Suppose that in the citric acid cycle, oxalacetate is labeled with 14C in the carboxyl carbon farthest from the keto group. Where would you expect to find the 14C label when alpha-ketoglutarate is converted to succinate?

Answers

Answer 1
Answer:

Answer:

All of alpha-ketoglutarate formed in the citric acid cycle would contain the radioactive ^(14)C while none of succinate would contain ^(14)C, and all of carbon dioxide released would contain ^(14)C.

Explanation:

When oxaloacetate in the citric acid cycle is labeled with ^(14)C in carboxyl carbon atom which is farthest from keto group, alpha ketoglutarate formed from this oxaloactetate has the full radioactive label. In the next step, the carboxylic group (that contains the ^(14)C) is eliminated in the form of the release of the carbon dioxide and succinate is formed. Succinate thus will not have radioactivity. CO_2 released had all the radioactivity.


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Essentially all of the mass of an atom is due to the ______.(A) electrons.(B) neutrons.(C) nucleons.(D) protons.

Answers

Answer:

Atom is made up of NUCLEUS and electrons revolving around nucleus.

Nucleus itself contains protons and neutrons. Protons and neutrons are approximately 1680 times heavier than the electrons. So the major contribution to the mass of an atom comes from the nucleus.

Final answer:

The mass of an atom is mostly carried by the nucleons, protons, and neutrons, in its nucleus. Electrons contribute very little to the overall mass of an atom because of their small mass.

Explanation:

Essentially, all of the mass of an atom is due to the nucleons. An atom is primarily composed of protons, neutrons, and electrons. However, the mass of an electron is so small that it contributes very little to the overall mass of an atom. The term 'nucleons' refers to the particles in the nucleus of an atom, namely the protons and neutrons. These particles carry most of the atom's mass given their relatively large mass compared to electrons.

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Develop expressions for the mole fraction of reacting species functions of the reaction coordinate for: A system initially containing 2 mol NH3 and 5 mol O2 and undergoing the reaction: 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H20 (g) A system initially containing 3 mol NO2, 4 mol NH3, and 1 mol N2 and undergoing the reaction: 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g)

Answers

Answer:

Individual mole fractions of all the species of the all reaction is as follows.

(a)

y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)

y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)

y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)

y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)

(b)

y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)

y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)

y_{H_(2)O}=(2\epsilon)/(8- \epsilon)

y_{SO_(2)}=(2\epsilon)/(8- \epsilon)

(c)

y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)

y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)

y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)

y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)

Explanation:

(a)

Initial number of moles of NH_(3) and O_(2) are 2 mol and 5 mol respectively.

The given chemical reaction is as follows.

4NH_(3)(g)+5O_(2)(g)\rightarrow 4NO(g)+6H_(2)O

The stoichiometric numbers are as follows.

v_{NH_(3)}=-4

v_{O_(2)}=-5

v_(NO)=4

v_{H_(2)O}=6

The total number of moles initially present -7

\Sigma v_(i)\epsilon = (-4-5+4+6)= 1\epsilon

The expression for the mole fraction of species"i" is as follows.

y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}

The individual mole fractions of all the species are as follows.

y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)

y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)

y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)

y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)

(b)

Initial number of moles of H_(2)S and O_(2) are 3 mol and 5 mol respectively.

The given chemical reaction is as follows.

2H_(2)S(g)+3O_(2)(g)\rightarrow 2H_(2)O(g)+2SO_(2)

The stoichiometric numbers are as follows.

v_{H_(2)S}=-2

v_{O_(2)}=-3

v_{H_(2)O}=2

v_{SO_(2)=2

The total number of moles initially present -8

\Sigma v_(i)\epsilon = (-2-3+2+2)= - \epsilon

The expression for the mole fraction of species"i" is as follows.

y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}

The individual mole fractions of all the species are as follows.

y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)

y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)

y_{H_(2)O}=(2\epsilon)/(8- \epsilon)

y_{SO_(2)}=(2\epsilon)/(8- \epsilon)

(c)

Initial number of moles of NO_(2), NH_(3)and N_(2) are 3 mol,4 mol and 1 mol respectively.

The given chemical reaction is as follows.

