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Alja [10]
2 years ago
14

Which of the following would you expect to see in the death of a star that is less than 0.5 solar mass

Chemistry
2 answers:
melamori03 [73]2 years ago
8 0

Answer:

The correct answer is option C: white dwarf .

Explanation:

 The white dwarf is a type of compact star that is generated when nuclear fuel with a mass of less than 9.10 solar masses has been used up.

This is a stage, and through this stage all the stars will pass, including the Sun (Remember that the sun is a star). This type of star is one of the most abundant.

Given this information we can say that the correct answer is option C.

Neko [114]2 years ago
5 0

The answer is; C

When such as star begins to exhaust its fuel, it expands into a red giant  because it begins to burn helium rather than hydrogen. When the helium is exhausted and it cannot burn carbon, a planetary nebula is formed when the fuel is exhausted. A white dwarf is left behind as remnants which begins to cool down over millions of years.

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The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

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\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+}  \;(aq)+ {\text{OH}}^{-} \; (aq).

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{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l).

\text{OH}^{-} ions from \text{NH}_4\text{OH} will shift the equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} to the right and reduce the amount of {\text{H}_3\text{O}}^{+} in the solution.

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The equilibrium between  \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions will shift to the left to produce more of both ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)

The indicator equilibrium will shift to the left as the concentration of {\text{H}_3\text{O}}^{+} increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

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