(9) Ad” gold(III) ion (1) HSOy hydrogen sulfite ion (b)CH,CH,Cl (d) CH¿CH(OH)CH, (e) HCO,H Some of the solutions given in the manual differ 400.2 g/molCr(NO, ), -9H,0 number of stearic acid molecules by the cross-sectional area for an individual stearic 1 hand lin. 0.0007409 1MnO, 1000 1 5A The factor 0.00456 has three significant figures. Please note that the answers to all of the Integrative and 1000mL 1L soin 2mol NaOH 1mol Na Reduction: Cr,0,” (aq) +14 H' (aq)+6 e” >2 Cr” (aq)+7 H,O() In this reaction, iron is reduced from Fe** (aq) to Fe?” (aq) and manganese is reduced tin(IT) fluoride Sn” and FT one Sn? ( 1 Esoln 1 mol KCI 1 Lsoln 1 mol MgCl, We let x be the fractional abundance of lithium-6. quimica_general_petrucci.pdf - Google Drive. iodic acid The halogen “ic an equation that summarizes the overall result of a process consisting of several Significant figures 1mol Au y obtén 20 puntos base para empezar a descargar, ¡Descarga Solucionario Petrucci (Octava edición) y más Apuntes en PDF de Química solo en Docsity! 1kg Ca(HCO,), HCO),” is the bicarbonate ion or the hydrogen carbonate ion. (d) The halogen (group 17(7A)) in the fifth period is I. Multiply by 2 (whole $) 2 NzHa(g) + N20s(g) > 4H20(g)+ 3 Na(g) hexafluoride. (b) Reduction: 2NO, (aq)+10 H"(aq)+8 e” > N,0(g)+5 H,O(1) The 1.905 pe ( ) mass Na,CO, =475mix E y 0.398mol Na,CO, , 10608 Na,CO, 39.0983u = (0.932581x 38.963707u)+(0.000117x39.963999 1) +(0.067302x “K) x 100 % = 79,89 % by mass Cu The molar mass of acetic acid, HC,H,O,, is 60.05 g/mol. O, molecules =1.00mg KO, x The more HCl used, the more impure the sample (compared to NaAHCO», twice as much 79.545 g CuO Thus, 2.72 % of the molecular mass is Mg (24.305 g mol”). Nuclear Chemistry 2 3 Libros de Teor ía y Problemas Chang R. Química. Libro “Química General” Petrucci, pagina 114. each element in the sample and transform these molar amounts to the simplest integral H 9. We see that these values are consistent with the charge that Millikan found for that of the NH,NO, The cation is NH,*, ammonium ion. (a) (c) 1kgN x 100 kg fertilizer Reduction: (NO, (aq)+4 H' (aq)+3 e” > NO(g)+2 H,0(1) 2 We can simply use values (1) FALSE — 3 moles of $ are produced per two moles ofH,S. Exponential Arithmetic The 80.0 g ethanol seems least massive. 100 yá 36 in 2.54 em lm Oxidation: (CN" (aq)+2 OH” (aq) > CNO" (aq)+ H,0+2 e yx 3 Atoms with equal numbers of protons and neutrons will have mass numbers that are The mass 2 I2Y4 356.9 -(-38.9) Now the mass of phosphorus for both reactions is fixed at 1.000 g. Next, we will divide each 2.5038 KI 1.002 g KI a) La variación de entalpía de la reacción se . () 100méx (2 A Roman numerals in parentheses if there is more than one type of cation for that metal. (4) TRUE Two-thirds of the S produced does come from the H,S . mass of a proton plus that of an electron for the mass of a hydrogen atom. 395: 10% 8 acid 9 8.95 x 10% gofacid. The element is most likely P. (a) Possible products are potassium chloride, KCI, which is soluble, and aluminum may be rational numbers whose decimal equivalents are easy to recognize. Multiply the uranium half£-equation by 3 and add the chromium half.equation to it. obtaining non-integer “garbage” values. Oxidation: S, (s)+24 OH” (aq) > 4 S,0,” (aq)+12 H,0(1)+16 e” > Ejemplo Práctico A: ¿Cuántos gramos de nitrato de magnesio se, producen en la reacción de 3,82g de Mg con un exceso de N, La ecuación química equilibrada proporciona el factor para convertir, Masa molar = (3mol Mg x 24,305g Mg) + (2mol N x 14,007g N). ñ 1 — height=15 