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Hitman42 [59]
2 years ago
9

A phytoplankton is continually exposed to an oil-soluble chemical. Over time, the concentration of the chemical within the phyto

plankton increases. This is an example of
Chemistry
1 answer:
Elis [28]2 years ago
3 0

The given statement is example of the bioaccumulation.

What is phytoplanktons?

Phytoplankton, a flora of freely floating, often minute organisms that drift with water currents.

Bioaccumulation is the accumulation of contaminants by species in concentrations that are orders of magnitude higher than in the surrounding environment.

Contuining contamination of chemical leads to bioaccumulation in phytoplanktons population.

Learn more about bioaccumulation here:

brainly.com/question/131305

#SPJ4

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Which is the best description of description of speed?
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A chemical test has determined the concentration of a solution of an unknown substance to be 2.41 M. a 100.0 mL volume of the so
Oksana_A [137]

Answer : The molar mass of unknown substance is, 39.7 g/mol

Explanation : Given,

Mass of unknown substance = 9.56 g

Volume of solution = 100.0 mL

Molarity = 2.41 M

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of unknown substance}\times 1000}{\text{Molar mass of unknown substance}\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

2.41M=\frac{9.56g\times 1000}{\text{Molar mass of unknown substance}\times 100.0mL}

\text{Molar mass of unknown substance}=39.7g/mol

Therefore, the molar mass of unknown substance is, 39.7 g/mol

5 0
3 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
An atom is the smallest identifiable unit of a compound true or false
notsponge [240]
 False that  atom  is the smallest  identifiable  unit  of  a compound.

The   smallest   identifiable  unit  of   a compound  is  the    Element.  Element  is the one which  make up the  compound  and element  is made up by atoms.  Example  of element  is  oxygen  and  hydrogen  which  make  up water (H2O) which is  a compound.
7 0
3 years ago
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