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Leya [2.2K]
3 years ago
15

Consider a closed containing a solid in equilibrium with its vapor. The volume of the solid is much less than that of the contai

ner. Let the partition function of the solid be given by Q-qss where Ns is the number of molecules in the solid phase. Minimize the total Helmholtz free energy at fixed N,V, and T, and show that the equilibrium condition is given by Ng=qo/q2 where Ng is the number of gas phase molecules and qg is the partition function of a gas phase molecule.

Chemistry
1 answer:
Furkat [3]3 years ago
5 0

Answer:

Explanation:

check the attachment below

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C10H22 + O2 -- CO2 + H2O balance
Tasya [4]

Answer:

1,31÷2 =10,11

Explanation:

c10h22+31÷2o2=10co2+11h2o

5 0
3 years ago
Unrestricted populations of organisms experience _____.
Fudgin [204]
Unrestricted populations of organisms experience Exponential growth. 

As a population reaches is carrying capacity there is an increase in competition for All of the above.
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5 0
3 years ago
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How many electrons in an atom could have these sets of quantum numbers n=7 l=3 ml=-1?
zysi [14]

Solution:

Since we have ml=-1

it shows that it has two 2e- i;e it fond in 2nd subshell in f orbital. And each subshell can hold 2 e-.

Thus the required answer is 2 electrons hold by an atom.

4 0
3 years ago
What is the mass number of 80Br
Amanda [17]
Mass number of an element write in bottom of it!

So, there 80 would be the mass number of ₈₀Br.

Hope this helps!
4 0
3 years ago
Easy Chem, Will Give brainliest
Crazy boy [7]

Answer:

3.94 L

Explanation:

From the question given above, the following data were obtained:

Mass of O₂ = 5.62 g

Volume of O₂ =?

Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:

Mass of O₂ = 5.62 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mole of O₂ =?

Mole = mass / molar mass

Mole of O₂ = 5.62 / 32

Mole of O₂ = 0.176 mole

Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.

Thus 5.62 g (i.e 0.176 mole) of O₂ occupied 3.94 L at STP

5 0
3 years ago
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