicates a solution with water as the solvent. (a) () 289%6mmx“"_ -289.60m (8) 0.086 cmx 2% 0.86 mm 1kg () %PO,= - x100% CINÉTICA Y EQUILIBRIO QUÍMICO, INGENIERIA DE LA REACCION QUIMICA FUNDAMENTOS Y TIPOS DE REACTORES, Diseño de reactores homogéneos Román Ramírez López Isaías Hernández Pérez I, Química Básica ALEJANDRINA GALLEGO PICÓ ROSA M.ª GARCINUÑO MARTÍNEZ M.ª JOSÉ MORCILLO ORTEGA MIGUEL ÁNGEL VÁZQUEZ SEGURA UNIVERSIDAD NACIONAL DE EDUCACIÓN A DISTANCIA, Catálisis enzimática Fundamentos químicos de la vida Aníbal R. Lodeiro (coordinador) Libros de Cátedra, Obtención y caracterización de óxido de titanio dopado con nitrógeno como fotocatalizador por el método de Pechini para uso en reactor solar (CPC). > =8.9919908 PCl, Both P and Cl are nonmetals. Harwood . the anion. (e) “C is the temperature of a substance expressed on a scale (the Celsius scale) where The atom described is neutral, magnesium nitride Mg? _6.94lu—7.0160lu Imol % 62.08 g The 9. 1 x 10% ug Rb Rb(natural) + "Rb(spiked) = =2.905 “Rb(natural) A ternary acid consists of equation for this reaction, 2K1+Pb(NO,), > 2KNO, + Pbl,, shows that Organic Chemistry of each O is —2 (rule 6). these contributions would add up to a precisely integral mass. Step 5: *Rb(natural)+"Rb(spiked) _ mass Mgl, required =(0.02500—0.0219) molI' x Y C= => 100% =75.71% C % m=219128H. =4,0x10' gMgCl, Most of the elements in 43, =2,73x10%C atoms The total for both calcium The ions are Ca” and CI”. The %, dichromate (e) NaAl(OH), CO, (s)+4 H (aq)> Al” (2aq)+ Na” (aq)+3 H,0(1)+ CO, (8) Oxidation: Fe” (aq) > Fe” (aq)+ e REVIEW QUESTIONS oxygen mass =100.00g- 73.27 g C-3.84g H-10.68g N=12.218g 0 determines whether the resulting solution is acidic or basic. Determine the mass of O in a mol of Cux(OM)»CO; and the molar mass of Cu,(OH)»CO». [er ] total =[ CI” ] from NaC1+[CI” ] from MgCl, = 0.438 M+0.102 M=0.540 M CT 70.905kgCl, 6kmolCl, combine the half-equations to obtain the net redox equation. (b) Li is in group 1(14); it should form a cation by losing one electron: Li*. two Li? Ralph H. Petrucci. 1mol Ag,CrO, % 331.73 g Ag,CrO, (9) 20,H, (8)+70, (8) >400, (8) +6H,0(1) number of necklaces = 10.0 kg beads x 5 mol € ¿2.0118 € *M and 356.9 *C = 100 "M. To find the mathematical relationship between these [a]. mL (0.200 L) of AgNO, Y hcotare=1 a 1kgN x 100 kg fertilizer CuCl copper(I) chloride Hg,Cl, mercury(I) chioride mass Na,CO, =475mix E y 0.398mol Na,CO, , 10608 Na,CO, most oxygen per gram of reactant. amount H =3.84g Hx——— =3,81mol H 0.7625 >5.00mol H Academia.edu no longer supports Internet Explorer. Y, 237mL mess en=2-228%2 702078 H, * 100.08 alloy 18.015 gH,0 lmol Al _ 1mol AICL, NO(g) = 1.00mol O =24.0gN (aq)+ VO,” (aq)+6 H'(aq)> Fe” (aq)+ VO” (aq)+3 H,O(1) Whereas a chemical formula is rather analogous to a "word," chemical equations parallel "sentences.". (e) be made from each quantity of beads. attraction. 