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marishachu
1 month ago
15

After passing through pyruvate dehydrogenase and the citric acid cycle, one mole of pyruvate will result in the formation of ___

_____ moles of carbon dioxide and ________ mole(s) of ATP (or GTP).
Chemistry
1 answer:
castortr0y [3K]1 month ago
7 0

Answer:

3 and 1

Explanation:

It is understood that the production of carbon dioxide molecules and adenosine triphosphate (ATP) molecules occurs in a ratio of 3:1, meaning for every three CO₂ molecules, one ATP molecule is generated.

Thus, we can surmise that one mole of pyruvate undergoing the citric acid cycle and the pyruvate dehydrogenase process yields CO₂ and ATP in the same ratio of 3:1.

 

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At the boiling point, the density of the liquid is 809 g/l and that of the gas is 4.566 g/l. how many liters of liquid nitrogen
KiRa [2933]

Result: 1.68 L of liquid nitrogen is generated during the gas liquefaction process.

Clarification:

This process involves transforming gaseous nitrogen into its liquid form.

The two states possess distinct densities, thus occupying varying volumes; however, the mass remains constant.

Step 1: Calculate the mass of nitrogen gas

Let’s determine the mass of nitrogen gas associated with 297 L.

The density formula is:

Density = \frac{Mass}{Volume}

With a density of nitrogen gas at 4.566 g/L and a volume of 297 L, we can compute the mass of nitrogen gas as follows:

Using these values yields:

Mass = Density \times Volume

Mass = \frac{4.566g}{L} \times 297L

Mass = 1356g

The mass of nitrogen gas calculates to be 1356 g.

Step 2: Derive the volume of liquid nitrogen from the mass obtained

The mass for liquid nitrogen remains the same.

With the density of liquid nitrogen at 809 g/L, we can substitute this into our formula to find the volume of liquid.

Volume = \frac{Mass}{Density}

Volume = \frac{1356g}{809g/L}

Therefore, the volume of liquid nitrogen is 1.68 L.


6 0
1 month ago
Primordial swamps decomposing under ancient seas and tons of rock layers gave rise to an important fuel used today. That fuel is
lorasvet [2795]
I think the right choice is C. Coal, as it's utilized in making a multitude of products across the globe. I trust this information is useful to you.:)
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1 month ago
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A sealed 10.0 L flask at 400 K contains equimolar amounts of ethane and propanol in gaseous form. Which of the following stateme
castortr0y [3046]

Answer:

(C) The average speed of molecules in ethane is the same as that of propanol.

Explanation:

In gas behavior, temperature is directly linked with speed. At a constant temperature, speed remains consistent. Also, we understand that ideal gases exhibit uniform behavior, irrespective of their type.

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1 month ago
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How much CO2 (L) is produced when 2.10 kg of sodium bicarbonate reacts with excess hydrochloric acid at 25.0 °C and 1.23 atm? A)
KiRa [2933]

The equation representing the reaction between sodium bicarbonate and hydrochloric acid is as follows:

NaHCO_3_(_s_) + HCl_(_a_q_) \implies NaCl_(_a_q_) + CO_2_(_g_) + H_2O_(_l_)

The substances NaHCO_3 and HCl combine in a 1:1 ratio. Therefore, we calculate the quantity of sodium bicarbonate and its molar mass to determine the moles formed.

NaHCO_3_M_r = 22.99 + 1.008 + 12.011+ 3 \times 16.0= 84.01 g/mol.

2.1kg\ NaHCO_3 \times \frac{1000g}{kg} \times \frac{mol}{84.01g/mol} = 24.997\ mol.

We also recognize that the stoichiometric proportions are 1:1:1:1:1, which leads to the conclusion that the moles of CO_2 equal 24.977 moles.

Next, we apply the ideal gas equation PV=nRT, where P denotes pressure, V refers to volume, R is the gas constant, and T represents the temperature in kelvins. We rearrange to solve for V

PV= nRT \implies V= \frac{nRT}{P}= \frac{ 24.997\ mol \times 8.2507m^3\ atm \times 298.15K }{mol \times K \times 1.23 atm} = 49967\ m^3

The final answer should be expressed in liters, 1L = 1000\ m^3, hence

49967\ m^3 \times\frac{L}{1000\ m^3} =49.97L\ CO_2\ produced

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1 month ago
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(58 g)/ (4L) reduce units to one
eduard [2782]
The result is 14.5 g L⁻¹. Here, the problem indicates to reduce the units to one. The existing units are g/L. To achieve a singular unit format, we can move L to the numerator, which can be executed as per the exponent laws; specifically, 1 / aˣ = a⁻ˣ. Thus, we can express 1 / L as L⁻¹. Consequently, the simplified unit remains g L⁻¹. However, remember to leave a space between two different units. This ultimately depicts a unit of density.
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