Since this equation is balanced, we know that the law of conversation of mass id applied, and we could calculate easily.
Na= 2
NO3= 2
Ca= 1
<span>Cl= 1</span>
It's lone a little distinction (103 degrees versus 104 degrees in water), and I trust the standard rationalization is that since F is more electronegative than H, the electrons in the O-F bond invest more energy far from the O (and near the F) than the electrons in the O-H bond. That moves the powerful focal point of the unpleasant constrain between the bonding sets far from the O, and thus far from each other. So the shock between the bonding sets is marginally less, while the repugnance between the solitary matches on the O is the same - the outcome is the edge between the bonds is somewhat less.
Answer:
1: At temperatures below 542.55 K
2: At temperatures above 660 K
Explanation:
Hello there!
In this case, according to the thermodynamic definition of the Gibbs free energy, it is possible to write the following expression:
Whereas ΔG=0 for the spontaneous transition. In such a way, we proceed as follows:
1:
It means that at temperatures lower than 542.55 K the reaction will be spontaneous.
2:
It means that at temperatures higher than 660 K the reaction will be spontaneous.
Best regards!
At the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
So Kc = [H2] [Br2] / [HBr]^2
7.04X10^-2 = [H2][Br] / [HBr]^2
at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root:
∴ √(1/7.04X10^-2)= [HBr] / [H2]^1/2*[Br]^1/2
∴ Kc for the second reaction = √(1/7.04X10^-2) = 3.769
