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300 g O 2 1 mol O x 2 3 g O 2 2 1 mol CO x 2 mol O 2 = 0469 mol CO The second assumption is correct Therefore, oxygen is the limiting reactant 2 2 440 g 0469 mol CO x = 6 g CO 1 mol 12 In an experiment with Zn and S, it was found that 307 g of ZnS was produced If the(16 kJ)/(05 mole H 2 (g)) = 3672 kJ per mole of H 2 (g);400 g C are 671 g H 533 g O mol C 11 g C mol H 1008 g H mol O 1600 g O = 333 mol C = 666 mol H = 333 mol O CH 2O empirical formula mass of CH 2O molar mass of lactate 9008 g 3003 g 3 C 3H 6O 3 is the molecular formula O 333 333 H 666 333 C 333 333 Combustion apparatus for determining formulas of organic compounds Figure 35

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"¯Œ^ ƒƒ"ƒO 40'ã ƒXƒgƒŒ[ƒg-O(Ans 600 g) (h) A mass of g of "KHP" (potassium hydrogen phthalate;Mass = (425 g x 0133 J/g deg x 754 o) / (418 J/g deg x 25 o) = 408 g mass = _41 g_____ 4 A chemical compound has a molecular weight of 05g/mole 1400 g of this compound underwent complete combustion under constant pressure in a calorimeter with a heat capacity of 2980 J o C1 The temperature went up by 1195 degrees Calculate

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About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How works Test new features Press Copyright Contact us CreatorsThe mole ratio of CHO in 100 g of ketonox is given by, € Moles C = 4615 g C 1 mol C 11 g C = 3843 mol C Moles H = 7746 g H 1 mol H 1008 g H = 7685 mol H Moles O = 4611 g C 1 mol O 1600 g O = 28 mol OStep 1 O 3 (g) NO 2 (g) NO 3 (g) O 2 (g) step 2 NO 3 (g) NO 2 (g) N2O5 (g) The experimental rate law is rate = kO 3NO 2 Because the rate law conforms to the molecularity of the first step, that must be the ratedetermining step The second step must be much faster than the first one The decomposition of Hydrogen Peroxide 2 H 2O2 (aq

2N2O5(g)→4NO2(g) O2(g) From the graph looking at t = 300 to 400 s 61 2 M Rate O = 9 10 Ms 540 550 560 570 0 50 100 150 0 250 300 time (s) ln P k = 26x103s1 Reaction HalfLife 12 Reaction Halflife Halflife is the time taken for the concentration ofA 15 kg block of Ni at 100 °C is placed into 500 mL of water that has a temperature of 21 °C What is the final temperature assuming the specific heat of Ni is 044 J/g°C and the specific heat of water is 4184 J/g°CCalculate the number of moles of O (atomic oxygen) in 400 grams of oxygen a) 250 moles of O b) 125 moles of O How many atoms of carbon are present in 345 g of caffeine, C 8 H 10 N 4 O 2?

G t2 (a) we solve for y = h which yields h = 518 m for y 0 = 0, v 0 = 4 m/s, q 0 = 600° and t = 550 s − 0 0y − 2 (b) The horizontal motion is steady, so v x = v 0x = v 0 cos θ 0, but the vertical component of velocity varies according the equations before Thus, the speedi id at impact is v = ()v 0 cosθ 0 2 v 0 sinθ 0 −gt 2742 x 1023 atoms O x (1 mol / 6022 x 1023) = 1232 mol O / 308 = 4 K2MnO4 – potassium manganate 2 Samples of a compound are found to be 2791% Fe, 2408% S, and 4801% O Can you figure out the empirical formula of the anion based on your knowledge of nomenclature?N2(g) 3H2(g) ⇔2NH3(g) ∆Go = 330 kJ K298 = 60 x 105 Thermodynamically favored at 298 K However, rate is slow at 298K Commercial production of NH3 is carried out at temperatures of 800 to 900 K, because the rate is faster even though K is

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Mass of a compound of a solution calculation Mass (g) = Concentration (mol/L) x Volume (L) x Molecular Weight (g/mol) The key here is to realize that you're dealing with a hydrocarbon, that is, a compound that contains only carbon and hydrogen Notice that the products of this combustion reaction are carbon dioxide, #"CO"_2#, and water, #"H"_2"O"# This tells you that all the carbon that was initially a part of the hydrocarbon will now be part of the carbon dioxide Likewise, all the hydrogen thatStarting with initial concentrations of 0040 mol/L of N 2 and 0040 mol/L of O 2, calculate the equilibrium concentration of NO in mol/L (a) mol/L (b) mol/L (c) 0011 mol/L (d) 0080 mol/L (e) 010 mol/L 9 K c = 0040 for the system below at 450 o C If a reaction is initiated with 040 mole of Cl 2 and 040 mole of PCl 3 in a liter container, what is the equilibrium

