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zimovet [89]
3 years ago
15

What materials formed the solar system?

Chemistry
2 answers:
laiz [17]3 years ago
7 0

The answer would be clouds of gas and dust.

Hoped this helps ;)

Paha777 [63]3 years ago
6 0

Scientists believe that the solar system was formed when a cloud of gas and dust in space was disturbed, maybe by the explosion of a nearby star(called a supernova). This explosion made waves in space which squeezed the cloud of gas and dust.

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If 5000 balloons are each filled with 0.5L then how many does each balloon have
Colt1911 [192]
ANSWER:

If 5000 balloons are each filled with 0.5L, then each balloon has 0.5L. The answer is in the question.

Please mark as brainliest if you found this helpful! :)
Thank you <3
5 0
3 years ago
What is the “Island of Stability”?
3241004551 [841]

Answer:

The island of stability is a term from nuclear physics that describes the possibility of elements with particularly stable "magic numbers" of protons and neutrons. This would allow certain isotopes of some transuranic elements to be far more stable than others, that is, decay much more slowly.

Explanation:

7 0
2 years ago
Which of the following lists contain only elements?
N76 [4]
In which elements are belonging to? I will grant that option B ; ti's the best answer.
                 In which thou might contain the elements zinc , gold , aluminium , and last by not least oxygen.

I truly hope ti's answer helps thou. 
7 0
3 years ago
You will need to prepare 12 mL of 25% Sodium Phosphate Buffer (pH 4) solution for Activity 2. What volume of the stock Sodium Ph
zhuklara [117]

Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>

The process of preparing solutions from stock solutions of higher concentration is known as dilution.

Dilution is done with the aid of the dilution formula given below:

  • C1V1 = C2V2

where

  • C1 is the concentration of stock solution
  • V1 is the volume of stock solution required to prepare a diluted solution
  • C2 is the concentration of the diluted solution prepared
  • V2 is the final volume of the diluted solution

From the data provided:

C1 is not given

V1 is unknown

C2 = 25%

V2 = 12 mL

  • Assuming C1 is 50% solution

Volume of stock, V1, required is calculated as follows:

V1 = C2V2/C1

V1 = 25 × 12 /50

V1 = 6 mL

Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

Learn more about dilution formula at: brainly.com/question/7208546

6 0
2 years ago
How many molecules of Mg3N2 (magnesium nitride) are formed when excess Mg (magnesium)
dybincka [34]

Explanation:

<em>3Mg(s) + N2(g) = Mg3N2(s)</em>

First check that the equation is balanced. In this case, it is.

Assuming that magnesium is the limiting reactant:

  1. First find the molecular weight using the Periodic Table.

       We find that the atomic mass of magnesium is approximately

       <em>24.3g</em>, so the molecular weight is just <em>24.3g\mol</em>

   

    2. Next we need the mole to mole ratio. As there are <em>3</em>

        magnesiums for <em>1</em> magnesium nitride (shown by the coefficients), the                    

        mole to mole ratio is<em> 1 mol Mg3N2\3 mol Mg.</em>

   

    3. We need the amount of the substance, in grams. Since you have not    

        stated it in the question, I'll just do <em>10g</em> AS AN EXAMPLE. Note that    

       depending on the amount, the LIMITING REAGENT MAY DIFFER.

   4.  Finally, we need the molecular weight of <em>Mg3N2</em>, which we can easily    

        calculate to be around <em>100.9\mol.</em>

<em />

   5.  Putting this all together, we have<em> 10gMg⋅ (mol Mg\24.3gMg) </em>

<em>         (1mol Mg3N2\ 3mol Mg) (100.9g Mg3N2\mol Mg3N2)</em>

     

        the units will cancel to leave <em>gMg3N2</em> (grams of magnesium nitride):

       

<em>        10gMg ⋅ (mol Mg\24.3gMg) (1mol Mg3N2\3mol Mg)</em>

<em>        (100.9g Mg3N2\mol Mg3N2)</em>

<em />

Doing the calculation yields approximately 13.84g.

Assuming that nitrogen is the limiting reactant:

Similarly, following the above steps but with <em>10g</em> of nitrogen yields <em>36.04g</em>

In conclusion, as we produce less amount of <em>Mg3N2</em> when we assumed that <em>Mg</em> was the limiting reagent, magnesium is the limiting reagent and nitrogen is the excess.

Note: This is in THIS CASE, where we have <em>10g</em> of both. The answer may vary depending on the amount of each substance.

7 0
2 years ago
Read 2 more answers
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