Answer:
Nitrogen (ii) oxide
Explanation:
To know the IUPAC name for NO, we shall determine the oxidation number of N in NO.
NOTE: The oxidation number of oxygen (O) is always – 2.
Thus the oxidation number of N in NO can be obtained as follow:
N + O = 0 (ground state)
N + (– 2) = 0
N – 2 = 0
Collect like terms
N = 0 + 2
N = +2
Thus, the oxidation number of Nitrogen (N) in NO is +2.
Therefore, the IUPAC name for NO is Nitrogen (ii) oxide
Answer:
Chemical formula.
Explanation:
A chemical formula have fix proportion of atoms of elements.
Chemical formula:
A chemical formula is the way of presenting the chemical proportion of atoms of those elements, that combine to form a compound.
For example:
Water consist of two atom of hydrogen and one atom of oxygen. The atoms of both elements combine to form a compound i.e water. The atoms of both elements always combine in a fixed ratio which is 2:1 in molecule of water.
H₂O
2:1
Carbon dioxide consist of two atom of oxygen and one atom of carbon. The atoms of both elements combine to form a compound i.e carbon dioxide. The atoms of both elements always combine in a fixed ratio which is 1:2 in molecule of carbon dioxide .
CO₂
1;2
<u>Answer:</u> The equilibrium concentration of 
is 1.285 M.
<u>Explanation:</u>
The chemical equation for the decomposition of phosphorus pentachloride follows:
The expression for equilibrium constant is given as:
![K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
We are given:
![[PCl_3]=0.18M](https://tex.z-dn.net/?f=%5BPCl_3%5D%3D0.18M)
![[Cl_2]=0.30M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D0.30M)
The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.
Putting values in above equation, we get:
![0.042=\frac{0.18\times 0.30}{[PCl_5]}](https://tex.z-dn.net/?f=0.042%3D%5Cfrac%7B0.18%5Ctimes%200.30%7D%7B%5BPCl_5%5D%7D)
![[PCl_5]=1.285](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D1.285)
Hence, the equilibrium concentration of is 1.285 M.