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1008 g H x 1 mol H O 2 mol H x 1802 g H O 1 mol H O 158 x 10 g H O x 2 2 2 23 = 1767 x 104g H The mass percentages of C and H can be calculated using the masses from the previous calculations Percent C = 387 mg 15 mg x 100% = 4090 = 409% C Percent H = 387 mg mg x 100% = = 457% H"to" 1 mol H 2 O "from" 1802 g H 2 O (g of H 2 O) 1 mol H 2 O = 101 mol H 2 O 1802 g H 2 O (00 mol H 2 O) 1802 g H 2 O = 3604 g H 2 O 1 mol H 2 O (4Molar mass of K = 391 g Molar mass of Mn = 549 g Molar mass of O = 160 g Molar mass of KMnO4 = 391 g 549 g (160 g x 4) Molar mass of KMnO4 = 1580 g What other calculations you can do with the molarity calculator?

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Q = m x C x DT q = 250g x 418J/g o C x 26 o C q = 37,6J or 38kJ 2 Calculate the specific heat capacity of copper given that 475 J of energy raises the temperature of 15g of copper from 25 o to 60 o q = m x C x DT C= q/m x DT C = 475J /(15g x 35 o C ) C= 039 J/g o C Type your answer here 40g means 40grams Wiki User ∙ 1931 This answer is979 g / 48 g > The moles of O in 306 g of M 2 O 3 48 g > There are 48 g in three moles of oxygen The three comes from oxygen's subscript in the formula (I suppose I should write 48 g/3mol 2x 48 > The molar mas of M 2 O 3 x is the atomic weight of M, note that is multiplied by two because of the subscripted two in the formula

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400 g NaOH Take either of these moles of reactants and calculate how much of the other reactant would be needed to react completely with it Let us take mol FeCl 3 3 mol NaOH (F) mol FeCl 3 x = 0185 mol NaOH needed 1 mol FeCl 3 Compare the moles needed (F) with the available moles from (E), aboveA) 857 x 10 23 b) 268 x 10 25 c) 108 x 10 24 d) 9 x 10 23 e) 4 x 10 23 16 Suppose you have a 100gram sample of each of the following240 mol I x x mol I 1 m 2 3 652 g Al o I 2 l AlI 2 = Now assume I 2 is limiting 11 2Al 3 I2> 2 AlI 3 Solution – Cont (b) Determine the yield of the AlI 3 if one starts with 1 g Al and 240 g I2 21 3 3 3 3 2 2 2 2 1 mol I mol AlI g AlI 240 g I x x x g I 2

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Equilibrium Calculations II CHM 152 1 At a particular temperature, Kc = × 10 −6 for the reaction 2 CO2 (g) U 2 CO (g) O2 (g) If mol CO2 is initially placed into a 50 L vessel, calculate the equilibrium concentrations of all species 2 CO2 (g) U 2 CO (g) O2 (g) Initial (M) 040 0 0 Change (M) −2x 2x xEquil (M) 040 − 2x 2x x2 2 c 2 2(414 g O/114 g) x 100 = 363% O with compounds, find molar mass of the compound and molar mass of the element you are trying to find and times by the subscript for that element within the formula, divide molar mass of element by total molar mass of compound and multiply by 100 to find1 Ca = 1 C = 111 3 O = 3 x Molar mass of CaCO 3 = g Iron(II) sulfate, FeSO 4 Molar mass of FeSO 4 = g Chapter 3 Solutions Solution a homogenous mixture in which the components are evenly distributed in each other Solute the component of a

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321 x 1024 H atoms ∣ 1 mol H ∣ 1 mol H2O ∣ 180 g H2O ∣ = 480 g H2O ∣ 602 x 1023 H atoms ∣ 2 mol H ∣ 1 mol H2O∣ 16 Calculate the number of C atoms in 2568 g of C10H8O 2568 g C10H8O ∣ 1 mol C10H8O ∣ 10 mol C ∣6022 x 10 23 C atoms ∣ =40 101 101 101 101 101 8 80 Bishop Redding San Francisco Eureka Fresno Susanville Ukiah Bakersfield Needles San Luis Obispo El Centro Nevada City D i e g o R i v e r r S u san Rive r S t a n i sl a u River C o s u m n e s R i v e r K n g s R i v e r M o k e l u m n e Riv e r Tuol u m n eRiv r A l a m o R i ve r Fr e s n o Riv e r Y u b a R2C 3 H 6 2NH 3 3O 2 2C 3 3 N 6H 2 O a 368 g b 735 g c 1470 g d 462 g e

