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zhuklara [117]
3 years ago
10

You are carrying out an experiment in the lab to study the glycolysis pathway. To do this, a liver extract (which is capable of

carrying out all of the normal metabolic reactions) is incubated with 14C-labeled glucose at carbon 1. Based on your knowledge of glycolysis, where does the 14C-label end up in pyruvate?

Chemistry
1 answer:
Virty [35]3 years ago
5 0

Answer:

When C1 is labeled in glucose, it ends up in the methyl group of pyruvate.

Aldolase cleaves a hexose into two trioses.

[See the image attached].

Asterisk indicates the label.

When C1 is labeled in glucose, it ends up in the carboxyl group of pyruvate.

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A phylogenetic tree may be built using morphological (body shape), biochemical, behavioral, or molecular features of species or other groups. In building a tree, we organize species into nested groups based on shared derived traits (traits different from those of the group's ancestor).
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3 years ago
Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w
Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619=  [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373  

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10  = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0
3 years ago
Substance P is carbon.
stepan [7]

Explanation:

substance Q could be <em><u>oxygen (O2)</u></em>

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3 years ago
What is the chemical symbol for iron
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Read 2 more answers
At the equivalence point of a titration of the [H+] concentration is equal to:
icang [17]

B. At the equivalence point of a titration of the [H+] concentration is equal to 7.

<h3>What is equivalence point of a titration?</h3>

The equivalence point of a titration is a point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution.

At the equivalence point in an acid-base titration, moles of base equals moles of acid and the solution only contains salt and water.

At the equivalence point, equal amounts of H+ and OH- ions combines as shown below;

H⁺ + OH⁻  → H₂O

The pH of resulting solution is 7.0 (neutral).

Thus, the pH at the equivalence point for this titration will always be 7.0.

Learn more about equivalence point here: brainly.com/question/23502649

#SPJ1

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