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Anika [276]
3 years ago
6

In the dissolution stage of a relationship, ________ comes after stagnation.

SAT
2 answers:
tatiyna3 years ago
8 0

Answer:

avoiding

Explanation:

In Knapp's relational development model, avoidance comes after stagnation when we discuss the coming apart stages. Stagnation is preceded by differentiating  and circumscribing. Avoidance is followed by termination.  

Alex_Xolod [135]3 years ago
8 0

<u>ANSWER: </u>

In the dissolution stage of a relationship avoiding comes after stagnation.  

<u>EXPLANATION: </u>

  • Dissolution in a relationship refers to the formal break up of a relationship. It involves several stages.
  • The stagnation stage is also known as the boring stage is the dissolution stage of a relationship where enthusiasm is being lost.
  • It is followed by avoiding where there occurs an emotional or physical separation and the relationship is pretty much over in this stage.
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Taking into account the definition of molarity, the concentration of CH₃OH in the solution is 3.12 \frac{moles}{liters}.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

Molarity=\frac{number of moles}{volume}

Molarity is expressed in units \frac{moles}{liters}.

<h3>Concentration of CH₃OH</h3>

In this case, you know:

  • number of moles= 50 g× \frac{1 mole}{32.04 g}= 1.56 moles, being 32.04 \frac{g}{mole} the molar mass of CH₃OH
  • volume= 500 mL= 0.5 L (being 1000 mL= 1 L)

Replacing in the definition of molarity:

Molarity=\frac{1.56 moles}{0.500 L}

Solving:

<u><em>Molarity= 3.12 </em></u>\frac{moles}{liters}

Finally, the concentration of CH₃OH in the solution is 3.12 \frac{moles}{liters}.

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<em>A function f (x) and g (x) then:</em>

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Like the number operations we do in real numbers, operations such as addition, installation, division or multiplication can also be done on two functions.

Suppose a function f (x) and g (x) then:

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(f + g) (x) is a new function of the sum of f (x) and g (x)

Likewise with other function operations:

(f-g) (x) = f (x) - g (x)

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