If three potatoes have a mass of 667 g, what will be the mass of 100 potatoes? A. 200 kg B. 22.2 kg C. 2223 g D. 20.0 kg

The isotopic mass of 47Z is calculated to be 46.96 amu. Isotopes of a single element differ in neutron count, and to ascertain the relative atomic mass, we consider each isotope's mass weighted by their natural abundance. This provided a computation to derive the mass of 47Z.

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.

M1V1 = M2V2
(2.50)(100.0) = (0.550)V2
V2 = 455mL

From 100.0 mL of 2.50 M KBr, you can prepare 455 mL of 0.550 M solution.

Answer:

• Please refer to the attached image for the graph with labeled axes and points.

Explanation:

This is a guide to fulfilling the instructions along with essential notes for understanding how to create such graphs:

1) The horizontal axis should indicate pressure ranging from 0 mb to 760 mb, while the vertical axis corresponds to volume ranging from 0 to 1 mL.

The x-axis captures the independent variable, and the y-axis records the dependent variable. Both axes must be accurately labeled, showing the variable names and their respective units.

In this context, the origin, (0,0), signifies the intersection of the axes at a pressure of 0 mb and a volume of 0.0 milliliters.

2) Allocate values for the divisions on the axes to maximize the usage of space on both.

An effective graph aims to utilize the entire space on both axes; for this, identify the maximum values for pressure and volume, and determine the corresponding marks.

The pressure range along the x-axis is [90, 760 mb], suggesting large divisions of 100 mb, with the farthest right mark at 800 mb. You can then subdivide each 100 mb interval into 10 smaller sections, using small divisions of 10 mb (my example employs 4 sections of 25 mb, but 10 mb is preferable).

The volume's range for the vertical axis is [0.1, 0.8], so it’s best to use divisions set at 0.1 ml.

3) Next, identify and label the points as follows:

• (90, 0.9) ⇒ 90 mb, 0.9 ml

• (100, 0.8) ⇒ 100 mb, 0.8 ml

• (400, 0.2) ⇒ 400 mb, 0.2 ml

• (600, 0.15) ⇒ 600 mb, 0.15 ml

• (760, 0.1) ⇒ 760 mb, 0.1 ml

The points represented as (x, y) are referred to as ordered pairs, indicating that the sequence is significant: the first number denotes the independent variable whereas the second denotes the dependent variable.

Thus, for the point (90, 0.9), 90 indicates a pressure of 90 mb and 0.9 indicates a volume of 0.9 ml.

To find (600, 0.15), since the horizontal increments are valued at 0.1, you should place the second coordinate of the point between the marks corresponding to 0.1 and 0.2 ml.

This allows you to accurately plot each point on the graph.

Respuesta:

D

Explicación:

Utilizamos la relación de moles para calcular la presión parcial. El número total de moles es 0.2 + 0.2 + 0.1 = 0.5 moles

Ahora, sabemos que la fracción molar del gas argón es 0.2/0.5

La presión parcial se determina así. Para calcular esto, simplemente multiplicamos el número de moles por la presión total.

0.2/0.5 * 5 = 1.0/0.5 = 2.00atm

D