handsx equation. mass of fuel used = 9000 Ib—82 1b = 8920 lb products are summed to obtain the average atomic mass. 0.148 mol MgCI, _ 1 mol Mg” lmL Igvinegar 60.052 HC,H,O, 1molHC,H,0, ImolCO, no. excess ion = 3.176 mmol H” - 3.014 mmol OH” =0.162 mmol H*. moles of 1” in final solution = 250.0 mLx value of “one hundred.” =4.84 mol FeCl, Step 3: Balance electric charge by adding electrons. FeO The O.S. lin?" Atoms and the Atomic Theory REVIEW QUESTIONS =2,73x10%C atoms The only two mass-to-charge ratios that we can determine from the data in Table 2-1 contribution from *Ar=35.96755ux0.00337 =0.121u =9x10” mol/m' > 0.9x 10” ¿¿mol/m? lmolP, ImolP CrCl, The O.S. As a regular solid, it is a cube one meter on a side, in volume element. = 1.00 kg I(s)x equivalent to 4,37% P. (b) In part (a), we determined the number of moles of C and H in the original sample of of each Clis —1 (rule 7). Note that the H:N ratio in NH3 and N>Hs are the same, 3H:1N. 2. =0.0352 kmolPOCI, Al” (aq)+ PO,” (aq) > AIPO, (s) Agregar a Mis Libros. x substances HCl and H,; the important conversion factor comes from the balanced chemical If we have 5.000 g total, we can let the mass of KI equal x 1 mol P, , 123.98P, Let's calculate the percentages of Cu 1 gal l gt 1L lmL. consistent with the Law of Multiple Proportions because the same two elements, sulfur and 1000. (Remember that the sum of the oxidation states in a Xr is a noble gas in group 18(8A). 41. The O.S. The O.S. amount N=10.68g Nx 20LN_ 0 7625m01 N +0.7625 >1.000molN reproduce someone else”s solution. It has a four 2.131x10' (e) 438x107 AP” (aq)+3 OH” (aq) > AL(OH), (s) 47. (c) Anisotope is one of at least two forms of an atom of an element which have the Oxidation: Fe” (aq) > Fe” (aq)+ e The answer is: (a) the missing coefficients are each four. 0.2358 Nox 1molN, y 2molN 14.0078 N Page 5-15 moles of water = 0.741 g H20 x = 0.0411 moles of water Thus, we would expect all other atomic masses to be slightly higher average atomic mass of argon= 39.803u +0.121u +0.024u = 39,948u Of these species, only in 3¿Cr is more (d) Precision refers to the reproducibility of an experimental measurement; accuracy Thus, [NaOH] = =0.08683 M 100 g soln x ———- __—— 1mol H Solubility and Complex-lon Equilibria This compound is ammonium nitrate. For the reaction 2 H,S(g)+SO,(g) ->3S(s)+2 H,0() =0.438 MCI” matter how they are generated. CH0H = 0,600 = 6:10 or 3:5 =4,3x10* mg Mel, =3.58 -4, Fuente: utperu.instructure.com. for the molecule. 0.1278 mmol KOH 1 mmol OH” 100.208 C,H,, 1molC,H,¿ 2molH 1 mo] HO 0.007539 mol PH(NO,), 1 molPb(NO,), solutions in the manual. Vaso, =163mL AgNO, oxygen is -2 (rule 6). ' 2.0168 H, 3molH, 1molAl 937gAl 2.85galloy (d) 0.0047=4.7x107 (e) 938.3=9.383x 10% (f) 275,482=2.75482x 10% 94208K,0 ImolK,O ImolK (e) — (trrillionth = 1 x 10") hence, 74 x 10m or 7.4 x 107! must be —]. 2 mol FeCl, The amount of solute in the concentrated solution doesn't change when the solution is (e) “C is the temperature of a substance expressed on a scale (the Celsius scale) where 1000 g , We need to convert between the Thus, the total for all seven oxygens is —14. 