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Rudiy27
2 years ago
14

A fish is removed from a contaminated lake. You determine that a particular toxin (X) is present in its cells at concentration X

= 1,500 μg/L. You place the fish in a tank full of clean water (X = 0 μg/L), and measure the toxin concentration in the fish cells each day for the next 10 days.
After making your hypothesis, you test it by measuring the toxin levels in the fish at various times during its 10 days in the tank. You observe that the level of toxin in the fish drops from 1,500 μg/L to 750 μg/L and then stabilizes at 750 μg/L. You test the water in the tank and find that after it stabilizes, toxin is present in the water at concentration 750 μg/L also. What can you do to continue to reduce the toxin level in the fish below 750 μg/L?
Chemistry
1 answer:
netineya [11]2 years ago
3 0
The level of toxins in the fish's cell is equivalent to the level of toxins in the water. Therefore, in order to reduce the toxins further, we should replace the now contaminated water with clean water. After the level of toxins in the fish's cell stops reducing, we replace the water with clean water once again.
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A block of aluminum occupies a volume of 15.0 cm and weighs 40.5 g. What is its density?
mafiozo [28]

Answer:

2.7 g/cm^3

Explanation:

D = m/V

D = 40.5 g / 15 cm^3

D =  2.7 g/cm^3

7 0
2 years ago
) determine the theoretical yield and the percent yield if 21.8 g of k2co3 is produced from reacting 27.9 g ko2 with 29.0 l of c
Airida [17]
The Balanced Chemical Equation is as follow;

                         4 KO₂  +  2 CO₂    →    2 K₂CO₃  +  3 O₂

First find out the Limiting Reagent,
According to equation,

         284 g (4 moles) KO₂ reacted with  =  44.8 L (2 moles) of CO₂
So,
                  27.9 g of KO₂ will react with  =  X  L of CO₂


Solving for X,
                          X  =  (44.8 L × 27.9 g) ÷ 284 g

                          X  =  4.40 L of CO₂

Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,

According to eq.

         284 g (4 moles) KO₂ formed  =  138.2 g of K₂CO₃
So,
         27.9 g of KO₂ will form  =  X g of K₂CO₃

Solving for X,
                        X  =  (138.2 g × 27.9 g) ÷ 284 g

                        X  =  13.57 g of K₂CO₃

So, 13.57 g of K₂CO₃ formed is the theoretical yield.

%age Yield  =  13.57 / 21.8 × 100

%age Yield  =  62.24 %
3 0
3 years ago
Read 2 more answers
Which of the molecules represented below contains carbon with sp2 hybridization?
Darya [45]

Answer:

Explanation:

sp² hybridization is found in those compounds having double bond .

Out of the given compounds only C₂H₂Cl₂ has double bond so this compound contains carbon with sp² hybridization .

Rest have sp³ hybridization because they are saturated compounds .

3 0
3 years ago
Difference between giant ionic structure,giant covalent structure and simple molecular structure and also covalent bonds and ion
Alex_Xolod [135]
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8 0
2 years ago
Classify each of the four compounds as a conjugated, isolated, or cumulated diene. Compound A: Two alkenes are joined by a sigma
Lina20 [59]

Explanation:

Conjugated diene is the one that contains alternate double bonds in its structure. That means both the double bonds are separated by a single bond.

Cumulated diene is the one that contains two double bonds on a single atom. This means it has two double bonds continuously.

Isolated double-bonded compound has a single bond isolated by two to three single bonds.

Compound A: Two alkenes are joined by a sigma bond.

For example:

-CH_2=CH-CH=CH2-

It is a conjugated diene.

Compound B: Two alkenes are joined by a C H 2 group.

It is a cumulative diene.

Compound C: Two alkenes are joined by C H 2 C H 2.

Then it is an isolated alkene.

Compound D:  A cyclohexene has a double bond between carbons 1 and 2. Carbon 3 is an sp 2 carbon that is bonded to another s p 2 carbon with an alkyl substituent.

Hence, compound D is a conjugated diene.

8 0
2 years ago
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