3Mg + N₂= Mg₃N₂
n(Mg)=12,2g÷<span>24,4g/mol=0,5mol-limiting reagent
</span>n(N₂)=5,16g÷28g/mol=0,18mol
n(Mg₃N₂):n(Mg)=1:3, n(Mg₃N₂)=0,166mol, m(Mg₃N₂)=0,166·101,2=16,8g.
%(N)= 2·Ar(N)÷Mr(Mg₃N₂) = 2·14÷101,2=27,66%=0,2766
%(Mg) = 3·Ar(Mg)÷Mr(Mg₃N₂)= 3·24,4÷101,2=72,34% or 100% - 27,66%= 72,34%.
This is a difficult task because zinc is much more active than copper and could hardly be passivated. ... The sur- face immediately turns white (the color of copper(I) iodide) and the yellow-brown color of iodine quickly fades. Rinse the coin with water, brighten it with polish and cloth and begin the whole process again.N
W = AB x F x Cos < AB, F
or just W= AB x F for short
Answer:
<u>One lone-Pair is present in Ammonia</u>
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Explanation:
The number of valence electron in N = 5
The number of Valence electron in H = 1
The formula of ammonia = NH3
Total valence electron in ammonia molecule = 5 +3(1) = 5+3 = 8
The lewis structure suggest that :
Nitrogen completes its octet by sharing the electron pair with 3 hydrogen atoms.
3 electron of Nitrogen are involved in sharing with Hydrogen
So,<u><em> remaining two electron are left non-bonded</em></u> . Hence they exist as lone- pair
So, there is only 1 lone pair in the ammonia molecule .
The shape of NH3 is bent according to VSEPR theory . This is so because the presence of 1 lone pair causes more repulsion and occupy more space.
Thus the lone pair is changing the shape of the ammonia molecule . It also increase the dipole moment of the molecule , which gives polarity to it.