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olga nikolaevna [1]
3 years ago
15

why is the ability of apple snails to survive an increase in the salinity of water considered an adaptation

Chemistry
1 answer:
masya89 [10]3 years ago
7 0

Apple snails inhabit a wide range of ecosystems from swamps, ditches and ponds to lakes and rivers. The majority of the species prefer lentic water above streaming water and only a few species have adapted to rivers with strong current.


<span>Breathing through the siphon when submerged (<span>Pomacea canaliculata</span>).</span>

The lung/gills combination in apple snails reflects their adaptation to habitats with oxygen poor water. This is often the case in swamps and shallow waters. The without thelung they would completely depend on their gills, which would decrease their ability to survive. 
Another advantage of air breathing in combination with a shell door (operculum) is the ability to survive periods of drought often common in these habitats during the dry season. 

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A high frequency sound wave 
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A student requires 2.00 L of 0.100 M NH4NO3 from a 1.75 M NH4NO3 stock solution. What is the correct way to get the solution?
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To solve this we use the equation,

 

M1V1 = M2V2

 

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

 

1.75 M x V1 = 0.100 M x 2.0 L

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3 years ago
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11. A chemical reaction happens when chemicals combine to form a new _______
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substances

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A chemical reaction happens when substances break apart or combine to form one or more new substances.

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A certain sample of a liquid has a mass of 42 grams and a volume of 35 centimeters3. What is the density of the liquid? The dens
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3 years ago
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At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard reaction Gibbs energy is -5798 kJ/mol.
natka813 [3]

Explanation:

The given data is as follows.

             T = 298 K,          \Delta H^{o} = -5645 kJ/mol

          \Delta G^{o} = -5798 kJ/mol

Relation between \Delta H and \Delta G are as follows.

          \Delta G^{o} = \Delta H^{o} - T \Delta S^{o}    

             -5798 kJ/mol = -5645 kJ/mol - 298 \times \Delta S^{o}

                       -153 kJ/mol = -298 \times \Delta S^{o}

                    \Delta S^{o} = 0.513 kJ/mol K

Now, temperature is 37^{o}C = (37 + 273) K = 310 K

Since,        \Delta G = \Delta H^{o} - T \Delta S^{o}

                            = -5645 kJ/mol - 310 K \times 0.513 kJ/mol K

                            = (-5645 kJ/mol - 159.03 kJ/mol)

                            = -5804.03 kJ/mol

As, change in Gibb's free energy = maximum non-expansion work

            \Delta G = \Delta G_{310 K} - \Delta G_{298 K}

                           = -5804.03 kJ/mol - (-5798 kJ/mol)

                           = -6.03 kJ/mol

Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.

5 0
3 years ago
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