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UkoKoshka
19 days ago
12

A sample of helium gas has a volume of 0.180L, a pressure of 0.800a and a temperature of 29 C. What is the new temp or gas of vo

lume of 90mL and pressure of 3.20 atm
Chemistry
1 answer:
Anarel [2.7K]19 days ago
7 0

Solution:

The gas's new temperature is 604K

Justification:

Assuming standard temperature and pressure, we can determine the gas's temperature using the ideal gas law;

Step 1: Formulate the general gas law equation

P1V1/T1 = P2V2/T2

Step 2: Insert the values, converting as needed to standard units.

P1 = 0.800 atm

V1 = 0.180 L

T1 = 29°C = 273 + 29 = 302K

P2 = 3.20 atm

V2 = 90 mL = 90 * 10^-3 L = 0.09 L

Step 3: Solve for T2

The new gas temperature T2 is calculated as:

T2 = P2V2T1/(P1V1)

T2 = 3.20 * 0.09 * 302 / (0.800 * 0.180)

T2 = 86.976 / 0.144

T2 = 604K

The gas's new temperature is 604K.

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In 2021, you were given a 100. g wine sample to verify its age. Using tritium dating you observe that the sample has 0.688 decay
lorasvet [2668]

Answer:

1984

Explanation:

Utilizing the equation;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half-life of the radioactive isotope

t= age of the wine

Ao= initial activity of the wine

A= activity at time = t

Substituting values, we have 0.693/12.3 = 2.303/t log (5.5/0.688)

0.693/12.3 = 2.079/t

0.056 = 2.079/t

t= 2.079/0.056

t= 37 years

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7 0
27 days ago
An experimenter studying the oxidation of fatty acids in extracts of liver found that when palmitate (16:0) was provided as subs
eduard [2645]

Answer:

Fatty acids with an even number of carbons, like palmitate, undergo complete β-oxidation in the liver mitochondria, resulting in CO₂, as acetyl-CoA, their end product, can enter the TCA cycle.

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.

Explanation:

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.

6 0
5 days ago
Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul
Anarel [2728]

Answer:

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Explanation:

Assuming all calculations occur at standard pressure and a temperature of -1.72°C :

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Where

n is the number of moles of hydrogen

n is the mass of hydrogen

\rho is the density of hydrogen

6 0
1 month ago
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castortr0y [2906]
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