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.
Answer:The Geocentric and the Heliocentric models are descriptions of the universe known at the time when these models were developed. The main difference between the two models is the perceived center of the known universe. According to the geocentric model, Earth is the center of the known universe
Explanation:I answerd the question before in Earth Science
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
