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:
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Answer:
The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²
Explanation:
The given parameters is as follows;
The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g
The required information = The number of formula units of potassium chloride (KCl)
The Molar Mass of KCl = 74.5513 g/mol
![The \ number \ of \ moles \ of \ a \ substance, n = \dfrac{The \ mass \ of \ the substance}{The \ Molar \ Mass \ of \ the \ substance}](https://tex.z-dn.net/?f=The%20%5C%20number%20%5C%20of%20%5C%20moles%20%5C%20of%20%5C%20a%20%5C%20substance%2C%20n%20%3D%20%5Cdfrac%7BThe%20%20%5C%20%20mass%20%20%5C%20%20of%20%20%5C%20%20the%20substance%7D%7BThe%20%20%20%5C%20%20Molar%20%20%5C%20%20%20Mass%20%20%20%5C%20%20of%20%20%5C%20%20%20the%20%20%20%5C%20%20substance%7D)
Therefore, we have;
![The \ number \ of \ moles \ of \ KCl= \dfrac{3.81 \ g}{74.5513 \ g/mol} \approx 0.051106 \ moles](https://tex.z-dn.net/?f=The%20%5C%20number%20%5C%20of%20%5C%20moles%20%5C%20of%20%5C%20KCl%3D%20%5Cdfrac%7B3.81%20%5C%20g%7D%7B74.5513%20%5C%20g%2Fmol%7D%20%5Capprox%200.051106%20%5C%20moles)
1 mole of a substance, contains Avogadro's number (6.022 × 10²³) of formula units
Therefore;
0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³ ≈ 3.077588 × 10²² formula units
From which we have, the number of formula units in 3.81 g of potassium chloride (KCl) ≈ 3.08 × 10²² formula units.
Answer:
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Explanation:
The mass of iron (III) oxide (Fe2O3) : 85.12 g
<h3>Further explanation</h3>
Given
3.20x10²³ formula units
Required
The mass
Solution
1 mole = 6.02.10²³ particles
Can be formulated :
N = n x No
N = number of particles
n = mol
No = 6.02.10²³ = Avogadro's number
mol of Fe₂O₃ :
![\tt n=\dfrac{3.2.10^{23}}{6.02.10^{23}}=0.532](https://tex.z-dn.net/?f=%5Ctt%20n%3D%5Cdfrac%7B3.2.10%5E%7B23%7D%7D%7B6.02.10%5E%7B23%7D%7D%3D0.532)
mass of Fe₂O₃ (MW=160 g/mol)
![\tt mass=mol\times MW=0.532\times 160=85.12~g](https://tex.z-dn.net/?f=%5Ctt%20mass%3Dmol%5Ctimes%20MW%3D0.532%5Ctimes%20160%3D85.12~g)