AK4 - = +111=740mmHg+30mm(hl)=770mmHg 55’ a M01 He = 7.41...

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Unformatted text preview: = +111=740mmHg+30mm(hl)=770mmHg 55’ a M01 He = 7.41 g He x (1 11161 Pie/4.003 g He) = 1.85 mol PV M=L68x10fi§ T=-“= Latm * R _ "r 1.85 molx 0.08206 mol K t(°C) = (1.68x103 — 273) =1.41><103 °c 38. (M) We first determine the molar mass of the gas. T= 24.3+273.2 = 297.5 K P = 742 mmHg x—iat—m— = 0.976 tm 760 mmHg = mRT _ 2.650 g x 0.08206 L atm mol'IIC1 x 297.5 K W = 155 g/mol 1000 mL Then we determine the empirical formula of the gas, based on a 100.0-g sa ple. mo1c=15.5 ngfl=L29 molC+0.649——)1.99 molC 12.01 g c mol C1 = 23.0 g c1x in—‘i’J—Ci = 0.649 mol C1+0.649 —> 1.00 mol c1 35.45 g C1 molF = 61.5 ng—Lnl’li = 3.24 molF+O.649 —+ 4.99 molF 19.00 g F Thus, the empirical formula is CZClF5 , which has a molar mass of 154.5 g/ 01. This is the same as the experimentally determined molar mass. Hence, the molecul r formula is CZCIFS. ' 0.976atmx[428mLx 11‘ J V___ ———v—m ALvaU VJ. luxv 5H0, Lllvll 1L3 Ulllpllluul .l - ,‘ pieces of information are combined to obtain the molecular formula of the gs. % a L atm 2.33 g/L x 0.08206 x 296 K M = dig}: = I r1103; = 57.7 g/mol 746 mmH h g x 760 mmHg 1 mol C 12.0 g C 1 mol H 1.01 g H Multiply both of these mole numbers by 2 to obtain the empirical formula, CZH5 which has an empirical molar mass of 29.0 g/mol. Since the molar mass (calcula 57.7 g/mol above) is twice the empirical molar mass, twice the empirical formul the molecular formula, namely, C 4H10 . mol C = 82.7 g C x = 6.89 mol C + 6.89 ——> 1.00 mol C mol’H=l7.3ng =17.1molH-:—6.89-—>2.48 molH \ ~\\ ' . M . . . C ‘_,<,,\ ,4 $11)) ' In this case, H2 is the llmltmg reagent 2LNH _ VolumeNH3 =313LH2>< 3 -209LNH3 3 L H2 _ (b) Moles of NH3 (@ 315 °C and 5‘25 am!) 2 mol NH3 01 5.25 atmx313L 23.41x101m01H2xm=2.27x1 = I 2 0.08206 L “m x (315 + 273)K _ mol K L atm -2.27x1o1 mol'NH3 x0.08206 x298 K °c, 727mmHg) = ___.____—————-—T—atE———-——— . V(@25 727 mmHg x R¥ _¥‘ \ 62. (D) M - “MM” ‘ ’ ’ “‘ l ‘ " (a) The %CO2 in ordinary air is 0.036%, %CO2 in expired air is 3.8%. P{CO2 expired air} = 3.8% CO2 _11X102 CO . . . P {CO2 ordinary air} 0.036% CO2 _ ' 2 (emed am to ordl while from the data of this probl calculation shows. 28.013 N Alma.air mol N2 x g 2 mol. ()2 g 02 1 mol O2 +[0.038 mol co2 x 44"” g C02)+[0.059 mol HZOX 39.9 Ar +{0.009 mol Ar x g )= 28.7 g/mol of expired air of or'dfnary air, EExpired air is less d densmes: 3M; , M T1 Ti - - l . 1‘. 2 = = —— = —— Square both Sides and solve for T,. I; = 4x 273 K= 102 K 3R7"2 T2 273 K M Alternatively, recall that 1 J = kg m2 3'2 um = =1,84 x103 m/s ‘ Solve this equation for temperature with ums do bled :.;§§% 2 -3 3 2 ' 3"" = = 2.016x10 kglnol(2xll.:4x120-12r$) =L09X103K _ 3x8.3145 x..