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Find local businesses, view maps and get driving directions in Google Maps Mass of water = 25 g, Mass of ethyl alcohol = 25 g and mass of acetic acid = 50 g Molecular mass of water (H 2 O) = 1 g x 2 16 g x 1 = 18 g mol1 Molecular mass of ethyl alcohol (C 2 H 5 OH) = 12 g x 2 1 g x 6 16g x 1 = 46 g mol1 Molecular mass of acetic acid (CH 3 COOH) = 12 g x 2 1 g x 4 16g x 2 = 60 g mol142 g/mole) was dissolved in water and made up to 1000 mL in a volumetric flask 1500 mL of the "KHP" solution required 2485 mL

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2791 g Fe x (1 mol / g) = / = 1 x 2 = 2Find (f g)(x) for f and g below f(x) = 3x 4 (6) g(x) = x2 1 x (7) When composing functions we always read from right to left So, rst, we will plug x into g (which is already done) and then g into f What this means, is that wherever we see an x in f we will plug in g That is, g acts as our new variable and we have f(g(x))Graph g (x)=x g(x) = x g ( x) = x Rewrite the function as an equation y = x y = x Use the slopeintercept form to find the slope and yintercept Tap for more steps The slopeintercept form is y = m x b y = m x b, where m m is the slope and b b is the yintercept y = m x b y = m x b Find the values of m m and b b using the form

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Consider as another example a sample of compound determined to contain 531 g Cl and 840 g O Following the same approach yields a tentative empirical formula of Cl 0150 O 0525 = Cl 0150 0150 O 0525 0150 = ClO 35 Cl 0150 O 0525 = Cl 0150 0150 O 0525 0150 = ClO 35O a 400 g b 160 g c 800 g d 640 g e 160 g ____ 27 Acrylonitrile, C 3 H 3 N, is a molecule used to produce a plastic called Orlon How many grams of acrylonitrile could be produced by reacting 5 g of propene, C 3 H 6 with excess ammonia, NH 3 and oxygen?40 A saturated solution of NaNO 3 is prepared at 60°C using 100 grams of water As this solution is cooled to 10°C, NaNO 3 precipitates (settles) out of the solution The resulting solution is saturated Approximately how many grams of NaNO 3 settled out of the original solution?

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 12 g iron / 100 g sample = x g iron / 250 g sample Crossmultiply and divide x= (12 x 250) / 100 = 30 grams of iron How to Calculate Volume Percent Concentration of a Solution Volume percent is the volume of solute per volume of solution This unit is used when mixing together volumes of two solutions to prepare a new solution25 g of aluminum 270 g/mL 1) Plan Calculate the volume for each metal and select the metal sample with the greatest volume 1) 25g x 1 mL = 93 mL aluminum 270 g 2) 45 g x 1 mL = 23 mL gold 193 g 3) 75 g x 1 mL = 66 mL lead 113 g Ref Timberlake, "Chemistry", Pearson/Benjamin Cummings, 06, 9th EdDetermine the masses of each element assuming 100 g 400 g C 671 g H 5329 g O Convert the masses into moles of each element 400 g C x 1 mol C = 333 mol C 11 g C 671 g Hx 1 mol H = 666 mol H 1008 g H 5329 g Ox 1 mol O = 333 mol O 1600 g O Express the moles as the smallest possible ratio CH 2O

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Chain of equalities 1802 g H 2 O = 1 mole of H 2 O The conversion factor How many grams are there in 00 moles of H 2 O?Example 2N2O5(g) →4NO2(g) O2(g) N 2 O 5 c d t b Concentration versus time from experiment a Rate b c d Rate = kN2O5 a N2O5 Experimental rate law k is a constant (slope Slope of a of line) Slope of b Slope of c Slope of d – For most reactions of the typeSkill 35 Lactic acid (M = 9008 g/mol) contains 400 mass % C, 671 mass % H, and 533 mass % O (a) Determine the empirical formula of lactic acid (b) Determine the molecular formula Plan Assume 100 g lactic acid to express each mass % as grams Convert grams to moles and find the empirical formula

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