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:
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Answer:
well in a solid pack together as tightly as possible in a neat and ordered arrangement.
and a liquid are close together (touching) but they are able to move/slide/flow past each other
Explanation:
Answer: The enthalpy change is 34.3 kJ
Explanation:
The conversions involved in this process are :
Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change = ?
m = mass of water = 72.0 g
= specific heat of ice = 
= specific heat of liquid water = 
n = number of moles of water = 
= enthalpy change for fusion = 6010 J/mole
Now put all the given values in the above expression, we get
![\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B72.0g%5Ctimes%202.09J%2Fg%5E0C%5Ctimes%20%280-%28-18%29%5E0C%5D%2B4.00mole%5Ctimes%206010J%2Fmole%2B%5B72.0g%5Ctimes%204.184J%2Fg%5E%29C%5Ctimes%20%2825-0%29%5E0C%5D)
(1 KJ = 1000 J)
Therefore, the enthalpy change is 34.3 kJ
Answer:
5Atm
Explanation:
I just guess and it’s right
Answer:
0.75 g/cm³
Explanation:
Given data:
Mass of wooden block = 180 g
Length of block = 10 cm
Width of block = 6 cm
Height or thickness = 4 cm
Density of block = ?
Solution:
Volume of block = height × length × width
Volume of block = 4 cm × 10 cm× 6 cm
Volume of block = 240 cm³
Density of block:
density = mass/ volume
d = 180 g/ 240 cm³
d = 0.75 g/cm³
