2 the answer is 2 because 1.0 * 2 =2
Answer: C) number of protons
Answer: pure substances.
Explanation:
The given substances are:
All what surrounds us, which has mass and occupies spaces, is matter. There are two kind of matter: pure substances and mixtures.
Pure substances have a uniform and constant composition. On the other hand, mixtures are combinations of two or more pure substances in any arbitratry ratio.
Pure substances may be elements or compounds. The elements are the substances conmposed by one only kind of atom. In the list of substances given, Li and O₂ are elements: all the atoms in Li are lithium, and all the atoms in O₂ are oxygen atoms.
Compounds are the chemical combination of two or more different kind of atoms. In the given list H₂O₂ and NaCl are compounds. As you see, H₂O₂ contains atoms of hydrogen and oxygen, chemically bonded, in a fixed ratio (2 atoms of hydrogen by 2 atoms of oxygen). And NaCl has atoms of Na (sodium) and Cl (chlorine), chemicaly bonded, in a fixed ratio (1:1).
There are only 118 known elements and you can find them in any modern periodic table. Therer are virtually infinitely many compounds since many different combinations of the elements can be attained.
Elements and compounds have in common that they are classified as pure substances.
Muscles bones no on organs and no on skin
Radio active decay reactions follow first order rate kinetics.
a) The half life and decay constant for radio active decay reactions are related by the equation:



Where k is the decay constant
b) Finding out the decay constant for the decay of C-14 isotope:



c) Finding the age of the sample :
35 % of the radiocarbon is present currently.
The first order rate equation is,
![[A] = [A_{0}]e^{-kt}](https://tex.z-dn.net/?f=%20%5BA%5D%20%3D%20%5BA_%7B0%7D%5De%5E%7B-kt%7D%20%20%20)
![\frac{[A]}{[A_{0}]} = e^{-kt}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BA%5D%7D%7B%5BA_%7B0%7D%5D%7D%20%3D%20e%5E%7B-kt%7D%20%20)


t = 7923 years
Therefore, age of the sample is 7923 years.