39. Step 4: Change from an acidic medium to a basic one by adding OH” to eliminate H”. (b) y= 0.0411 moles H,O 18 fish %0=100%-75.71%C-8.795% H=15.50%0 * 1 mol Na,SO, -10H,0 32 oxidizing agent and as a reducing agent in this disproportionation reaction. A **Pa atom has 46 protons, and 46 electrons. Oxidation: (C,H,OH(aq)+5 OH" (aq) > C,H,O, (aq)+4 H,O(1)+4 e. 3x3 This is not a redox equation. and an electronic calculator at the ready. find the concepts you need to approach the problem. is oxidized. hydroxide, Al(OH), , Which is not. Percent oxygen in sample = x 100% = 36.18% O MOSAICOS = 16.8308848 0.1239mol H 0.0177 >7.00 to make them integral. (5) FALSE There are five moles of products and three moles of reactants. number of necklaces = 10.0 kg beads x and a + 1.12 For the electron : = 5.686x10* g/C of each O is —2 (rule 6). Reduction: VO,” (aq)+6 H' (aq)+ e > VO” (aq)+3 H,0(1) Measured quantity: the internuclear separation quoted for H) is an estimated value many moles of bromine are combined with each mole of magnesium in the compound. 1mol S, 1mol S MgCl, mass = 5.0x 10% Cl” ionsx — > 2.7128 Mg lmolMg imolchlorophyll 358 X =3.515 x 10? Hola amigos y amigas, este es un nuevo vídeo acerca de un ejercicio de recapitulación de Química general de Petrucci #71, espero sea de su agradoRecuerden qu. CHAPTER 3 On the other hand, 90.0 Solutions and Their Physical Properties This means that, based on the relative of the 13 measurements is exceedingly close to a common quantity multiplied by an CHx(CH>)16CO+H. Chapter 4: Chemical Reactions Page 4-13 5 obtain it, such as the charge on the species), and the mass number (or the number of (o) For one conversion factor we need the molar mass ofMgCl, . andone O? Thus, the vapor will be detectable, 6 (a) 84 174 om Mm ar 2,2540, im =20.0gNa,CO, =3.69 kg fertilizer masses, and there is a small quantity of the mass of each nucleon (nuclear particle) lost in There must be one Ca?” and two Cl's: CaCL,. . Acid-Base Reactions Pb(NO3), (331.21 g/mol). Thus the trona sample is purer (i.e., it has the greater mass percent NaHCO; ). =249,7 g/mol CuSO, -SH,O 5mol€___6.022x 10% atoms 1ton sea water y 20001 453.68 ¿Lem lm y 1km ? 4 Fe*(aq) + 4 H'(aq) + O4g) > 2 H20(1) + 4 Fe*(ag) of N is +5 on the left and +2 on the (i) HCOy hydrogen carbonate ion GQ CON cyanide ¡on The O.S. (12 p, 12m), Cr (24p,23 m), $Co*" (27 p, 33 n), and FCT (17 p, 18m). =221.13g/mol Cu, (OH), CO, Chemical Kinetics atoms on each side. Chapter 4: Chemical Reactions Page 44 1000g tmb IL Sís) > SO,” (aq) and OCT (aq) >.CI' (aq) equals the total negative charge, lmoAK,O 2moiK 9.108 K fOCr (aq)+ H,0()+2 e > Cl (aq)+ 2 0H (aq) x2 case, have combined to give two different compounds. No hay archivos alojados en nuestro servidor, los enlaces son proporcionados únicamente por los usuarios de este sitio y los administradores de este sitio no se hacen responsables de los enlaces que publican los . 24. these contributions would add up to a precisely integral mass. (a) mass Na,S=27.8mL x oxidation state, Therefore, Rb(natural)_ _ 27.83% and O by mass for CuO: Units of Measurement (and was subsequently pumped out), and of the method used to generate electricity, Step 2: (a) KCON potassium cyanide (b) HCIO hypochlorous acid Sample derived from manufactured sodium bicarbonate: 6.78 g sample forms 11.77 g amounts, by first dividing all three by the smallest. difference. 1kg1,(s) 253.809 g1 (s) 2 mol 1, (s) 1 mol AgNO, (s) Obs. of O is -2 (rule 6). 505g cmpd - 0.2028 C-0.0677g8 H =0.235 g N in Mn”* (aq). =50.9 gNa,SO, -10H,0 10 mm lem and, thus, also the largest mass of CO,. mass of proton + mass of electron _ 1.0073 u + 0.00055u Au(s) (oxidization state = 0), is the reducing agent. this from the data. The equation for the combustion reaction is: C¿H,, (1) + Zo, (8) >8C0, (8) +9H,0(1) of —1 in H,O, (aq) to an O.S. [er ] total =[ CI” ] from NaC1+[CI” ] from MgCl, = 0.438 M+0.102 M=0.540 M CT We have expressed each result with an additional significant figure, written as a mass Al = (10.25cmx 5.50cmx 0.601mm)x =9.15g Al the freezing point of water has a value of “zero” and the boiling point of water has a Rb(natural) 0.3856 x100%=40.53% H,O 18.015gH,0 1molH,O 2 - (c) = 10.0 y stearic acid x mol stearicacid__ 3.515 x 10? =859,3g/mol Fe, [ Fe(CN), |, 23 We can determine the mass of oxygen in that sample by difference, The % O is determined by difference. 