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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.

You might be interested in
Is milk a solution, suspension, compound , colloid ?
kakasveta [241]

Answer:

Colloids (heterogeneous)

The difference between a colloid and a suspension is that the particles will not settle to the bottom over a period of time, they will stay suspended or float. An example of a colloid is milk. Milk is a mixture of liquid butterfat globules dispersed and suspended in water.

3 0
2 years ago
CAN SOMEONE PLEASE HELP ME ASAP PLEASEE ANYBODY LITERALLY ANYONE OUT THERE PLEASE
IgorC [24]

Answer:

Electrons

Explanation:

Atoms can combine together and share electrons between them. Atoms that share electrons are linked to each other in a form, called a molecule that is in a lower energy state than either of the separate atoms alone.

Pls Brainliest! It would mean a lot! ;)

4 0
3 years ago
What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

3 0
3 years ago
Respuesta funcion oxido de Mg+2 + O-2
Effectus [21]

Answer:

Mg2 + O2 → 2MgO

Explanation:

Hope this helps!! I got it right.

7 0
2 years ago
Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo
Leto [7]

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

CH_4+2O_2\rightarrow CO_2+2H_2O

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

left\ over=0g

Regards.

7 0
3 years ago
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