At STP, copper (Cu) would be the only substance here that will exist in the solid state.
So we look equation for the free Gibbs free energy (ΔG) which depends on entalpy (ΔH), temperature (T) and entropy (ΔS):
ΔG = ΔH - TΔS
ΔG is negative (-) because the water absorption on the silica gel surface is a spontaneous process.
ΔH is negative (-) because the water absorption on the silica gel surface is a exothermic process (it releases heat and if you want to desorb the water form the silica gen you need to add heat which is a endothermic process).
ΔS is negative (-) because the water is adsorbed, so from disorderly state you take the water molecules and put them in a orderly state and by doing that you decrease the entropy.
<>"On addition to water the Na+ section of NaCl is attracted to the oxygen side of the water molecules, while the Cl- side is attracted to the hydrogens' side of the water molecule. This causes the sodium chloride to split in water, and the NaCl dissolves into separate Na+ and Cl- atoms."<>
Answer:
0.78 M
Explanation:
First, we need to know which is the value of Kc of this reaction. In order to know this, we should take the innitial values of N2, O2 and NO and write the equilibrium constant expression according to the reaction. Doing this we have the following:
N2(g) + O2(g) <------> 2NO(g) Kc = ?
Writting Kc:
Kc = [NO]² / [N2] * [O2]
Replacing the given values we have then:
Kc = (0.6)² / (0.2)*(0.2)
Kc = 9
Now that we have the Kc, let's see what happens next.
We add more NO, until it's concentration is 0.9 M, this means that we are actually altering the reaction to get more reactants than product, which means that the equilibrium is being affected. If this is true, in the reaction when is re established the equilibrium, we'll see a loss in the concentration of NO and a gaining in concentrations of the reactants. This can be easily watched by doing an ICE chart:
N2(g) + O2(g) <------> 2NO(g)
I: 0.2 0.2 0.9
C: +x +x -2x
E: 0.2+x 0.2+x 0.9-2x
Replacing in the Kc expression we have:
Kc = [NO]² / [N2] * [O2]
9 = (0.9-2x)² / (0.2+x)*(0.2+x) ----> (this can be expressed as 0.2+x)²
Here, we solve for x:
9 = (0.9-2x)² / (0.2+x)²
√9 = (0.9-2x) / (0.2+x)
3(0.2+x) = 0.9-2x
0.6 + 3x = 0.9 - 2x
3x + 2x = 0.9 - 0.6
5x = 0.3
x = 0.06 M
This means that the final concentration of NO will be:
[NO] = 0.9 - (2*0.06)
[NO] = 0.78 M
Answer:
d. n = 4, l = 2, m = 0
Explanation:
A d-orbital in the fourth energy level can only be represented by n = 4, l = 2, m = 0.
n is the principal quantum number(n) which signifies the energy level, here it is 4;
l is the azimuthal quantum number or l which signifies the shape of the subshell, it takes a value of n - 1, 0,1,2 or 3
m is the magnetic quantum number(m) which can be -2, -1, 0, 1 , 2
So, the right and fitting solution is n = 4, l = 2, m = 0
