EM 1411 Exam 3 Summer 1 2018 Which of the following bonds would be considered no
ID: 704832 • Letter: E
Question
EM 1411 Exam 3 Summer 1 2018 Which of the following bonds would be considered non-polar A) N-H ?) ?? c) c-o 11) covalent? E C-C -12) Determining lattice energy from Bom-Haber eycle data requires the use or A) B) Coulomb's law the octet rule Periodic law Aes's lavmdetermuna aicer enery Avogadro's number What is the rms speed (m/s) of NO2 at 28.4 °C? A) 124 B) 40.2 C) 404 D) 233 E) 12.8 13) Consider the following species when answering the following questions: BClh (ii) CCl (ii) TeCl (iv) XeF (v) SF 14) Which of the molecules has a see-saw shape? B) (v) c) 0) D) (iv) E) (iii) A 15)of the hydrogen halides, only- -is a weak acid. A) HF (aq) HBr (aq) C) HI (ag) D) HCI (aq) E) They are all weak acids. A 16) Of the halogens, which are gases at room temperature and atmospheric pressure? fluorine and chlorine fluorine, chlorine, and iodine c) fluorine, chlorine, bromine, and iodine D) fluorine, chlorine, and bromine E) fluorine, bromine, and iodineExplanation / Answer
11) if the electronegativity difference is lessthan 0.5, it is non-polar covalent bond.
answer: D) C-H (electronegativity difference = C(2.55)-H(2.2) = 0.35)
12) answer: D) hess law
13) answer: c) 404
rms velocity(Vrms) of NO2 = sqrt(3RT/M)
R = 8.314 j.k-1.mol-1
T = 28.4 C = 301.55 k
M = molar mass of NO2 = 46*10^-3 Kg/mol
Vrms = sqrt(3*8.314*301.55/(46*10^-3))
= 404 m/s
14) answer: E) iii
Tecl4 = AB4L type molecules has see-saw shape.
15) answer: A) HF , (because F-has more electronegativity)
16) answer: A) F2 and Cl2
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