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:
Explanation:
d =
m
V
m = d×V
V =
m
d
DENSITY
Density is defined as mass per unit volume.
d =
m
V
Example:
A brick of salt measuring 10.0 cm x 10.0 cm x 2.00 cm has a mass of 433 g. What is its density?
Step 1: Calculate the volume
V = lwh = 10.0 cm × 10.0 cm × 2.00 cm = 200 cm³
Step 2: Calculate the density
d =
m
V
=
433
g
200
c
m
³
= 2.16 g/cm³
MASS
d =
m
V
We can rearrange this to get the expression for the mass.
m = d×V
Example:
If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass?
m = d×V = 500 mL ×
1.11
g
1
m
L
= 555 g
VOLUME
d =
m
V
We can rearrange this to get the expression for the volume.
V =
m
d
Example:
What is the volume of a bar of gold that has a mass of 14.83 kg. The density of gold is 19.32 g/cm³.
Step 1: Convert kilograms to grams.
14.83 kg ×
1000
g
1
k
g
= 14 830 g
Step 2: Calculate the volume.
V =
m
d
= 14 830 g ×
1
c
m
³
19.32
g
= 767.6 cm³
Answer:
Carbon Tetrachloride
Explanation:
1 Carbon atom, 4 chlorine atoms (hence "tetra" prefix)
