"Only electrons are involved in chemical reactions" is the statement among the following choices given in the question that is the <span>best explanation for chemical reactions not to produce radioactive particles. The correct option among all the options that are given in the question is the third option or option "C". </span>
Answer is: 5,75·10⁻¹.
Kf = 2,3·10⁶ 1/s.
K = 4,0·10⁸ 1/s.
Kr = ?
Kf - <span>forward rate constant.
K - </span><span>equilibrium constant.
Kr - </span><span>reverse rate constant.
</span>Since both Kf and Kr are constants at a given temperature, their ratio is also a constant that
is equal to the equilibrium constant K.<span>
K = Kf/Kr.
Kr = Kf/K = </span>2,3·10⁶ 1/s ÷ 4,0·10⁸ 1/s = 5,75·10⁻¹.
Answer;
-The charges of the positive and negative copper ions cancel each other out.
Explanation;
-A normal atom has a neutral charge with equal numbers of positive and negative particles.That means an atom with a neutral charge is one where the number of electrons is equal to the atomic number.
-Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged). If an atom gains or loses one or more electrons, it becomes an ion. If it gains one or more electrons, it now carries a net negative charge, and is thus anionic.
Answer:
The percent composition of fluorine is 65.67%
Explanation:
Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.
That is, the percentage composition is the percentage by mass of each of the elements present in a compound.
The calculation of the percentage composition of an element is made by:

In this case, the percent composition of fluorine is:

percent composition of fluorine= 65.67%
<u><em>The percent composition of fluorine is 65.67%</em></u>