§_m_i— molK V l] 8;; (M) x» a ' ' ‘ "I aw“ (:1) 29329121: {£922 = £2.92 =LO7 (b) MEL M(D20) _ rate (02) M(N2) 28.01 rate (D20) - M(H20) '- 14 12 (c) rate (IZCOZ) = M(M = = rate (235 UFs) = M(233 UFG) ‘ rate ( C02) M ( C02) V 46.0 rate (238UF6) M (235 l nRT nza (M) For (:12 (g), nza = 6.49 L2 atm and nb = 0.0562 L. PW: V —nb _ V2 At 0°C, Pvdw = 9.90 atm and Pidea1= 11.2 atm, off by 1.3 atm or + 13% RT 1.00m01x9M—+II:“£><373K )1 mo ° . = _. = = 15.3 atm (a) At 100 C13de V ZIOOL 0.08206Latm l.00mol><———~————XT 2 P = mOIK —6'—49L———az“3=004222"atm—162a “W (2.00 — 0.0562)L (2.0%) =0.0422 x 373K — 1.62 = 14.1atm Race, is offby 1.2 atm or +85% . RT 1.00molx9m—fiéflx473K 11 m0 0 : -— = v = atm _ (b) At 200 C 19.46., V ZOOL Pvdw =0.O4222 x 473 K —1.623 =18.35 atm Pideal is offby 1.0 atm or +55% RT 1.00 mol x W x 673K n mo ° _ =—= =27.6atm Rd“, =0.0422 x 673 K — 1.62 = 26.8 atm 1'?deal is off by 08' atm or +3.0% 1 1, (D) First, balance the equations: ,5 Fe + 2 HCI -—> 2 FeClz +H2 ,2A1+6HC1—->2A1Cl3+3H2_ . 1 "Then, we calculate the partial pressure of 02: P02 =PTOT - PHI0 =841 - 16.5 = 824.5 torr molHZ = (1.0849 atm)(0.159 L) M (0.082058 L - atrn-K-‘)(292 K) Now, we will try to e metals and H2: = 0.007192 mol xpress the calculated moles of H2 using the mole relationship b tween the lmolH 3molH ~ molH = ———2. 1 + —-———2 1 =0.007198 1 2 (momejmo Fe) (2molA1)(mo’”) m0 1 mol Fe 1 mol A1 = 1.0 “~— + 1. . _ ( )(maSSFe)[55.85 g Fe] ( 5X0 1924 maSSFE)(26'98 g Al] 0.007192 = 0.017905 (massFe ) + 0.01070 — 0.055597(massFe) Solving for massFe yields 0.09307 g of Fe. %Fe = (0.09307/0. 1924) x100 = 48.4% A liquid boils when its vapor pressure is equal to the atmospheric pressure; the v‘ por pressure of a liquid increases directly with temperature. Atmospheric pressure decreases with increasing altitude because, at higher elevations, there is less air pushing down on the surface of the liq id. To picture this, image that atmospheric pressure results from a column of air extending from the sur ace of the liquid all the way out to the edge of the earth’s atmosphere-the weight of this column c ‘uses the pressure. Traveling to higher altitudes makes this column of air "shorter", resulting in less pr ssure. Thus, since Denver is at a higher elevation than New York, it experiences reduced atmospheric ressure and water needs to be heated to a lower temperature before its vapor pressure is equal to at ospheric pressure and it can boil. ...
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This note was uploaded on 06/10/2011 for the course CHEM 2090 taught by Professor Zax,d during the Spring '07 term at Cornell University (Engineering School).

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AK4 - = +111=740mmHg+30mm(hl)=770mmHg 55’ a M01 He = 7.41...

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