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Afina-wow [57]
3 years ago
8

What is the connection between diamonds and stardust?

Chemistry
1 answer:
Ede4ka [16]3 years ago
8 0

According to an article dated back in February 8, 1992 which is entitled, “Science: Stardust is made of diamonds” on a website called newscientist (https://www.newscientist.com/article/mg13318073-000-science-stardust-is-made-of-diamonds/), American astronomers believed that diamonds are made in supernova explosions. It was said that the diamonds were the foundation of uncommon combinations of isotopes found in some meteorites. Donald Clayton of Clemson University in South Carolina suggested that the weightiest isotopes were more common in meteorites for the reason that the rare gases shaped in the neutron-rich outcome of a supernova explosion. Clayton also said, “the observed mixture of isotopes could have been produced only during the collapse of a massive star to form a neutron star”. This happens in a Type II explosion, for example the Supernova 1987A in the Large Magellanic Cloud. And rare gases like xenon become stuck in both weighty and light isotopes after the ejected gas from such a supernova cools down enough to create dust. The existence of the diamonds with these unusual gases in meteorites infers an alike source. Some of the carbon in the supernova fragments produces ordinary graphite dust, whereas some produces diamond dust. Considerable amount of stardust may be made of diamonds, if Clayton was not mistaken.

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Delicious77 [7]

Answer:

Explanation:P = 1 mol x 30.97 g/mol = 30.97 g N = 2 mol x 14.01 g/mol = 28.02 g. O = 4 mol x ... 5.68 g O2 f. Molecules of water produced? 2.46 g H2O ×. 1 mol H2O. 18.02 g H2O. × ... How many grams of sodium sulfate will be produced by the complete ... 2 mol BF3. 3 mol H2. = 1.65 mol BF3 required. Calculate moles H2 required:.

4 0
3 years ago
Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.1
inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V

Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

This is less than zero[0]. Therefore,  it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.  

7 0
3 years ago
The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?
Alex Ar [27]

____________________________________________________

Answer:

Your answer would be a). 2.0 × 10-9

____________________________________________________

Work:

In your question the "ph" of a 0.55 m aqueous solution of hypobromous acid temperature is at 25 degrees C, and it's "ph" is 4.48.

You would use the ph (4.48) to find the ka for "hbro"

[H+]

=

10^-4.48

=

3.31 x 10^-5 M

=

[BrO-]

or: [H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-]

Then you would find ka:

(3.31 x 10^-5)^2/0.55 =2 x 10^-9

____________________________________________________

<em>-Julie</em>

6 0
3 years ago
Most important factor in determining an element's place in the periodic table
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The number of Protons.
(which is also the ATOMIC NUMBER)
3 0
3 years ago
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Answer:

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Explanation:

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In reactant side we have 1 S, 4F, 6 H and 3 O

In product side we have 1 S, 4F, 6 H and 3 O

6 0
3 years ago
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