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Nookie1986 [14]
2 years ago
10

Elements with atomic numbers of 104 and greater are known as super-heavy elements. None of these elements have been found in nat

ure but instead have been made in a laboratory. They are very difficult and expensive to create, and they break down into other elements quickly. There currently are no practical applications for any of these elements. However, some scientists believe that, with further work, they may discover some isotopes of super-heavy elements that are more stable and that could possibly have practical implications. Do you think that scientists should continue to try to create super-heavy elements and expand the periodic table? Explain why or why not.
Chemistry
2 answers:
Lemur [1.5K]2 years ago
8 0

Here we have to justify the work of scientists to generate more unknown elements after atomic number 104 which are so unstable and cannot be obtained in nature.

It is fact that the stability of the super heavy element is hard to obtain, as when there are so much proton  and neutron at the nucleus of such element makes it so unstable.

So the try to make more unknown element to enlarge the periodic table is not justified as per the stability and utility of such elements are concerned. But in fact it helps to get the new elements which may exist and may have some radioactivity which can be used.

But apparently the making of new material in laboratory to expand the periodic table is not justified.  

yan [13]2 years ago
8 0

Answer:

Here we have to justify the work of scientists to generate more unknown elements after atomic number 104 which are so unstable and cannot be obtained in nature. So the try to make more unknown element to enlarge the periodic table is not justified as per the stability and utility of such elements are concerned. But in fact it helps to get the new elements which may exist and may have some radioactivity which can be used.

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Determine the empirical formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72% oxygen.
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Answer:

The answer to your question is:       C₁₈ H₂₇ N O₃

Explanation:

Data

Carbon = 70.79 g

Hydrogen = 8.91 g

Nitrogen = 4.58 g

Oxygen = 15.72 g

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AT C = 12 g

AT H = 1 g

AT N = 14 g

AT O = 16 g

                   Carbon

                                  12 g ------------------------  1 mol

                           70.79  g -------------------------   x

                                 x = (70.79 x 1) / 12

                                 x = 5.9 mol of C

                  Hydrogen

                                   1 g -----------------------  1 mol

                                8.91 g ---------------------   x

                                  x = (8.91 x 1) / 1

                                  x = 8.91 mol of H

                  Nitrogen

                                 14 g ---------------------- 1 mol

                                4.58 g -------------------   x

                                   x = (4.58 x 1) / 14

                                   x = 0.33 mol

                  Oxygen

                               16 g ------------------------  1 mol

                               15.72 g --------------------   x

                                  x = (15.72 x 1)/16

                                 x = 0.98

Divide by the lowest number of moles

Carbon               5.9 / 0.33     =  17.9   ≈ 18

Hydrogen           8.91 /  0.33  =   27

Nitrogen             0.33 / 0.33  =    1

Oxygen               0.98 / 0.33 =    2.9  ≈ 3

                             C₁₈ H₂₇ N O₃

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