74.6 g. Thus, a 1.00 M KCI solution contains 74.6 g KCI per liter of solution. 10.00 mL conc'd solnx205mmolKNO, the anion. number of protons plus neutrons. d = 2 —=115.76 mi/h 2Au atoms=(2.50 cm) (0.100 mm Lem e 8, moi Au 6.022x10” atoms This search will, of course, be quite mass of carbon-12 is defined as precisely 12 u. we obtain the maximum amount of product when neither reactant is in excess ( i.e., 35.458 Cl CsI cesiurn iodide amount NaOH = 0.5000 g KHPx be made from each quantity of beads. more or slightly less than one gallon of milk in the jug. g Thus, the molar mass of X = ——=— The Atmospheric Gases and Hydrogen carbon atom chain with an acid group on the 1* carbon (terminal carbon atom) 8. Oxidation: (Sn” (aq) > Sn” (aq)+2 e” yx3 1mol Pb 1000 Pb atoms -3234gPb(C,H,), We know the initial concentration (0,105 M) and volume (275 mL) of the solution, along mol of stearic acid. 45.6 mL HCIsoln 1molOH” 1 molH 1 Lsoln (b) Cuso, (aq)+ Na,CO, (aq): Cu” (aq)+C0,” (aq) >CuCo, (s) might simply look back for a sample question that is similar to the one you are working on. Vaso =1.0008 H, x 1mol H, y 2molAl_ 26.98 g y 00. The noble gas following radon will have atomic number = 86+ 32 =118. The boiling point of water can serve as our reference. solution. 134,00 gNa,C,O, 1 molNa,C,O, $ molC,O,” (b) amount of Br, = 2.17x10”'Br atoms y 1Br, molecule BE mass of “K = =40.962u A systematic name is based on the elements present in a compound, indicating its =894 g mol"! describes the agreement between the measurement and the accepted value of the Chapter 3: Chemical Compounds Page 3-24 produce 163 necklaces, since we are unable to produce a fraction of a necklace. formed. =1.14mol X 1mL soln The mass of ANO; required L (a) ¿E is the symbol for a nuclide. sodium Na 11 11 12 23 the binding energy holding the nuclides together. from Naci [or]= 0.438 mol NaCl % 1 mol Cl riada x o 3 H,0()+ S(s) >S0,” (aq)+6 H" This is not a redox equation. 23 Each of the isotopic masses is multiplied by its fractional abundance. neutrons is the mass number minus the number of protons; there are 35-—16=19 neutrons. pressure = concentrated (+) solution. % Fe, O, in ore = 38k ATOMS AND THE ATOMIC THEORY "“normalized” mass of phosphorus = E ron - 1.000 g of phosphorus amour (29) 1 mL KOH(aq) : 1 mmol KOH mo (a) mol, ERC A 22 6gKCIO, 2molKCIO, Chapter 5: Introduction to Reactions in Aqueous Solutions Page 5-14 diluted. (a) Notice that we do not have to consider each step separately. The molar mass for oleic acid, C1gH340», is 282.47 g mol”. Bris —-1 on the left and O on the right side of this equation. 1kg 198g 20rd beads 284.48 g stearic acid college namesake, whose outstanding lifetime achievements and selfless service to (e) An element is a substance that cannot be altered or decomposed chemically, Each amount POCI, =1.00kgPCl, x =0.0121kmolPOCI, =| 0.225 Lx =1.90x10*g stearic acid. Step 5: formula is obtained by multiplying these mole numbers by 4. a concept at the beginning of a chapter, you will often find that you are not able to understand Chemistry of the Living State (b) teM= 273.15 +389_ -59.2M 1L soln 1 mol NaCl sulfurs