Question

15) For a particular process that is carried out at constant pressure. A AE = 140 kJ and AH = 170 kJ. C) AE = 110 kJ and AH = 140 kJ. 140 kJ and w*-30 kl. Therefore, B) AE = 170 kJ and AH = 140 kJ. D) AE - 140 kJ and AH110 kl. 16) How much energy is evolved during the formation of 296 g of Fe, according to the reaction below? Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) AH*rxn--852 kJ A) 5.60 x 103 kJ B) 8.80 x 102 kJ C) 5.50103 kJ (2.26 103 E) 6,60 x 102 kJ 17) Determine the oxidation state of Pin K3PO3. A) O B) +6 C) 3 D)-3 E) +2 18) Calculate the wavelength (in nm) of a the red light emitted by a neon sign with a frequency of 4.74 x 104 Hz. A) 633 nm B) 158 nm C) 466 nm D) 704 nm E) 142 nm L . An n a ntained in minder with a movable niston is increased from 30.0°C to 300.0

Answer 1

(15) Given Pressure = constant q= 140kJ Wa-pdv = -30kJ So, According to first low of thermodynamic AE =q+W = 140+(-30) = 140-30 IDE = 110kJ AH = AE + Alev) = AE + PAV + VAP Since Pa constant so , AP=O! Now, AH = AE+ PAV = 110 + 30 W =-PAV=-30kJ PAV = 30 kJ [AH = 140 kJ | Correct option is c AE = 110 kJ and AH2 140 kJ (16) Fe₂O3(s) + 2Al(s) - AdzO3(s) + 2Fe (s) Atomic mass of iron = 55.845 gmolt Amount of iron formed = 2969 So, Moles of iron formed = 296 g = 5.3 mol 55. 845gmelt

Energy evdued in formation of amol = 852 kJ so, energy evolved in formation of 5.3 mole 352 X 5.3 = 2257.8 kJ 2 - 2.26x103 kJ Correct option is e (17) K₃ PO₂ oxidation oxidation Oxidation state of K-atom= +1 state of Pratom = x state of Bratom = -2 So, 3x(+1) + x + 3xl-2) = 0 egual to zero because not net charge) 3+2-6=0 0-3 =0 SC = +3 so oxidation state of Pratom = +3 correct answer is D (18) Frequency = 4.7481014 H₂ = 4.7481014 st Frequency (0) = c(speed of light) Frequency - Twavelength) . So, A= G = 3x108 mst = 0.6329x186 m 4.748101451 A = 632.9810-9 m correct option is Ą IA ~ 633nm