The predicted order of ionization energies is Li > Na > K > Rb
<em>Atomic size increases as you go down a Group</em> (see image). We are adding electrons to increasingly larger shells.
The valence electrons are further from the attraction of the nucleus, so they are less tightly held.
Thus, <em>Li has the highest ionization energy</em> and <em>Rb the lowest</em>.
Answer:
Keq = 0.053
7.3 kJ/mol
Explanation:
Let's consider the following isomerization reaction.
glucose 6‑phosphate ⇄ glucose 1 - phosphate
The concentrations at equilibrium are:
[G6P] = 0.19 M
[G1P] = 0.01 M
The concentration equilibrium constant (Keq) is:
Keq = [G1P] / [G6P]
Keq = 0.01 / 0.19
Keq = 0.053
We can find the standard free energy change, ΔG°, of the reaction mixture using the following expression.
ΔG° = -R × T × lnKeq
ΔG° = -8.314 J/mol.K × 298 K × ln0.053
ΔG° = 7.3 × 10³ J/mol = 7.3 kJ/mol
Answer:
0.75 g/cm³
Explanation:
Given data:
Mass of wooden block = 180 g
Length of block = 10 cm
Width of block = 6 cm
Height or thickness = 4 cm
Density of block = ?
Solution:
Volume of block = height × length × width
Volume of block = 4 cm × 10 cm× 6 cm
Volume of block = 240 cm³
Density of block:
density = mass/ volume
d = 180 g/ 240 cm³
d = 0.75 g/cm³
A) eventually they will be in thermal equilibrium
