Argon atom can exist freely in nature, why

Chemistry

1 answer:

lana [24]2 years ago

7 0

Answer:

Argon can exist freely in nature because it has a full octet of electrons the way its found in the nature is the same way its found in periodic table of element in vast amouts of stabilization.

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If the fugacity of a pure component at the conditions of an ideal solution is 40 bar and its mole fraction is 0.4, what is its f

yaroslaw [1]

Answer : The fugacity in the solution is, 16 bar.

Explanation : Given,

Fugacity of a pure component = 40 bar

Mole fraction of component = 0.4

Lewis-Randall rule : It states that in an ideal solution, the fugacity of a component is directly proportional to the mole fraction of the component in the solution.

Now we have to calculate the fugacity in the solution.

Formula used :

$f_i=X_i\times f_i^o$

where,

$f_i$ = fugacity in the solution

$f_i^o$ = fugacity of a pure component

$X_1$ = mole fraction of component

Now put all the give values in the above formula, we get:

$f_i=0.4\times 40\text{ bar}$

$f_i=16\text{ bar}$

Therefore, the fugacity in the solution is, 16 bar.

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2 years ago

Explain the difference between an endothermic and an exothermic reaction using the concepts of potential energy, stability, and

kakasveta [241]

An Exothermic reaction releases energy into the surroundings and so the products have more potential energy then the reactants. The enthalpy change is a negative value. Whereas, an endothermic reaction involves the absorption of energy into the system and so the reactants have more potential energy than the products. The enthalpy change is a positive value. This is clearly represented in energy profile diagrams.

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2 years ago

Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.

Svetradugi [14.3K]

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$