CHMI 1007 Midterm review 1. A water sample is found to contain the pollutant chl

Question

CHMI 1007 Midterm review 1. A water sample is found to contain the pollutant chlorobenzene with a concentration of 15 ppb (by mass). What volume of this water contains 5.00x 10 mg chlorobenzene? (Assume density of 1.00 g/ml.) Initial rate data a 25°C are listed in the table for the reaction: 2. NH+(g) + N20g)+ Initial [NO2 NO2 (aq) H2O(n Initial [NH Initial rate of of NH 0.24 0.12 0.12 0.1 0.15 7.2 x 10 3.6 x 10 5.4x a) What is the rate law? b) What is the rate constant? c) What is the reaction rate when the concentrations of NH and NO5are 0.39 M and 0.052M respectively? 3. At 550°C, the reaction follows first order kinetics, in an experiment, it is determined that 75% of a sample of HCOOH has decomposed in 72 seconds. Find the half-life for this reaction. NH.NCO(aq) (NHz)2co,aq) The rate law equation for this process is rates k[NH4NCO], where k-o.01 13 L.mol ,min 1-lf the original concentration of NHNCO in solution is 0.229 mol/L, how long will it take for the concentration to decrease to 0.180 mol/L 5. In the synthesis of ammonia, if-siH,lats 4.5 x 104 mol··min 1, what is NHlat? N2(g) + 3H2(g) 2NH3(g) 6. Dinitrogen trioxide, NOio, decomposes to NO (g) and NO2 (g) in an endothermic process (AH 40.5 Kj/mol). Predict, by comparison of Q and K, the effect of the following changes on the relative concentrations of reactants and products as equilibrium is re-established: a. More N Os is added b. More NO2 is added C. The volume of the reaction flask is increased d. The temperature is lowered An equilibrium mixture at unspecified temperature consists of 3.120 g of PCs, 3.845g of PClh and 1.787 g of Cl in a 1.OOL flask. If you add to this 1.418g of Cla, how will the reaction be affected? What will the concentrations of PCs, PCh, and Cla be when equilibrium is re-established?

Explanation / Answer

Q1.

concentration of chlorobenzene = 15 ppb by mass, that is, 15 micrograms per kg

find volume of water, containing 5*10^2 mg of chlorobenzene

first, calculate total mass of solution

C = mass/volume

15 micrograms per kg = 5*10^2 mg / kg of solution

kg of solution = 5*10^2 mg / 15 micrograms

change to mg:

kg of solution = 5*10^2 mg / 15 *10^-3 mg = (5*10^2 )/(15 *10^-3) = 33333.333 kg of solution

33333.333*10^3 grams, so 33333.333*10^3 mL = 33333.333 L

we need

V = 33333.333 Liters of solution, in order to get a 5*10^2 mg of chlorobenzene

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