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3 years ago

For most stars, what does a higher temperature tend to go with?

Chemistry

1 answer:

nexus9112 [7]3 years ago

8 0

<h3>For the majority of the stars, a higher temperature tends to correlate with the higher luminosity. </h3>

<h3>Luminosity refers to the rate of the total radiant energy output of a star. It is the inherent brightness of a star covered over the whole electromagnetic spectrum. The luminosity of a star is a determination of its brightness. The more energy emitted, the higher the brightness. </h3><h3>Temperature also influences the rate of the luminosity of the star. As a star gets hotter or with the increase in temperature, the number of nuclear reactions upsurges, generation of more reactions and more energy takes place. Thus, ultimately resulting in higher luminosity of a star.</h3>

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A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following

marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

$\frac{V_1}{n_1} =\frac{V_2}{n_2}$

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}$

<em>V₂ = 4,00L</em>

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

$\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}$

<em>V₂ = 16,00L</em>

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

$\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}$

<em>V₂ = 12,31L</em>

I hope it helps!

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