5.079 Hx 22 =5.02molH +0.6288 >7.98molH | formulais [NaOH] = =0.08683 M These results are consistent with the Law of Multiple Proportions because the masses =1.80 mol Br, normally produced in chemical reactions; O, ( 8) is more thermodynamically stable () 100méx (2 A Net: 4 Fe(OH), (s)+ O, (g)+2 H,0(1)>4 Fe(OH), (s) 0.0671mol H +0.0168 > 3.99mol H x100%=40.53% H,O We can simply use values 3mol F 6.022x10%F atoms Reduction: (MnO,” (aq)+8 H'(aq)+5 e” > Mn” (aq)+4 H,O(1) 3x2 mass before reaction = 0.382 g magnesium +2.652g nitrogen =3.034g 1mol Ag, Cro, Most halides are soluble in water; CuCl, is soluble in water. The sum of the oxidation numbers of the two oxygens 1mL soln PRACTICE EXAMPLES (d) Mg(0H), (s) +2 H' (aq) ¿0.0% P¿Os CHAPTER 2 Total time = 216.000 h +0.050h +0.012h = 216.062 h C+0,01968 = 1.000 mol C; 0.02460 mol H=+0.01968 =1.250 mol H. The empirical (e) A hypothesis is a tentative explanation of a natural law. number of stearic acid molecules by the cross-sectional area for an individual stearic 505g cmpd - 0.2028 C-0.0677g8 H =0.235 g N (b) The calculation is performed as follows: each arrow in the o | steari i . mass Fe = 0,04125 L tierantx 902140 molMnO, _ 5 molFe” 55847 gFe =0,246 g Fe “normalized” mass of chlorine = E - 5.723 g of chlorine mA 331.73gAg,Cr0, 1molAg,CrO, 0.250molK,CrO, MgCh(ag) > Mg” (aq) +2 CI'(a9) Reduction: Cr,0,” (aq) +14 H' (aq)+6 e” >2 Cr” (aq)+7 H,O() This is not a redox equation. Chapter 2: Atoms and the Atomic Theory Page 2-8 ImolZnO 2molions 6,022x10”ioms (reaction 1) (20 Ejercicios), DOCX, PDF, TXT or read online from Scribd, 100% found this document useful (2 votes), 100% found this document useful, Mark this document as useful, 0% found this document not useful, Mark this document as not useful, Save Ejercicios de Estequiometría - &quot;Química General&quot; P... For Later. H,CO The O.S. lm ) =0,177g Na,5 6.022x10” Br, molecules of 23 IkmolP,Oy _ 10kmolPOCI, 2 275 ml soln LL 3155 Ba(OH),-8H,O 1 molBa(OH), -8 H,O x ImolKI__x Each molecule of C,H, contains 6 H atoms and 2 C atoms, 8 atoms total. 100cm ¡ ¿Ar < y K < Co < ¿Cu < ¿Cd < 39Sn < “¿Te 4B The empirical formula is obtained by dividing the number of moles of water by the To determine the average atomic mass, we use the following expression: 0.01032 moles CuSO, Use (b) NHx(aq): NH; affords OH' ions necessary for the precipitation of Mg(OH)> 11b 100.0 g vinegar (a) 321x107” =0.0321 (b) 5.08x10"* =0.000508 S, For an atom of a free element, the oxidation state is O (rule 1). second for time, and the mole for amount of substance. =149.2u/C,H,,NO,S molecule =9x10” mol/m' > 0.9x 10” ¿¿mol/m? water that were initially present together in the original hydrate sample. =50.9 gNa,SO, -10H,0 the anion. Balance O atoms: N¿Ha(g) + 1/2 N204(g) > 2 H20(g)+ Na(g) Robert W. Hilts . SO) (40.05% S) and S¿0 (80.0 % S) (2 O atoms — 1 S atom in terms of atomic masses) table indicates that 18 is the atomic number of the element argon.
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