Answer:
C + O2 → CO2
Explanation:
C + O2 → CO ----------------- (1)
from equ (1) on reactant side, C has 1 mole, O has 2 moles
from equ (1) on product side, C has 1 mole, O has 1 mole
Thus, to balance the equation, O should have 2 moles
C + O2 → CO2
The response would become spontaneous if the value of ΔG° was negative.
According to the estimated value of ΔG°, it is shown that ΔG° value decreases as temperature value increases. The value shifts from being more favorable to being less favorable. It would appear that the value of ΔG° would be negative at a specific temperature, causing the reaction to occur spontaneously.
The reaction is in an equilibrium state if ΔG = 0. If ΔG < 0, the reaction is spontaneous in the direction written. The relationship between terms from the equilibrium is paralleled by the relevance of the sign of a change in the Gibbs free energy.
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Answer: Partial pressure of 
at a depth of 132 ft below sea level is 2964 mm Hg.
Explanation:
It is known that 1 atm = 760 mm Hg.
Also, 
where, 
= partial pressure of
P = atmospheric pressure
= mole fraction of
Putting the given values into the above formula as follows.
= 0.780
Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of is as follows.
= 
= 2964 mm Hg
Therefore, we can conclude that partial pressure of at a depth of 132 ft below sea level is 2964 mm Hg.
A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a stock solution whose concentration is known.
Answer:
The activation energy for this reaction = 23 kJ/mol.
Explanation:
Using the expression,
Where,
is the activation energy
R is Gas constant having value = 8.314×10⁻³ kJ / K mol
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (280 + 273.15) K = 553.15 K
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (376 + 273.15) K = 649.15 K
So,
<u>The activation energy for this reaction = 23 kJ/mol.</u>
