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:
If we assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively, Vmix = 20.5 cm³.
Explanation:
The molar volume of a substance is the ratio between the volume and the number of moles of the substance. It represents the volume that 1 mol of it occupies. Because we don't have access to page 24, let's assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively.
The volume of mixture (Vmix) is the sum of the volume of each substance, which is the number of moles multiplied by molar volume, so:
Vmix = 0.300*57 + 0.200*17
Vmix = 17.1 + 3.4
Vmix = 20.5 cm³
Answer and Explanation:
The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation
R = molar gas constant
K = A(e^(-Ea/RT))
Taking natural log of both sides
In K = In A - (Ea/RT)
In K = (-Ea/R)(1/T) + In A
Comparing this to the equation of a straight line; y = mx + c
y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A
a) From the question, m = (-Ea/R) = -1.10 × (10^4) K
(-Ea/R) = -1.10 × (10^4) = -11000
R = 8.314 J/K.mol
Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol
b) c = In A = 33.5
A = e^33.5 = (3.54 × (10^14))/s
c) K = A(e^(-Ea/RT))
A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol
K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s
QED!
Answer:
The empirical formula is PCl3
Explanation:
Mass of P is 30.97 g, thus 1.523 g of P equivalent to 0.05 moles of P
Mass of Cl is 35.45 g, thus 5.228 g of Cl equivalent to 0.15 moles of Cl
Therefore moles of P : moles of Cl = 0.05:0.15 = 1:3
Therefore the empirical formula, PCl3
