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3 years ago

How does the density of protostars compare to other stars?

Chemistry

1 answer:

lorasvet [3.4K]3 years ago

5 0

Protostars are less dense than other stars.

Explanation:

Protostars are very young ‘stars’ made from hydrogen clouds that are beginning to coalesce and collapse under their weight. The hydrogen has not even begun fusing. Therefore, they are mainly made of hydrogen which is the lightest element in the universe.

Stars, however, have begun fusing hydrogen to other heavier elements like helium, carbon, oxygen, and iron. The elements are much heavier than hydrogen making other stars much denser than protostars.

Learn More:

For more on protostars vs stars check out;

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Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)

allsm [11]

Answer:

$\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}$

Explanation:

1. Density from mass and volume

$\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}$

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$V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}$

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$\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}$

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water = 12.8 mL

Volume of object = 11.8 mL

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3 years ago

the specific heat capacity of water is 1 cal/gc how much heat in calories is released when 25.7 of water is cooled from 85 to 49

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Answer:

The amount of heat that is released is -925.2 cal

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is the amount of heat that a body can receive or release without affecting its molecular structure, that is, it does not change the state (solid, liquid, gaseous). In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state.

The equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

c= 1 $\frac{cal}{g*C}$
m= 25.7 g
ΔT= Tfinal - Tinitial= 49 °C - 85 °C= -36 °C

Replacing:

Q= 1 $\frac{cal}{g*C}$ *25.7 g* (-36 C)

Solving:

Q= -925.2 cal

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