Answer:
No one is correct. The correct expression is:
Keq = [H₂]² . [O₂]² / [H₂O]²
Explanation:
To build the Keq expression in a chemical equilibrium you must consider the molar concentrations of reactants / products, and they must be elevated to the stoichiometric coefficient.
The balance reaction is:
<u>2</u> H₂O (g) ⇄ <u>2</u> H₂ (g) + O₂ (g)
Keq = [H₂]² . [O₂] / [H₂O]²
In opposite side: <u>2</u> H₂ (g) + O₂ (g) ⇄ <u>2</u> H₂O (g)
Keq = [H₂O]² / [H₂]² . [O₂]
D=m/v ⇒ m=d*v
d=density
m=mass
v=volume
d(ether)=0.71 gr/cm³=0.71 gr/ ml
v=130 ml
m=d*v
m=0.71 gr/ml*(130 ml)=92.3 g
Solution: m=92.3 g
0.01772288- Calculators do come in handy for this haha
Answer:
Correc option: 
Explanation:
size of atom : it says somthing about how many shell present in a particular atom or ion and it can also be evaluated on the basis of radius of atom.
Br^- and Kr has highest number of shell as compared to other group of species .
Na ,S , Mg ,P all are from 3rd period but Kr and Br^- in the 4th period so size of species of this group will more,
Size increases on increasring the shell number
Doping Se (group VI elements) with P(group V)elements would produce a P-TYPE semiconductor with HIGHER conductivity compared to pure Se
the reason is P dopant will introduce holes in the Se as P has lesser valence electron