1) The answer is: 1.75·10²³ are in a sample of copper.
m(Cu) = 18.46 g; mass of copper.
n(Cu) = m(Cu) ÷ M(Cu).
n(Cu) = 18.46 g ÷ 63.55 g/mol.
n(Cu) = 0.29 mol; amount of copper.
Na = 6.022·10²³ 1/mol; Avogadro number.
N(Cu) = n(Cu) · Na.
N(Cu) = 0.29 mol · 6.022·10²³ 1/mol.
N(Cu) = 1.75·10²³; number of copper atoms.
2) The answer is: number of formula units is 6.975·10²³.
m(NaCl) = 67.69 g; mass of salt.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 67.69 g ÷ 58.44 g/mol.
n(NaCl) = 1.16 mol; amount of salt.
Na = 6.022·10²³ 1/mol; Avogadro number.
N(NaCl) = n(NaCl) · Na.
N(NaCl) = 1.16 mol · 6.022·10²³ 1/mol.
N(NaCl) = 6.975·10²³; number of formula units.
3) The answer is: 1.506·10²⁴ molecules of water.
m(H₂O) = 44 g; mass of water.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 44.99 g ÷ 18 g/mol.
n(H₂O) = 2.5 mol; amount of water.
Na = 6.022·10²³ 1/mol; Avogadro number.
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 2.5 mol · 6.022·10²³ 1/mol.
N(H₂O) = 1.506·10²⁴; number of water molecules.
4) The answer is: the mass of magnesium is 6.08 grams.
n(Mg) = 0.250 mol; amount of magnesium.
M(Mg) = 24.305 g/mol; molar mass of magnesium.
m(Mg) = n(Mg) · M(Mg).
m(Mg) = 0.250 mol · 24.305 g/mol.
m(Mg) = 6.076 g; mass of magnesium.
Magnesium (Mg) is metal from 2. group of Periodic table of elements with atomic number 12.
5) The answer is: there are 1.5 moles of sulfur.
m(S) = 48.096 g; mass of sulfur.
M(S) = 32.065 g/mol; molar mass of sulfur.
n(S) = m(S) ÷ M(S).
n(S) = 48.096 g ÷ 32.065 g/mol.
n(S) = 1.50 mol; amount of sulfur.
Sulfur is a chemical element with symbol S and atomic number 16.
6) The answer is: the mass is 1.65·10⁻²¹ grams.
Ar(MgO) = 40.3; relative atomic mass of magnesium oxide.
m(MgO) = 24.6 · 40.3 · 1.66·10⁻²⁷ kg.
m(MgO) = 1645.7·10⁻²⁷ kg.
m(MgO) = 1.65·10⁻²¹ g; mass og magnesium oxide.
Ar is relative atomic mass (the ratio of the average mass of atoms of a chemical element to one unified atomic mass unit) of an element.
The unified atomic mass unit (amu) is a standard unit of atom mass.
One unified atomic mass unit is approximately the mass of one nucleon (1.66·10⁻²⁷ kg).
7) The answer is: 6.88·10²³ molecules of dinitrogen pentoxide.
m(N₂O₅) = 123.46 g; mass of dinitrogen pentoxide.
M(N₂O₅) = 108.01 g/mol; molar mass of dinitrogen pentoxide.
n(N₂O₅) = m(N₂O₅) ÷ M(N₂O₅).
n(N₂O₅) = 123.46 g ÷ 108.01 g/mol.
n(N₂O₅) = 1.14 mol; amount of dinitrogen pentoxide.
N(N₂O₅) = 1.14 mol · 6.022·10²³ 1/mol.
N(N₂O₅) = 6.88·10²³; number of molecules.