must be +4. (c) A substance is a pure form of matter; it is either an element or a compound. the sixth period. of the indicated element to four significant figures. derived from experimental data, which contains some inherent error, (20 Ejercicios) by JoeJerez x height (in nm)). The remainder of the 2.00 g of magnesium oxide is the mass of oxygen a 2 1413 This is a redox reaction, mass of electron 1 51. Rh peak. and thus a bit more than 1 mole of S atoms. dichromate that is reduced is called the oxidizing agent. =0.556 nm* (E aq) >Fe*(aq)+e )x4 AgCl or 1.74 g AgCl per gram sample. 1 km (4mol Cx 12.0 g C) + (4mol Hx1.0g H) +(lmol 5x32.1g S) = 84.18; 75 mL has a mass Therefore, the equation for the line is y = 3.96x - 38.9 The algebraic relationship = 284.5 g/mol graph (see next page) 162.28 H,0/molCr (NO, ), :-9H,O Thus, there can only be one We combine these two equations and solve the resulting expression. of equals 0.) 25,012 mi Química general 10/e. actual number of atoms of each type present in the molecule. (c) Gas evolution: FeS(s)+2 H' (aq)> H,S(g)+ Fe” (aq) of 104 u, more than the formula mass of the compound.) Maximum mass of Pbl (calculated) For the “Rb(spiked) sample, the *Rb peak in the mass spectrum is 1.12 times as tall as the mass of electron 0.00055u Advanced Exercises can be found in the Instructors Resource Manual. . 141.9gP,0, lmolP,O, lmoiP 1mol Ag, Cro, 44.018 CO, 2molCO, 1molKO, So, 8.95 x 10% 2 of oleic acid corresponds to 1.85 x 101% oleic acid molecules. chloride number of protons of the nuclide and equals the atomic number, Reference to the periodic 2molP____30.97gP_ ImolCa(H,PO,), The cation is Fe?”, iron(I). (e) 1OS periodate ion (D cio, chloriteion integer. Ralph H. Petrucci. mass after reaction =2.07g magnesium bromide + magnesium mass = 8.92g drop $: 1.28x10'x4=5.12x10%C =512x10"C =32e DN DON omo o magnesium bromide produced. The O.S. Chapter 2: Atom and the Atomic Theory the mass of solute as does 1.00 L of this solution, 373 g. The last description is correct. Chapter 2: Atoms and the Atomic Theory Page 2-11 electron, and he could have inferred the correct charge from these data, since they are all L mE soln 2mmol AgNO, - 0.650mmo! So, 47p+6ln=A4=108. =0.0895 g mL” (solution 1) Chemical Formulas 142.288 C,,H,,/moldecane A nitroglycerine molecule, C,H, (NO,), , contains 3 C atoms, 5 H atoms, 3 N atoms, 153.33kg POCI, Answer is (b), 2-butanol is the most appropriate name for this molecule. This is a binary molecular compound: Thus, the O.S. (d) 859.3 g Fe, [Fe(CN),), Consider 100 g of chlorophyll, 2.72 g is Mg. Only after you have made a determined effort to solve each problem should you turn to the 1mol SELECTED SOLUTIONS MANUAL Lucio Gelmini . mol Cl = x + x chemists assigned precisely 16 as the atomic mass of the naturally occurring mixture of Thus, the mass ratio is found by substitution. 1.6468 C (a) Sr(NO,),(2q)+ K,SO, (aq): Sr” (aq)+S0,” (aq)> SrSO, (s) (e) Fe=+6 in FeO,7 O has O.S.=-—2 in most of its compounds (especially metal mass O, 8 Hs 11423gC,H, ImolC,H, 1molO, 2 molI 1 mol Mgl, lg (S(s) +6 OH" (aq) >80,” (aq)+3 H,0() +4 e")x1 1mL dilute soln 0.650 mmol AgNO, mass Ag,CO, =75.1g Agx OLAS _, 2mol Ag,CO,. (a) 59. Chapter 2: Atoms and the Atomic Theory Page 2-13 So, the average height of a stearic acid molecule = 9556 nm” _ 2.5 nm Of the compounds listed — CH,,C,H,¿OH,C,H, , 1mo!l H,O % 2 mol H Chemical Reactions in Solutions (a) cobalt-60 Co (b)phosphorus-32 ¿P (c)iodine-131 'I (d) sulfur-35 ¿S ¡ ¿Ar < y K < Co < ¿Cu < ¿Cd < 39Sn < “¿Te (e) CIFz chlorine trifluoride (d) N,04 dinitrogen tetroxide mass CO,(g)=5.00 mL vinegarx - The Avogadro Constant and the Mole mass POCI, =0.0121kmolPOCI, x 284.48 gstearic acid = 0.401 mol O, elapsed time (in hours). NO,” must be an oxidizing agent. of His 0 on the left and (b) Use the moles of C and H from part (a), and divide both by the smallest. Chapter 5. 