CHEM 1110 REVIEW oUIZ Bubble in the circle (·) on Scantru. Sbrerot pr-ided DO NO
ID: 696992 • Letter: C
Question
CHEM 1110 REVIEW oUIZ Bubble in the circle (·) on Scantru. Sbrerot pr-ided DO NOT MAKE ANY MARKS IN THE QUIZ BOOKLET! Use only the scratch paper that is prvied TURN IN EVERYTHING WHEN YOU FINISH R-0.0821 L atm/mol K-8.314/ol K -1.987 cal/mol K 760 mm Hg 760 torr-76 enn Hg 1 mol of gas accupies 224 L a ST 1. How many atoms of hydrogen are contained in 12 gof CSI. IM-44. a. 44 x 10 b. 18 x 105 c. 13 x 10 d. 1.6 x 10 1.25 mL of SO: gas effuses through a pinbole in 135 sec. Whust is the molecular weigh of a second gas if 1.25 ml of it requires 185 see at the same T and P a. 4 b. 34 . 120 d. 88 Determine the heat of reaction for the combustion of 1 mole of C.HO compound C21 102 COmp -454 394 a. 2384 b +226 c. +2384 d. 226 What is the energy of the transition from the N-2 energy level to the N 3 energy evel? a. -3.63 x 10-19 J/atom b. +3.03 x 10-19 J/atom c. 3.03 x 10-19 J/atom d. +3.63 x 10',Jatom 4. The energy of the ground electronic state in the hydrogen atom is 2.18 x 10atomExplanation / Answer
1)
number of moles of C3H8 = 12g/44g/mol = 0.2727 moles
1 mole of C3H8 contains 8 moles of hydrogen atoms
0.2727 moles of C3H8 contains 0.2727*8 moles of H atoms
number of atoms of Hydrogen = 0.2727 * 8 * 6.023 * 10^23 = 1.3 * 10^24
Answer : C
2)
r1/r2 = square root of (M2/M1) according to Graham's law of effusion = t2/t1
where r1 = rate of effusion o SO2
r2 = rate of effusion of another gas
t1 = time taken for effusion of SO2
t2 = time taken for effusion of unknown gas
M1 = molecular weight of SO2
M2 = molecular weight of unKnown
t2/t1 = square root of (M1/M2)
185/135 = square root of (64/M2)
M2 = molecular weight of unknown = 34 g/mol
Answer : B
3)
2 C2H6O2 + 5 O2 ------> 4 CO2 + 6 H2O
dH rxn = dH products - dH reactants
dH rxn = 4*dH CO2 + 6*dH H2O - 2 * dH C2H6O2 - 5 dH O2
dH rxn = 4*(-394) + 6*(-286) - 2*(-454) - 5*0
dH rxn = -2384 kJ/mol
Answer : A
4)
E = -Rh(1/n1^2)
2.18 * 10^-18 = - Rh(1/1^2)
Rh = -2.18 * 10^-18
E = -Rh(1/n1^2-1/n2^2)
where Rh = rydberg constant = -2.18 * 10^-18
n1 = initial state of electron
n2 = final state of electron
E = -(-2.18 * 10^-18) (1/(2)^2 - (1/3)^2)
E = +3.03 * 10^-19 J/atom
Which indicate the energy released from atom when the electron excited from 2 state to 3 state.
Answer : B