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1 year ago

___________ is a dangerous concept because toxic chemicals concentrate as they pass up the food chain. group of answer choices

1 answer:

3 0

<u>Biomagnification</u> is a dangerous concept because toxic chemicals concentrate as they pass up the food chain. group of answer choices

What is biomagnification?

Biomagnification is defined as the accumulation of a particular substance in the body of the organisms at different trophic levels of a food chain.
An example for biomagnification is the accumulation of insecticide DDT which gets accumulated in zooplanktons.
Zooplanktons are those which consumes small fishes.

In the entire food chain or in the predators, the chemical toxic will gets increased.
It makes all the animals to get more impact and also it makes the food chain to the higher toxic.
Biomagnification can be also known as the biological magnification.
Also in plants, if a farmer sprays pesticides on the crops which enters the food chain from crops these pesticides enter into the organisms that feed on it.

Learn more about Biomagnification,

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What is the molar solubility of nickel(II) sulfide in 0.091 M KCN? For NiS, Ksp = 3.0 × 10 –19; for Ni(CN) 4 2–, Kf = 1.0 × 10 3

Marta_Voda [28]

Answer:

The value is $x = 0.0227 \ M$

Explanation:

From the question we are told that

The concentration of $KCN \ \ i.e \ \ CN^{-}$ is $M_1 = 0.091 \ M$

The solubility product constant for $NiS$ is $K_{sp} = 3.0 *10^{-19}$

The stability constant for $Ni(CN)_4 ^{2-}$ is $K_f = 1.0 *10^{31}$

Generally the dissociation reaction for NiS is

$Ni S \underset{}{\stackrel{}{\rightleftharpoons}} Ni^{2+} + S^{2-}$

Generally the formation reaction for $Ni(CN)_4 ^{2-}$ is

$4CN^- + N_i ^{2+} \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_{4}$

Combining both reaction we have

$4CN^ - + NiS \ \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_4 + S^{2-}$

Gnerally the equilibrium constant for this reaction is

$K_c = K_{sp} * K_f$

=> $K_c = 3.0 *10^{-19 } * 1.0 *10^{31}$

=> $K_c = 3.0*10^{12}$

Generally the I C E table for the above reaction is

$4CN^ - \ \ \ + \ \ \ NiS \ \ \ \ \ \ \ \underset{}{\stackrel{}{\rightleftharpoons}} \ \ \ \ \ Ni(CN)^{2-}_4 \ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \ S^{2-}$

initial [ I] 0.091 0 0

Change [C] -4x +x + x

Equilibrium [E ] 0.091 - 4x x x

Here is x is the amount in term of concentration that is lost by $CN^-$ and gained by $Ni(CN)_4 ^{2-}$ and $S^{2-}$

Gnerally the equilibrium constant for this reaction is mathematically represented as

$K_c = \frac{[Ni (CN)_4^{2-} ] [S^{2-} ] }{ [CN^{-}]^4}$

=> $3.0*10^{12} = \frac{x * x}{ [0.091 - 4x ]^4}$

=> $3.0*10^{12}* [0.091 - 4x ]^4 = x^2$

=> $[0.091 - 4x ]^4 = \frac{x^2}{3.0*10^{12}}$

=> $[0.091 - 4x ] = \sqrt[4]{ \frac{x^2}{3.0*10^{12}}}$

=> $[0.091 - 4x ] = \frac{\sqrt{x} }{1316}$

=> $119.8 - 5264x =\sqrt{x}$

Square both sides

$(119.8 - 5264x)^2 =x$

=> $14352.04 - 1261255 x + 27709696x^2 = 0$

=> $27709696x^2 - 1261255 x + 14352.04 = 0$

Solving using quadratic equation

The value of x is $x = 0.0227 \ M$

Hence the amount in terms of molarity (concentration) of $Ni(CN)_4 ^{2-}$ and $S^{2-}$ produced at equilibrium is $x = 0.0227 \ M$ it then means that the amount of NiS (nickel(II) sulfide) lost at equilibrium is $x = 0.0227 \ M$

So the molar solubility of nickel(II) sulfide at equilibrium is

$x = 0.0227 \ M$

3 0

3 years ago

what would be the final volume when 2.20 M solution is made from 25.0 mL of a 12.0 M solution? plzz show work

cupoosta [38]

Answer:

136.36 mL

Explanation:

Here we have to use the dilution formula

From C1V1= C2V2

Where;

C1= initial concentration of the solution= 12.0 M

C2= final concentration of the solution= 2.20 M

V1 = initial volume of the solution= 25.0 ml

V2= final volume of the solution= ?????

Then recall;

C1V1=C2V2

V2 = C1V1/C2

Substituting values from the parameters given;

V2= 12.0 × 25.0 / 2.20

V2= 136.36 mL

7 0

3 years ago

Calculate the pH of a 0.0150 M HNO3 solution. Assume HNO3

stellarik [79]

Answer: 1.824

Explanation: facts

4 0

4 years ago

How many moles in 2 h2o molecules

alexandr1967 [171]

Answer:

2 molecules of water represents 3.32 x 10^-24 moles of water.

Explanation:

To find the solution to this problem, you have to use the concept of Avogadro´s number, that is in 1 mol of any element o compound there are 6.022 x 10^23 molecules. Then,

1 mol H2O ------------- 6.022 x 10^23 molecules

x= 3.32 x 10^-24 ---- 2 molecules.

2 molecules of water represents 3.32 x 10^-24 moles of water.

5 0

3 years ago

Eq-43 which type of fire extinguisher should be onboard a vessel with a permanently installed fuel tank

liq [111]

I think a type B fire extinguisher should be on board a vessel with a permanently installed fuel tank.
All vessels are required to have a type B fire extinguisher on board if one or more of the following conditions exists. That is; Inboard engine, vessel length of 26 feet or longer, enclosed living spaces, closed storage compartments in which flammable or combustible materials may be stored, permanently installed fuel tanks, and also closed compartments where portable fuel tanks may be stored.

7 0

3 years ago