The Nernst equation allows us to predict the cell potential for voltaic cells under conditions other than the standard conditions of 1M, 1 atm, 25°C. The effects of different temperatures and concentrations may be tracked in terms of the Gibbs energy change ΔG. This free energy change depends upon the temperature & concentrations according to ΔG = ΔG° + RTInQ where ΔG° is the free energy change under conditions and Q is the thermodynamic reaction quotient. The free energy change is related to the cell potential Ecell by ΔG= nFEcell
so for non-standard conditions
-nFEcell = -nFE°cell + RT InQ
or
Ecell = E°cell - RT/nF (InQ)
which is called Nernst equation.
Answer:
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
Explanation:
Given data:
Atomic mass of silicon= ?
Percent abundance of Si-28 = 92.21%
Atomic mass of Si-28 = 27.98 amu
Percent abundance of Si-29 = 4.70%
Atomic mass of Si-29 = 28.98 amu
Percent abundance of Si-30 = 3.09%
Atomic mass of Si-30 = 29.97 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100
Average atomic mass = 2580.04 +136.21+92.61 / 100
Average atomic mass = 2808.86 / 100
Average atomic mass = 28.08amu.
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
Answer:
The products are carbon dioxide and water
Explanation:
Step 1: Data given
Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant.
Step 2: The complete combustion of C3H7OH:
For the combustion of 1-propanol, we need O2.
The products of this combustion are CO2 and H2O.
C3H7OH + O2→ CO2 + H2O
On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3
C3H7OH + O2→ 3CO2 + H2O
On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.
C3H7OH + O2→ 3CO2 + 4H2O
On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).
To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.
2C3H7OH + 9O2→ 6CO2 + 8H2O
For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O
The products are carbon dioxide and water
Answer:
The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Explanation:
Given data:
[OH⁻] = 9.50 ×10⁻⁹M
pOH = ?
Solution:
pOH = -log[OH⁻]
Now we will put the value of OH⁻ concentration.
pOH = -log[9.50 ×10⁻⁹M]
pOH = 8
Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.