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Harman [31]
3 years ago
11

YALL I NEED HELP ASAP

Chemistry
1 answer:
masya89 [10]3 years ago
6 0

Answer:

Yes

Explanation:

 

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M/ABCI<br>m.CADC = 112<br>BK D 117​
raketka [301]

Answer:

what do u mean????? i dont get the question

4 0
3 years ago
Which of the following would you except to see in the death of a star that is less than 0.5 solar mass
Ket [755]

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.

7 0
3 years ago
Percent yield ? how do i do this
Naddik [55]

Answer:Is this from the stochiometry unit?

Explanation:

4 0
3 years ago
A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constant-pressure calorimeter that has a heat cap
san4es73 [151]

Answer:

There is a 1:1 neutralization reaction,  so, number of moles:

0.085 l * 0.900 mol/ L = 0.0765 mol HCL

The heat produced:  0.0765 mol * -56.2 kJ/mol = -4.229 kJ (this is the heat of neutralization)

Change in temperature:    The mass to use in the equation Q=cmT.  

4229 J / (4.186 J/gC*170 g) = 6.042 C

Add to the initial temperature:

18.24 + 6.042 = 24.29 C°

3 0
3 years ago
How much NaAlO2 (sodium aluminate) is required to produce 2.15 kg of Na3AlF6?
Harrizon [31]

Answer:

839 g

Explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

M_r:              81.97                209.94

                   NaAlO₂ + … ⟶ Na₃AlF₆ + …

Mass/g:                                   2150

1. Calculate the <em>moles of Na₃AlF₆ </em>

Moles of Na₃AlF₆ = 2150 × 1/209.94       Do the operation

Moles of Na₃AlF₆ = 10.24 mol Na₃AlF₆

2. Calculate the <em>moles of NaAlO</em>₂

The molar ratio is 1 mol NaAlO₂:1 mol Na₃AlF₆

Moles of NaAlO₂ = 10.24 × 1/1 = 10.24 mol NaAlO₂

3. Calculate the <em>mass of NaAlO₂ </em>

Mass of NaAlO₂ = 10.24 × 81.97             Do the multiplication

Mass of NaAlO₂ = 839 g

5 0
3 years ago
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