T-Mobile LTE 2:49 PM 0 99% Done 6 of 6 CHM3,SECT.\'s 009 THURSDAY,OCTOBER 19.201
ID: 544919 • Letter: T
Question
T-Mobile LTE 2:49 PM 0 99% Done 6 of 6 CHM3,SECT.'s 009 THURSDAY,OCTOBER 19.2017 Answer all eauestions correctly in pen in the namination booklet (10 pekew l) How many grams of argon gas are presem in a 7.5 liter sample that eserts a peessure of 575 mm Hg at 35 C? 2) A 1.58 gram sample of C,H,X,has a volume of 297 mL at 769 the element X Hg and 35'C.e 3) What mass of NaCN is needed to form 8.5L of HCN at 22'C, 750 mmm Hg 4) A 355 mL sample of O, and water vapor is collected at 27 C. The mixture everts a total peessure of 775 mm Hg, with water having a vapor pressure of 27 mm Hg. How many grams of O, are in the mixture? 5) An effusion experiment takes 32 seconds to complete using NH y How long woudád iz take to perform the experiment using HBO 6)A45 gram sample ofironmetal kses 3,412 Joules of heat when cooling from 35°Cto! What is the specific heat (c) of iron? 7) A 2.80 gram sample of CaCI, is dissolvod in a calorimeter containing 100 grams of water i c 4.18 J/gC). The temperature of the water rises from 20.5 C to 25.4 C. What is the heat of the reaction: CaCl." Car-+ 2cr 8) How much heat is generated in the production od 31.6 grams of AL,0, AI+30,-2 A1,O H-33514k 9) What is the enthalpy of the reaction 20+H, -CH, given 11-"2999 AH-393.5 J AI 1-285.8 kJ 2C,H,+50,-4CO,+2H,O C+0,-CO 10) Using the data provided, determine the enthalpy of the reaction: NH,+30, 2N,+61,0Explanation / Answer
1.
Assuming the ideal gas beahviour of Ar,
Ideal gas equation is written as,
P V = n R T ------------- (1)
Given that,
P = 575 mmHg = 575 / 760 atm
V = 7.5 L
T = 35 + 273.15 = 308.15 K
R = 0.0821 L.atm.K-1.mol-1
Therefore,
(1) =>
n = P V / R T
n = (575/760) * 7.5 / (0.0821 * 308.15)
n = 0.224 mol
But,
amount of substance = mass / molar mass
0.224 = Mass of Ar / 40.0
Mass of Ar = 8.97 g.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.