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Bezzdna
2 months ago
7

Calculate the pressure of O2 (in atm) over a sample of NiO at 25.00°C if ΔG o = 212 kJ/mol for the reaction. For this calculatio

n, use the value R = 8.3144 J/K·mol. NiO(s) ⇌ Ni(s) + 1 2 O2(g)
Chemistry
1 answer:
Anarel [2.9K]2 months ago
7 0

Based on the given information:
Δ G° = 212 KJ/mol, the temperature is calculated as 25+273 = 298 K, and the gas constant R equals 0.008314 KJ/mol.

The reaction can be characterized as NiO(s) ⇌ Ni(s) + \frac{1}{2} O_{2 _{(g)}. To determine the oxygen pressure, we can apply Gibb's free energy equation with the quotient remaining, which is expressed as: ΔG = ΔG° +RT lnQ. At equilibrium, Q equals K, leading ΔG to be 0;
therefore, it simplifies to 0 = ΔG° + RT ln K. By rearranging, we arrive at ln K = ΔG° / (RT). Calculating gives, ln K = 212 / (0.00831 X 298) = 85.6.
Thus, we find K = 1.51 X 10^{37}.
Consequently, with K = 1.51 X 10^{37},

Ultimately, K = (PO_{2})^ \frac{1}{2}. Therefore, (1.51 X 10^{37}) ^ \frac{1}{2} results in 3.87 X 10^{18}.
The resulting pressure of oxygen will be = 3.87 X 10^{18} Pa.

To convert Pa to atm, we find 3.756 X
10^{13} atm.
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The response to your inquiry is: option B. 0.25 atm

Explanation:

To solve this issue, the combined gas law must be applied:

P₁V₁ = P₂V₂ / T₁T₂

The data is as follows: P1 = 0.99 atm, V1 = 2 L, T1 = 273 K, P2 =?, V2 = 4 L, T2 = 137 K.

By isolating P2 in the equation, you find

P2 = P1V1T2 / T1V2. Substituting in the numbers gives: P2 = (2 x 0.99 x 137)/(273 x 4). The resulting P2 equates to approximately 0.25 atm.

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To determine the specific heat capacity of the metal and assist in its identification, the heat absorbed by the calorimeter can be computed using: Energy = mass * specific heat capacity * temperature change Q = 250 * 1.035 * (11.08 - 10) Q = 279.45 cal/g. Next, we employ the same formula for the metal because the heat taken in by the calorimeter should equal the heat expelled by the metal. -279.45 = 50 * c * (11.08 - 45) [the minus sign indicates energy release] solving for c gives us 0.165. Therefore, the specific heat capacity of the metal amounts to 0.165 cal/g°C.
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Answer:

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On the other hand, odd-number fatty acids such as undecanoic acid generate acetyl-CoA and propionyl-CoA during their final pass. To allow entry into the TCA cycle, propionyl-CoA must go through additional processes, including carboxylation.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is facilitated by propionyl-CoA carboxylase, a biotin-dependent enzyme that is inhibited by avidin. In contrast, the oxidation of palmitate does not require carboxylation.

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Fatty acids with an even number of carbons, such as palmitate, are completely oxidized to CO₂ in the liver mitochondria due to the ability of their oxidation product, acetyl-CoA, to enter the TCA cycle where it is further oxidized to CO₂.

Undecanoic acid is classified as an odd-number fatty acid, consisting of 11 carbon atoms. The last stage of β-oxidation for odd-number fatty acids, like undecanoic acid, produces a five-carbon fatty acyl substrate that is oxidized and split into acetyl-CoA and propionyl-CoA. To enter the TCA cycle, propionyl-CoA needs additional reactions such as carboxylation. Since the oxidation occurs using a liver extract, CO₂ must be supplied externally for propionyl-CoA carboxylation, enabling the complete oxidation of undecanoic acid.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, which contains biotin. The function of biotin is to activate CO₂ before it is transferred to the propionate group. The addition of avidin obstructs the complete oxidation of undecanoic acid as it binds very tightly to biotin, thereby hindering the activation and transfer of CO₂ to propionate.

In contrast, palmitate oxidation does not require carboxylation, meaning that the presence of avidin doesn't influence its oxidation.

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