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svet-max [94.6K]
3 years ago
12

Activated carbon binds to most substances through van der waals or london dispersion forces. how did you use ethanol to confirm

that the pollutant had been removed from the water by adhering to the activated carbon?
Chemistry
1 answer:
sattari [20]3 years ago
5 0

Answer:

The answer is a reaction of aqueous or absolute ethanol with the pollutant

Explanation:

Activated carbon filtration is a commonly used water treatment technology based on the adsorption of contaminants onto the surface of a filter. This method is effective for the removal of certain organics (such as unwanted taste and odours, micro-pollutants and synthetic organic chemicals), chlorine, fluorine or radon from drinking water or wastewater.

Carbonaceous resin have been known to be cleansed by ethanol 20% more effectively than steam.

Becauseof the hydrogen and van der waals force in a mixture of ethanol and water, releasing the pollutants into aqueous or absolute ethanol, and testing the resulting purity of the alcohol would determine whether or not purification of the water sample has occurred.

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3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con
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Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

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Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}

c) Half life for first-order kinetics is computed by:

t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}}  =53.3min

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M

Best regards.

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