Answer:
Molar mass of unknown solute is 679 g/mol
Explanation:
Let us assume that the solute is a non-electrolyte.
For a solution with non-electrolyte solute remains dissolved in it -
Depression in freezing point of solution, 
where, m is molality of solute in solution and
is cryogenoscopic constant of solvent.
Here 
If molar mass of unknown solute is M g/mol then-

So, 
so, M = 679 g/mol
In Lewis dot structures, you draw the atom in the center and then surround the atom with its valence electrons. The Lewis structure for O is as shown in the attached diagram.
<h3>What is the Lewis structure of O ?</h3>
Lewis Structure of an atom of oxygen contains 6 electrons in the valence shell. Four valence electrons exist in lone pairs. It implies that oxygen atom must participate in two single bonds or one double bond in order to have an octet configuration.
A simplified representation of the valence shell electrons in a molecule is called Lewis Structure. It shows how electrons are arranged around individual atoms in the molecule.
To know more about Lewis structure, refer
brainly.com/question/1525249
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Producers (autotroph)
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Answer:
Explanation:
Nitrogen gas in nitrogen cylinder is in gaseous form only . So when gas is released , pressure decreases because moles of gas in the cylinder decreases .
Propane gas in cylinder is in liquified form . The gas is partially in gaseous and liquid form . So when gas is released outside , pressure decreases inside but due to low pressure , some liquid form is converted into gaseous form which restores the pressure of gas inside cylinder . Hence , the pressure of gas remains unchanged inside cylinder.
Graduated cylinder
Stopwatch
Thermometer