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Crazy boy [7]
2 years ago
8

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

Chemistry
1 answer:
Ket [755]2 years ago
7 0

B. White Dwarf.

<h3>Explanation</h3>

The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.

As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:

Chandrasekhar Limit: 1.4 \;M_\odot.

A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its 0.5 \;M_\odot, which is much smaller than the Chandrasekhar Limit.

As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.

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Find the number of grams <br><br>4.00 moles of CU(CN)2 ​
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Answer:

462g

Explanation:

First, let us calculate the molar mass of Cu(CN)2. This is illustrated below:

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4 0
2 years ago
The enthalpy of vaporization (ΔH°vap) of benzene is 30.7 kJ/mol at its normal boiling point of 353.3 K. What is ΔS°vap at this t
vazorg [7]

<u>Answer:</u> The correct answer is Option c.

<u>Explanation:</u>

Vaporization is defined as the physical process in which liquid particles get converted to gaseous particles.

Liquid\rightleftharpoons Gas

The value of standard Gibbs free energy is 0 for equilibrium reactions.

To calculate \Delta S^o_{vap} for the reaction, we use the equation:

\Delta S^o_{vap}=\frac{\Delta H^o_{vap}}{T}

where,

\Delta S^o_{vap} = standard entropy change of vaporization

\Delta H^o_{vap} = standard enthalpy change of vaporization = 30.7 kJ/mol = 30700 J/mol    (Conversion factor: 1 kJ = 1000 J)

T = temperature of the reaction = 353.3 K

Putting values in above equation, we get:

\Delta S^o_{vap}=\frac{30700J/mol}{353.3K}=86.9J/(mol.K)

Hence, the correct answer is Option c.

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3 years ago
A BLANK can be separated by filtration
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A solid from the liquid it suspends can be separated by filtration.

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