Burning Mg in the air and reacting with O2 forming a white powder of MnO
So the equation is going to be:
Mn + O2 ⇒ MnO (this equation is not conserved)
to make it equilibrium:
1- First we should put 2Mno to equal the O2 on both sides.
So it will be:
Mg + O2⇒ 2MgO
2- Second we should put 2Mn to equal the Mn on both sides.
2Mg + O2⇒ 2MgO (this equation is conserved)
After putting the physical states the final equilibrium equation is going to be:
Δ
2Mg(s) + O2(g)⇒ 2MgO(s)
Sorry I don’t get the question Buh I’m sure you’ll get it eventually
Answer:
Option "B" is correct.
Explanation:
According to VSEPR theory, There are repulsion forces exists among the bond pair - bond pair or bond pair - lone pair of electrons. In 
and 
, the number of electron pairs are same but methane has all the four bond pairs where in ammonia, three bond pairs and one lone pair exists. And thus there are repulsion forces possible in between the lone pair and bond pair of electrons thus the arrangement of electron pairs around both the molecules is same or different depending up on the conditions leading to maximum repulsion.
I think it is C. coefficients but I don't know what subscripts are
