Why can salt and sugar both dissolve in water, even though one is ionic and the other is covalent?
2 answers:
Answer:
Explanation:
The water molecules, , are polar, with positive partial charges on the hydrogen and a negative partial charge on the oxygen. This makes it possible to interact with ionic compounds such as salt (NaCl). This interaction occurs through the partially positive and negative ends of the water, so that the positive charges attract the negative ones. The NaCl salt in water dissociates (separates) into
and
ions.
ions are surrounded by negative partial charges of water oxygen, while
ions are surrounded by hydrogen ends with positive partial charge. In this way the salt is dissolved in water.
Sugar is a molecular compound formed by covalent bonds. In a polar covalent bond, electrons are shared unevenly. This creates a slightly negatively charged end and a slightly positively charged end. This is what happens with sugar. These extremes are those that interact with the positive and negative extremes of water, mentioned previously. Then it is possible to say that sugar dissolves in water because both substances are polar substances.
In short, <em>water dissolves most of the substances that are polar or ionic, as in the case of sugar and salt. </em>
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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]