What name should be used for the ionic compound
LiI?
The name that should be used for
the ionic compound LiI is Lithium iodide. When naming ionic compounds, you have
to take into consideration pairing of two components.
As the atomic mass of iron is 55.847u
We say that it is the mass of one mole of iron.
By formula it can be find by
No.of mole=mass in g/molar mass
Mass in gram = no.of mole x molar mass
No.of mole =1
Molar mass = 55.847g
Mass in gram = 1x55.847
= 55.847g of Fe
The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation of an atom in a chemical compound.
<u>Explanation:</u>
The oxidation number of an atom is the charge that atom would have if the compound was composed of ions. 1. The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. The oxidation number of simple ions is equal to the charge on the ion.
The oxidation number of a mono atomic ion equals the charge of the ion. The oxidation number of H is +1, but it is -1 in when combined with less electro negative elements. The oxidation number of O in compounds is usually -2, but it is -1 in peroxides. The oxidation number of a Group 1 element in a compound is +1.
Answer:
9.63 L of NO
Explanation:
We'll begin by calculating the number of mole in 50.0 g of NH₄ClO₄. This can be obtained as follow:
Mass of NH₄ClO₄ = 50 g
Molar mass of NH₄ClO₄ = 14 + (4×1) + 35.5 + (16×4)
= 14 + 4 + 35.5 + 64
= 117.5 g/mol
Mole of NH₄ClO₄ =?
Mole = mass /molar mass
Mole of NH₄ClO₄ = 50/117.5
Mole of NH₄ClO₄ = 0.43 mole
Next, we shall determine the number of mole of NO produced by the reaction of 50 g (i.e 0.43 mole) of NH₄ClO₄. This can be obtained as follow:
3Al + 3NH₄ClO₄ –> Al₂O₃ + AlCl₃ + 3NO + 6H₂O
From the balanced equation above,
3 moles of NH₄ClO₄ reacted to produce 3 moles of NO.
Therefore, 0.43 mole of NH₄ClO₄ will also react to produce 0.43 mole of NO.
Finally, we shall determine the volume occupied by 0.43 mole of NO. This can be obtained as follow:
1 mole of NO = 22.4 L
Therefore,
0.43 mole of NO = 0.43 × 22.4
0.43 mole of NO = 9.63 L
Thus, 9.63 L of NO were obtained from the reaction.
