QUESTION 1 The following data may be used for the questions that follow: NH 3(g) + BF 3(g) →F 3BNH 3(s). Initial Rate Experiment [BF3] [NH3](M) (M/s) 1 0.250 0.250 0.2130 0.250 0.125 0.1065 0.200 0.100 0.0682 0.350 0.100 0.1193 0.175 0.100 0.0596 Based on the data in the preceding table, what is the reaction order with respect to NH 3? Zero order First order Second order Cannot be determined QUESTION 2 The following graphs can be used for the questions that follow:
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- Trial [?] (?)[A] (M) [?] (?)[B] (M) Rate (M/s) 1 0.310 0.370 0.0139 2 0.310 0.740 0.0139 3 0.620 0.370 0.0556 Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D ?= Units:Rate Law find the initial rate N2O4 - 2 NO2 experiment initial (N2O4) (M) initial rate (M/sec) 1 0.50 0.050 2 1.00 0.200 3 1.50 0.450Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D Trial [?] (?) [?] (?) Rate (M/s) 1 0.360 0.290 0.0144 2 0.360 0.580 0.0144 3 0.720 0.290 0.0576 k=
- Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D Trial [?] (?)[A] (M) [?] (?)[B] (M) Rate (M/s) 1 0.360 0.330 0.0164 2 0.360 0.660 0.0164 3 0.720 0.330 0.0656 ?=? Units =?Consider the following complex reaction: A + 2B → C whose initial rate at 25 °C was measured using three different sets of initial concentrations as listed in the following table: Trial [A](M) [B](M) Rate(M/s) 1 0.400 0.050 9.60 × 10−3 2 0.400 0.100 1.92 × 10−2 3 0.800 0.050 3.84 × 10−2 A) Write the rate law for this reaction. Use k to represent the rate constant. B) Solve for k in the above rate law. Express your answer to three significant figures. Include units for k.These questions are based on a reaction rate lab for my Chemistry homework. I attached my data values that I have already calculated. I think I am supposed to show more work for 4 and 5. I am genuinely confused on how to plot the graph too (6-7). Lastly, I am not sure if I got 8-10 correct? I think the rate law might be Rate = k [HCl] [Na2S2O3] but I am not sure if it needs an exponent or not?
- Rate = k[acetone]1[I2]0[H+]1 rate = 0.001M/ 332.26s = 3.01 x 10-6 my question is what is the S.I unit of this answer? 3.01 x 10-6 = k[0.80M]1 [0.001M]0 [0.20]1 k= (3.01 x 10-6)/ (0.16) k = 1.88 x 10-5 also S.I unit of this answer with clear explanation and stepsCalculate the full rate law using the information below: 2 NACHOS (g) → 12 CHIPS (g) + 3 CHEESE (g) [NACHOS]0 (M) Initial Rate (M–1s–1) 0.101 0.00484 0.0910 0.00437 Group of answer choices Rate = 5.61 × 10–2 M–1 s–1 [NACHOS]2 Rate = 6.03 × 10–1 M–2 s–1 [NACHOS]3 Rate = 4.79 × 10–2 s–1 [NACHOS] Rate = 1.66 × 10–3 M1/2 s–1 [NACHOS]2/3 Rate = 1.78 × 10–2 M–1/2 s–1 [NACHOS]3/2Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D Trial [?] (?) [?] (?) Rate (M/s) 1 0.340 0.210 0.0204 2 0.340 0.420 0.0204 3 0.680 0.210 0.0816 ?=
- Hello, I am having trouble with this lab question. 3. Calculate the rate for 40 years (from 1980 to 2020) – change of the rate in a year.( Y2-Y1) / (X2-X1) = (CO2_end – CO2_begin) / (Yearend – Yearbegin)What is the rate? Im not sure if I am doing it correctly ( 412.46 - 338.91) 40 years - 0 ? year mean 1980 338.91 1981 340.11 1982 340.86 1983 342.53 1984 344.08 1985 345.55 1986 346.96 1987 348.68 1988 351.16 1989 352.79 1990 354.05 1991 355.39 1992 356.1 1993 356.83 1994 358.33 1995 360.18 1996 361.93 1997 363.05 1998 365.7 1999 367.8 2000 368.98 2001 370.57 2002 372.59 2003 375.14 2004 376.95 2005 378.97 2006 381.13 2007 382.9 2008 385.01 2009 386.5 2010 388.76 2011 390.64 2012 392.65 2013 395.39 2014 397.34 2015 399.65 2016 403.09 2017 405.22 2018 407.61 2019 410.07 2020 412.46Determine the missing initial rate: 2N2O5 → 4NO2 + O2 [N2O5] Initial rate (Ms-1) 0.093 4.84x10-4 0.084 4.37x10-4 0.224 ??Determine the average rate of change of BB from ?=0 st=0 s to ?=272 s.t=272 s. A⟶2BA⟶2B Time (s) Concentration of A (M) 0 0.7300.730 136136 0.4450.445 272272 0.1600.160 rateB= __________M/s