What is mandrake tuber
1 answer:
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Answer:
Explanation:
When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,
Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C
Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H
Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O
Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.
Moles C = 3.41/3.41 = 1
Moles H = 4.58/3.41 = 1.34
Moles O = 3.41/3.41 = 1
Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3
Moles C = 1x3 = 3
Moles H = 1.34x3 = 4
Moles O = 1x3 = 3
Empirical Formula
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:
where:
- [A] is the concentration of species A,
- x is the order with respect to species A.
- [B] is the concentration of species B,
- 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:
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]