Please solve the three multiple choice problems 21-23. Please complete all of th
ID: 567870 • Letter: P
Question
Please solve the three multiple choice problems 21-23. Please complete all of them. Thank you. 21- vapor pressure 22- molecules in O2 C) increased nitrogen concentration in body tissues and fluids D) nitrogen starvation 21) Define vapor pressure. A) partial pressure of water in a liquid mixture Page Ref: 5.6 B) partial pressure of water in a gaseous mixture C) condensation of water E) water molecules D) water dissolved in a liquid Page Ref: 5.6 22) A gas mixture consists of N2, 02, and Ne, where the mole fraction of N2 is 0.55 and the mole fraction of Ne is 0.25. If the mixture is at STP in a 5.0 L container, how many molecules of O2 are present? B) 2.7 x 1022 molecules 02 A) 4.5 1022 molecules 02 C) 3.7 x 1023 molecules 02 E) 9.3 1024 molecules O2 D) 1.1 x 1023 molecules 02 Page Ref: 5.6 23) The following reaction is used to generate hydrogen gas in the laboratory. If 243 mL of gas is collected at 25 C and has a total pressure of 745 mm Hg, what mass of hydrogen is produced? A possibly useful table of water vapor pressures is provided below Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) 17.55 23.78 31.86 20 25 30 A) 0.0196 g H2 B) 0.0717 g H2Explanation / Answer
21) B , (pressure exterted by vapors on surface of solution/solvent)
22) total no of mol of mixture(n) = PV/RT
at STP , P = 1atm , T = 273.15 k
= (1*5)/(0.0821*273.15)
= 0.223 mol
molefraction of O2 = 1-(0.55+0.25) = 0.2
molefraction of O2 = nO2/nTotal
0.2 = x/0.223
x= nO2 = 0.0446 mol
No of molecules of O2 = nO2*6.023*10^23
= 0.0446*6.023*10^23
= 2.7*10^22
answer: B
23) partial pressure of H2 = 745-23.78 = 721.2 mmhg
No of mol of H2 gas = PV/RT
= ((721.22/760)*0.243/(0.0821*298))
= 0.009425 mol
mass of H2 gas = 0.009425*2 = 0.019 g
answer: A