6NO_(2)(g)+8NH_(3)(g)\rightarrow 7N_(2)(g)+12H_(2)O

The stoichiometric numbers are as follows.

v_{NO_(2)}=-6

v_{NH_(3)}=-8

v_{N_(2)}=7

v_{H_(2)O}=12

The total number of moles initially present -8

\Sigma v_(i)\epsilon = (-6-8+7+12)= 5 \epsilon

The expression for the mole fraction of species"i" is as follows.

y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}

The individual mole fractions of all the species are as follows.

y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)

y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)

y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)

y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)

Final answer:

Expressions for the mole fractions of reacting species are determined using stoichiometry and the initial molar amounts, taking into account the stoichiometric coefficients of the chemical reactions.

Explanation:

To develop expressions for the mole fraction of reacting species as functions of the reaction coordinate for the given systems, we will examine each reaction individually. For the reaction 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H2O (g), we can use stoichiometry to correlate the molar amounts of each species with reaction progress. Given the initial amounts, we will track how the molar amount changes for each mole of NH3 reacted.

Starting with 2 mol NH3 and 5 mol O2, the mole ratio from NH3 to NO and H2O is 1:1 and 1:1.5, respectively. The mole ratio from NH3 to O2 is 4:5. If x moles of NH3 react, the mole fractions for each species at any point in the reaction can be expressed as follows:

  • Mole fraction of NH3: (2 - x)/(Total moles)
  • Mole fraction of O2: (5 - 5x/4)/(Total moles)
  • Mole fraction of NO: (4x)/(Total moles)
  • Mole fraction of H2O: (6x)/(Total moles)

Note that 'Total moles' is the sum of the ongoing moles of all species. The mole fractions must always add up to 1 at any point during the reaction.

For the second reaction 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g), with initial amounts of 3 mol NO2, 4 mol NH3, and 1 mol N2, similar steps are taken. For every mole of NH3 reacted, the corresponding changes in molar amounts can be calculated from the stoichiometry of the balanced equation.

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What type of matter is pepperoni pizza

Answers

Answer:

Heterogeneous Mixture. Have a good day! =)

Explanation:

To create an image, the lens of an eye ________focuses light on the cornea
reflects light away from the retina
bends the cornea to correct vision
focuses light on the retina

Answers

It focuses light on the retina.

Answer:

focuses light on the retina

Explanation:

The figure below represents a reaction.What type of reaction is shown?SO3+ H2SO4 →→ HSO4
synthesis
decomposition
single displacement
double displacement

Answers

SO₃+ H₂SO₄ → H₂SO₄ This reaction is synthesis type of reaction, because that would be combination (synthesis) A + B → AB. Therefore, option A is correct.

What are the types of reaction ?

There are five basic types of chemical reactions are combination, decomposition, single-replacement, double-replacement, and combustion.

Biochemical reactions are chemical reactions that occur within living things. Metabolism refers to the sum of all biochemical reactions in an organism. Exothermic and endothermic chemical reactions are both part of metabolism.

Hydration is the process of combining water molecules with another substance to form a single, new compound. SO3 is an acidic oxide that reacts with and dissolves in water to form sulfuric acid, H2SO4.

Thus, option A is correct.

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Are you sure it isn’t SO3+H2O = H2SO4 because that would be combination (synthesis) A+ B=AB

Or SO3 + H2SO4 = H2S2O7
Because that would also be synthesis

When NaOH is added to water, the (OH) = 0.04 M. What is the [H30*]?What is the PH of the solution?

Answers

Answer:

[H₃O⁺] = 2.5 × 10⁻¹³ M

pH = 12.6

Explanation:

Step 1: Given data

Concentration of OH⁻: 0.04 M

Step 2: Calculate the concentration of H₃O⁺

Let's consider the self-ionization of water reaction.

2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

The ionic product of water is:

Kw = [OH⁻] × [H₃O⁺] = 10⁻¹⁴

[H₃O⁺] = 10⁻¹⁴ / [OH⁻]

[H₃O⁺] = 10⁻¹⁴ / 0.04

[H₃O⁺] = 2.5 × 10⁻¹³ M

Step 3: Calculate the pH

The pH is:

pH = -log [H₃O⁺] = -log 2.5 × 10⁻¹³ = 12.6