Answer:
The statement that best describes the trend in first ionization enery of elements on the periodic table is:
It generally decreases down a group because valence electrons are farther from the nucleus.
The first ionization energy measures how difficult is to release an electron from the outermost shell. The higher the ionization energy the more difficult it is to release an electron, the lower the ionication energy the easier to release an electron.
As the atomic number of the atom increases (which is what happens when you go down a group) the furthest the outermost shell of electrons will be (the size of the atoms increases) and so those electrons require less energy to be released, which means that the ionization energy decreases.
Hope it helps!
O3 + M2+(aq) + H2O(l) => O2(g) + MO2(s) + 2 H+
Eo(cell) = Eo(O3/O2) - Eo(MO2/M2+)
0.44 = 2.07 - Eo(MO2/M2+)
Eo(MO2/M2+) = 1.59 V
Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
Answer: for 3. I would say it means that the amplitude is greater because if a sound is louder that means the amplitude is larger, same with the light.So that would mean the softer the sound the smaller the amplitude same with the dim of the light so for 4. It would be the amplitude is smaller. Hope this helps! :) ( if you have any questions feel free to ask)
