Consider two identical 10.0 L one containing and the other containing O_2(g) Bot
ID: 528447 • Letter: C
Question
Consider two identical 10.0 L one containing and the other containing O_2(g) Both flasks are at 0 degree C and 4.0 atm. The density of the gas in the flask with CH_4 is _____the flask with O_2. (A) the same as (B) less then (C) greater than (D) not related to What accounts for the limitations of the ideal gas law? I. Real gas particles occupy space II. Real gas particles are attracted to each other. (A) Only I (B) Only II (C) Neither I II (D) Both I and II The graph shown have is for the first order decomposition of N_2O_3 what is the average rate of the reactions is the interval t = 0 to t = 3 hours? (A) 0.27 M-h^-1 (B) 0.60 m-h^-1 (C) 0.67 M-h^-1 (D) 0.90 M-h^-1 The table gives initial rates for the reaction 2A + B rightarrow C + D starting with different concentrations of A and B. What is the rate law for the reaction? (A) Rate = k[A][B] (B) Rate = k[A]^2[B] (C) Rate = k[A][B]^2 (D) Rate = k[A]^2[B]^2 What is the rate of decomposition of chromate ions (CrO_4^2-) if the rate of formation of dichromate ions (Cr_2O_7^2-) is 0.30 M middot s^-1? 2CrO_4^2- (aq) + 2H^+ (aq) rightarrow Cr_2O_7^2- (aq) + H_2O(l) (A) 1.20 M middot s^-1 (B) 0.60 M middot s^-1 (C) 0.30 M middot s^-1 (D) 0.15 M middot s^-1 What do X and Y represent in this diagram? (A) X + Y is the enthalpy of reaction. (B) X is the activation energy and Y is the enthalpy of reaction. (C) X is the energy of the transition state and Y is the activation energy (D) X is the enthalpy of reaction and Y is the energy of the transition state. The reaction of ozone with NO_2, 2NO_2 + O_3 rightarrow N_2O_5 + O_2, contributes to the is depletion of ozone in the atmosphere. Which rate law is consistent with this proposed mechanism for the reaction? NO_2 + O_3 rightarrow NO_3 + O_2(slow) NO_3 + NO_2 rightarrow N_2O_5 (fast) (A) Rate = k[NO_3][NO_2] (B) Rate = k[NO_3][O_2] [C] Rate = k[NO_2][O_3] (D) Rate = k[NO_2]^2[O_3]Explanation / Answer
Q35
V = 18 L of N2 when P = 1 atm, T = 27°C = 300 K, then find mass of NaN3 required
2 mol of NaN3 = 3 mol of N2
so
mol of N2 required
P V= nRT
n = PV/(RT) = 1*18/(300*0.082) = 0.731707 moles of N2
so
mol of NaN3 = 2/3*0.731707 = 0.4878046 mol of NaN3
mass = mol*MW = 0.4878046 *65 = 31.7 g
best answer is C