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zloy xaker [14]
2 years ago
12

If you repeat an experiment and the results are very different from the results you got the first time, the next step would be t

o
Select one:


a. trust the second results.

b. decide which results are better.

c. repeat the experiment again.

d. try a different experiment
Chemistry
2 answers:
Yuliya22 [10]2 years ago
5 0
Answer: option c. repeat the experiment again.

Because, the first approach is to assume that there was a mistake in one or both ot the two trials. Only if the results continue being different in several trials, you have to think in a design error and then modify the experiment.
shutvik [7]2 years ago
4 0
C , you need a definite answer , if you get a completely different results again , id say take the mean / average so you have some results
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Question 3. A batch chemical reactor achieves a reduction in
kotykmax [81]

Answer:

Rate constant for zero-order kinetics: 1, 58 [mg/L.s]

Rate constant for first-order kinetics: 0,05 [1/s]

Explanation:

The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:

r = k [A]^{x} [B]^{y}

where:

  • [A] is the concentration of species A,
  • x is the order with respect to species A.
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  • y is the order with respect to species B
  • k is the rate constant

The concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:

v(t) = -\frac{d[A]}{dt} = k [A]^{n}

For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.

<em>Rate Law:                                    rate = k</em>

<em>Concentration-time Equation:   [A]=[A]o - kt</em>

where

  • k: rate constant [M/s]
  • [A]: concentration in the time <em>t</em> [M]
  • [A]o: initial concentration [M]
  • t: elapsed reaction time [s]

For first-order kinetics, we have:

<em>Rate Law:                                        rate= k[A]</em>

<em>Concentration -Time Equation:      ln[A]=ln[A]o - kt</em>

where:

  • K: rate constant [1/s]
  • ln[A]: natural logarithm of the concentration in the time <em>t </em>[M]
  • ln[A]o: natural logarithm of the initial concentration [M]
  • t: elapsed reaction time [s]

To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

Zero-order kinetics

we use:                        [A]=[A]o - Kt

we replace the data:   5 = 100 - K (60)

we clear K:                 K = [100 - 5 ] (mg/L) /60 (s)  = 1, 583 [mg/L.s]

First-order kinetics

we use:                                  ln[A]=ln[A]o - Kt

we replace the data:               ln(5)  = ln(100) - K (60)

we clear K:                                   K = [ln(100) - ln(5)] /60 (s)  = 0,05 [1/s]

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