Answer:
Uniformitarianism is the idea that the natural processes operating today are the same as the natural processes that operated in the past
Answer:
Please help!!!
When a piston with a volume of 35 mL is heated from 298 K to 596 Kit
expands. Assuming the pressure on the piston remains constant, determine
the new volume of the cylinder. (round to the nearest whole number)
no
Explanation:
n0? = °^° hhaa jk
Answer:
<em>2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ </em>
Explanation:
If we know the ΔG° of a chemical reaction it is possible to calculate the equilibrium constant (k) of this procedure with the next equation:
ln Keq = -ΔG° / RT (1)
Where: Keq is equilibrium contant, ΔG° is standard state free energy change, R is gas constant and T is temperature.
Watching (1), it is possible to know that the large negative ΔG° the largest equilibrium constant. That is because R and T are always positive and to cancel the negative of equation it is necessary that ΔG° be negative.
Knowing this, is the oxidation of Hg the reaction that has the largely negative ΔG°. So, this reaction will have the largest equilibrium constant.
<em>2 Hg(g) + O₂(g) → 2 HgO(s) ΔG° = -180.8 kJ </em>
CaCO₃(s) → CaO(s) + CO₂(g) ΔG° =+131.1 kJ
3 O₂(g) → 2 O₃(g) ΔG° = +326 kJ
Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g) ΔG° = -28.0 kJ
I hope it helps!
To answer this item, we must take note that the ligand that binds the tightest is the one with the lowest dissociation constant, Kd. Kd's for both A and B are already given so, we only need to solve Kds for C and D.
Kd of C
0.3 = (1x10⁻⁶)/(1x10⁻⁶ + Kd) ; Kd = 2.3x10⁻⁶
Kd of D
0.8 = (1x10⁻⁹)/(1x10⁻⁹ + Kd) ; Kd = 2.5x10⁻10
Since Ligand D has the least value of dissociation constant then, it can be concluded that it binds the tightest.
