To get the number of gold atoms, you have to divide the mass of the gold by the mass of the gold atom. It follows this simple equation
.
Let x be the number of gold atoms. Plug in the values to a calculator.
x =
Both have the same units so the unit gram(g) can be cancelled.
x then would be equal to 1.53x10^22. So there are 1.53x10^22 atoms of gold in 5 g of gold
heating is repeated to ensure all water molecules have evaporated
Onization energy is the energy required to lose an electron and form an ion. The stronger is the attraction of the atom and the electron the higher the ionization energy, and the weaker is the attraction of the atom and the electron the higher the ionization energy. This leads to a clear trend in the periodic table. Given that the larger the atom the weaker the attraction of the atom to the valence electrons, the easier they will be released, and the lower the ionization energy. This is, as you go downward in a group, the ionization energy decreases. So, the element at the top of the group will exhibit the largest ionization energy. <span>Therefore, the answer is that of the four elements of group 7A, fluorine will have the largest first ionization energy.</span>
2H₂ + O₂ = 2H₂O
n(H₂)=m(H₂)/M(H₂)
n(H₂)=5g/2.0g/mol=2.5 mol
n(O₂)=m(O₂)/M(O₂)
n(O₂)=40g/32.0g/mol=1.25 mol
H₂ : O₂ = 2 : 1
2.5 : 1.25 = 2 : 1
n(H₂O)=n(H₂)=2n(O₂)=2.5 mol
m(H₂O)=n(H₂O)M(H₂O)
m(H₂O)=2.5mol*18.0g/mol=45.0 g
