ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
Learn more about ΔG° here:
brainly.com/question/17214066
#SPJ4
Answer:
The minimum amount of energy needed the the cell to perform various cellular,biochemical and physiological activities is known is Gibbs free energy.
Explanation:
The change in gibbs free energy of is very much important to determine whether a given reaction is spontaneous,non spontaneous or equilibrium.
1 If gibbs free energy change of a reaction is negative then the reaction is spontaneous.
2 If the free energy change is 0 then the reaction is in equilibrium stage.
3 If free energy change is positive then the reaction is non spontaneous.
Answer:
% weight of nickle = 24 %
Explanation:
molar mass of Nickel Sulfamate (Ni(SO₃NH₂)₂) = 250.87 g/mol
Solution
1st we write down the molar mass of Ni
molar mass of Ni = 59 g/mol
now we write down the number of moles of Ni in (Ni(SO₃NH₂)₂)
number of moles of Ni = 1 mol
Now we calculate the mass of nickle present in (Ni(SO₃NH₂)₂)
<em> mass = moles × molar mass</em>
mass = 1 mol × 59 g/mol
mass = 59 g
now we calculate the % weight of nickle in (Ni(SO₃NH₂)₂)
<em> % weight = (weight of element ÷ total weight) × 100</em>
% weight of nickle = (59 ÷ 250.87) × 100
% weight of nickle = 0.24 × 100
% weight of nickle = 24 %
Answer: The hydrogen ion concentration in molarity is 0.013
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pOH is calculated by taking negative logarithm of hydroxide ion concentration and pH is calculated by taking negative logarithm of hydrogen ion concentration
Putting in the values:
![pOH=-\log[7.609\times 10^{-13}]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B7.609%5Ctimes%2010%5E%7B-13%7D%5D)
Now , 
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![[H^+]=0.013M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.013M)
The hydrogen ion concentration in molarity is 0.013
