The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol
<h3>Further explanation</h3>
Given
Ratio of the concentrations of the products to the concentrations of the reactants is 22.3
Temperature = 37 C = 310 K
ΔG°=-16.7 kJ/mol
Required
the free energy change
Solution
Ratio of the concentration : equilbrium constant = K = 22.3
We can use Gibbs free energy :
ΔG = ΔG°+ RT ln K
R=8.314 .10⁻³ kJ/mol K
![\tt \Delta G=-16.7~kJ/mol+8.314.10^{-3}\times 310\times ln~22.3\\\\\Delta G=-8.698~kJ/mol](https://tex.z-dn.net/?f=%5Ctt%20%5CDelta%20G%3D-16.7~kJ%2Fmol%2B8.314.10%5E%7B-3%7D%5Ctimes%20310%5Ctimes%20ln~22.3%5C%5C%5C%5C%5CDelta%20G%3D-8.698~kJ%2Fmol)
Answer:
10.0 ml and 35.0ml respectively
Explanation:
if A=10.0ml let B be x
Now,
Lets assume the pan is balanced
then,
x = 10.0 ml
Therefore B= 10.0 ml
[Repeat the same process in next question]
What? That docent make any scene?
In order to compute this, we must first take a couple of assumptions of:
1) The laboratory size so we can calculate its volume
2) The number of students working in the lab so we know the total gas produced
Let the lab be
11 m × 9 m × 6 m
The volume then computes to be:
594 m³
We know that
1 Liter is 1 dm³
1 m = 10 dm
1 m³ = 1000 dm³
Therefore, the room volume in liters is:
594,000 Liters
Let there be 30 students in the laboratory
Total gas being produced:
6 × 30
= 180 Liters
This works out to be:
0.03% of Hydrogen by volume
Therefore, there is no risk of explosion given our assumption of size and students.
Answer:
Boron
Explanation:
The group III A is also known in the periodic table as group 13. The members of the group begin to appear in the second period.
At the beginning of this group III A is boron. Boron is a member of group III A as well as a member or period 2.
Boron is a metalloid and has only three valence electrons as other group III A elements.