Answer:
The 3rd answer down.
Na²O (sodium oxide) will be a base when exposed to water H²O
Explanation:
Sodium Oxide Na²O, will become Sodium Hydroxide after being exposed to water (at 80% I believe).
The oxygen ion in Na²O has 2 extra electrons which makes it highly charged and very attractive to hydrogen ions. The attraction is so strong that when Na²O comes in contact with H²O, the O(-2) strips off a hydrogen from water, forming 2 x OH ions which of course are still strongly basic.
Explanation:
The net equation will be as follows.
So, we are required to find
for this reaction.
Therefore, steps involved for the above process are as follows.
Step 1: Convert K from solid state to gaseous state
,
= 89 kJ
Step 2: Ionization of gaseous K
,
= 418 KJ
Step 3: Dissociation of
gas into chlorine atom
.
,
= 122 KJ
Step 4: Iozination of chlorine atom.
,
= -349 KJ
Step 5: Add
ion and
ion formed above to get KCl
.
,
= -717 KJ
Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.
= 89 + 418 + 122 - 349 - 717
= - 437 KJ/mol
Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.
Answer: The answer would be B. Hope this helps.
Explanation: