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Firlakuza [10]
4 years ago
6

Which answer best explains how the modern periodic table was derived? the sole work of mendeleev, the father of the modern perio

dic table a select group of scientific investigations that all had similar outcomes through federal funding so that the modern periodic table was quickly completed much scientific examination, investigation, and argumentation of data by many scientists?
Chemistry
1 answer:
saveliy_v [14]4 years ago
4 0
I think the correct answer would be the last option. The statement that would best explain the derivation of the  modern periodic table would be that it was made from much scientific examination, investigation, and argumentation of data by many scientists. It was indeed finalized by Mendeleev however his works are based on the works of previous scientists who focused on the elements in the periodic table. So, it was not solely a product of effort of only one scientist but rather a number of scientists. Without the the previous researches, Mendeleev would not be able to come to the modern periodic table.
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Write the balanced nuclear equation for the following. (Use the lowest possible coefficients. Omit states-of-matter in your answ
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Answer:

_{93}^{232}\text{Np} + _{-1}^{0}\text{e} \longrightarrow _{92}^{232}\text{U}

Explanation:

The unbalanced nuclear equation is

_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow ?

It is convenient to replace the question by an atomic symbol, _{x}^{y}\text{Z}, where <em>x </em>= the atomic number, <em>y</em> = the mass number, and Z = the symbol of the element.

_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow _{x}^{y}\text{Z}

Then your equation becomes

_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow _{x}^{y}\text{Z}

The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts and of the subscripts</em> must be the same on each side of the equation.  

Then

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Element 92 is uranium, so the nuclear equation becomes

_{93}^{232}\text{Np} + _{-1}^{0}\text{e} \longrightarrow _{92}^{232}\text{U}

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