determine the number of protons, electrons, neutrons, valence electrons and valency in oxygen atom (o) and oxide ion ( compare the stability of an atom and an ion. (atomic number of oxygen is 8).

                   Atom (O)           Ion (O²⁻)

Protons             8                       8

Electrons          8                      10

Neutrons          8                       8        ← only for the oxygen-16 isotope.

* The number of neutrons depends on the specific isotope. It is equal to the mass number of the isotope less the number of protons.

Explanation:

Oxygen is the element with atomic number 8. That means that the oxygen atoms have 8 protons.

Protons have relative positive charge equal to +1 and electrons have relative negative charge -1: same magnitude opposite sign.

Thus, the neutral oxygen, O, has also 8 electrons and 8 protons.

As per the number neutrons it depends on the specific isotope.

The isotopes are identified by the mass number, which is the number of protons plus neutrons. For instance, oxygen-16 isotope has mass number 16, 8 protons and 8 neutrons, while oxygen-15 isotope has mass number 15, 8 protons and 7 neutrons.

Oxide ion O²⁻ is the oxygen ion with -2 charge, meaning that it has 2 more electrons than protons. Then, it has 8 protons and 10 electrons, again the number of neutrons depend of the specific isotope.

The stability of an atoms or ion refers to its trend to remain in the same state or react to form another species. The ions O²⁻ are less stable than the neutral atoms O, because the negative charge will be attracted to any positive ion so ti will react faster than the neutral oxygen atom (O).

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol