Answer:
»Climatic change leads to malnutrition.
e.g when its cold season, osmoregulation decreases and rate of metabolism declines.
This is because the enzymes lack that optimum temperature for their catalytic activity to take place hence improper digestion.
»Leads to promotion of predation in ecosystem.
such as reptiles e.g snakes, lizards get away from shelters on cold days since they are cold blooded, hence their predators such as animals, birds increase.
Answer:
Explanation:
The pressure at the bottom of the tank is:
The force exerted on the circular bottom is:
![F=(73581.921\,Pa)\cdot (\frac{\pi}{4} )\cdot [(12\,ft)\cdot (\frac{0.305\,m}{1\,ft} )]^{2}](https://tex.z-dn.net/?f=F%3D%2873581.921%5C%2CPa%29%5Ccdot%20%28%5Cfrac%7B%5Cpi%7D%7B4%7D%20%29%5Ccdot%20%5B%2812%5C%2Cft%29%5Ccdot%20%28%5Cfrac%7B0.305%5C%2Cm%7D%7B1%5C%2Cft%7D%20%29%5D%5E%7B2%7D)
You need to find which intermolecular forces are between the molecules
dipole-dipole,h bonds, etc.
I'm not very good at explaining but this is what my prof said to help us
Identify the class of the molecule or molecules you are given. Are they nonpolar species, ions or
do they have permanent dipoles? Is there only one species or are there two?
In the case of ONE species (i.e., a pure substance), the intermolecular forces will be between
molecules of the same type. So if you are dealing with ions, the intermolecular forces will be ION-
ION or IONIC. If you are dealing with dipoles, then the intermolecular forces will be DIPOLE-
DIPOLE. If you are dealing with nonpolar species, the intermolecular forces will be DISPERSION
or VAN DER WAALS or INDUCED DIPOLE-INDUCED DIPOLE (the last three are desciptions
of the same interaction; regrettably we cannot call them nonpolar-nonpolar!).
In the case of TWO species (i.e., a mixture), the intermolecular forces will be between molecules of
one type with molecules of the second type. For example, ION-DIPOLE interactions exist between
ions dissolved in a dipolar fluid such as water.
Answer:
ionic
Explanation:
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.
The minimum volume of 0.2M sodium sulfide that will precipitate out aluminum from 50.0 mL, 0.25 M aluminum nitrate would be 0.094 L or 94 mL
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
Mole ratio of Na2S and Al(NO3)3 = 3:2
Mole of 50.0 mL, 0.25 M Al(NO3)3 = 50/1000 x 0.25
= 0.0125 mole
Equivalent mole of Na2S = 3/2 x 0.0125
= 0.0188 mole
Volume of 0.2M, 0.0188 mole Na2S = 0.0188/0.2
= 0.094 L or 94 mL
More on stoichiometric calculations can be found here: brainly.com/question/8062886
