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Answer:
Attached below
Explanation:
Free energy of mixing = ΔGmix = Gf - Gi
attached below is the required derivation of the
<u>a) Molar Gibbs energy of mixing</u>
ΔGmix = Gf - Gi
hence : ΔGmix = ∩RT ( X1 In X1 + X2 In X2 + X3 In X3 + ------- )
<u>b) molar excess Gibbs energy of mixing</u>
Ni = chemical potential of gas
fi = Fugacity
N°i = Chemical potential of gas when Fugacity = 1
ΔG = RT In ( a2 / a1 )
Answer:
Molecular formula
Explanation:
Molecular formula in the first place is required to understand which compound we have. We then should refer to the periodic table and find the molecular weight for each atom. Adding individual molecular weights together would yield the molar mass of a compound.
Then, dividing the total molar mass of a specific atom by the molar mass of a compound and converting into percentage will provide us with the percentage of that specific atom.
E. g., calculate the percent composition of water:
- molecular formula is
;
- calculate its molar mass: [tex]M = 2M_H + M_O = 2\cdot 1.00784 g/mol + 16.00 g/mol = 18.016 g/mol;
- find the percentage of hydrogen: [tex]\omega_H = \frac{2\cdot 1.00784 g/mol}{18.016 g/mol}\cdot 100 \% = 11.19 %;
- find the percentage of oxygen: [tex]\omega_O = \frac{16.00 g/mol}{18.016 g/mol}\cdot 100 \% = 88.81 %.
Answer:
A.
B.
Explanation:
Hello!
In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:
Thus we proceed as follows:
A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:
Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:
B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:
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