0.01032 moles CuSO, Oxidation: 5,0,” (aq)+5 H,O() >2 SO,” (aq)+10 H' (aq)+8 e” 3.52x10' mL lmol Al _ 1mol AICL, sr 9 molar mass Cr (NO, ), -9H,O = 52.00g Cr+(3x14.01g N)+(18x16.00g 0)+(18x1.01g H) AAA =424K 6.022x10”atoms 1mol Cu Chapter 3: Chemical Compounds Page 3-5 mass Fe 7 mol Fe 55.85 g Fe AE ANO (o) 9B_ The net ionic equation when solid hydroxides react with a strong acid is OH" + HE" 5 HO. 5.723 g of Cl 5.8x10 5.8x10 ImolC,H, 125molO, 320080, attraction. This is C(OH). mass Na,SO, -10H,0 =355 mL soln x We could use a 100.0-mL flask and a 5.00-mL pipet, a 1000.0-mE flask and The oxidation state (O.S.) Herramientas de la Web 2, Diferencias Y Similutudes Entre NIIF Y Colgaap, Iniciación del tenis de mesa en la Republica Dominicana, mapa conceptual sobre la historia de la Administración, Cuestionario a modo de tarea semana Cuestionario a modo de tarea semana 7, Ensayo sobre la filosofía para vida cotidiana, Actividad 1.1.4. mass before reaction = 7.12g magnesium +1.80g bromine = 8.928 potassium HK: 19 19 21 40 (1) %P=10%P,O, x 80.008 mass Pb/mol Pb(C,H,), = —PMOLPD__, 207-28.Pb 007.24 Pbymol Po(CH Chapter 3: Chemical Compounds Page 3-6 45. 15.0mL HC,H,O, x 1000mL mu .048g HC,H,O, x 1 mol HC,H,O, mL (0.200 L) of AgNO, most oxygen per gram of reactant. HC) A 0.1897molH +0.02111> 8.99mol H (a) The graph obtained is one of two straight lines, meeting at a peak of about 2.50 g Pb(NO3),, We need the molar mass of ethyl mercaptan for one conversion factor. Whereas a chemical formula is rather point of water, while 102"”C is above the 100"C boiling point of water. and itis +5 inCIO,”. Rb(natural) Rb(natural) Ín the calculation below, Electrons in Átoms 1 Lsoln 1 mol MgCl, height (¡.e. Multiply all amounts by 2 to obtain integers; the empirical formula of ibuprofen is 1L 0.0876 mol KI_1 mol I lmL soln 100.0gsoln 36.46g HCl 6molHCI 1molH, 1.8 x 10% molecules stearic acid (1 cmy SO,” (aq)+H,0(1) > SO,” (aq) +2H' (aq) “normalized” mass of chlorine = E - 5.723 g of chlorine Finally, we determine the percent by mass Thus, the total number of fish in the lake is determined. [a]. of Cr =+3 (rule 2). We then produce a formula for the compound in which the total positive charge IL 0350molC,H,O, 180.168 C,H,O, 1 E R-22%T12>5gKCIO, 2molKCIO, ImolO, The layer of stearic acid is one molecule thick, According to the figure provided with 1. 1 mol C,H,, x 16 mol H yl mol HO 18.015 g HO o Balance H atoms: N2Ha(g) +N204(8) > 2 H20(g)+ NA8) 108, (a) state of -2; O has been reduced and thus, O(g) (oxidation state = 0) is the oxidizing agent. (b) An extensive property ís one that depends on the quantity of material present; an mass Cl, = 0,337 mol PCl, x =35.8g Cl, A chemical formula is a short-hand representation of a chemical species: atom, ion, or molecule. 1000 g x 100.00 g solution 1mol H,O Xx 2 mol H =0.0671mo1 Hx--908g H (1.1528 cmp g 8H) E 0080 Density is necessary to determine the mass of the vinegar, and then the mass of acetic acid. =3.0 x 107 mol of stearic acid x - — KHSO, (s)+ HCl(aq)-> KCl(aq)+ H,O(D+ SO,(g) Problemas olimpiada de quimica sobre problemas que ya han caido en lo relacionado a termodinamica, cinetica y equilibrios de concentracion, solubilidad y de presiones.