1) Compund Ir (x) O(y)
2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g
3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g
4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g
5) Convert grams to moles
moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles
moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131
6) Find the proportion of moles
Divide by the least of the number of moles, i.e. 0.00656
Ir: 0.00656 / 0.00656 = 1
O: 0.0131 / 0.00656 = 2
=> Empirical formula = Ir O2 (where 2 is the superscript for O)
Answer: Ir O2
When sodium amide i.e.
reacts with water i.e. 
results in the formation of sodium hydroxide i.e. 
and ammonia 
.
The chemical reaction is given by:
Now, when ammonia i.e. reacts with water results in the formation of ammonium hydroxide i.e. 
The chemical reaction is given by:
Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).
The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.
Answer:
2 only is a pretty accurate answer the others dont make sense to me.
Explanation:
Answer:
Reduction
Explanation:
The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
In given reaction fluorine gas gain two electron and form fluoride ions.
F₂(g) + 2e⁻ → 2F⁻(aq)
The given reaction is reduction because oxidation state is decreased from zero to -1.
