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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Find the initial velocity for an enzymatic reaction when Vmax = 6.5 x 10–5 mol•sec–1 , [S] = 3.0 x 10–3 M, and KM = 4.5 x 10–3 M. A) not enough information is given to make this calculation B) 2.6 x 10–5 mol•sec–1 C) 1.4 x 10–2 mol•sec–1 D) 8.7 x 10–3 mol•sec–1 E) 3.9 x 10–5 mol•sec–1
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What is the percent yield if 23.1 grams of FeCl3 (162.2 g/mol) is created from 10.61 grams of iron (55.85 g/mol) in the following reaction Report your answer with three significantfigures
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Answers

The percentage yield of the reaction if 23.1 grams of FeCl₃ is created by 10.61g of Fe is 74.9 %.

What is percent yield?

Percent yield of a reaction is the ratio of actual yield to the theoretical yield multiplied by 100.

As per the given balanced reaction, one mole or 55.85 g of Fe is needed to produce 1 mole or 162.2 g of FeCl₃.

Then theoretically, the mass of FeCl₃ which can be produced by 10.61 g of Fe is calculated as follows:

= (10.61 × 162.2) / 55.85

=30.81 g

This is the theoretical yield.

The actual yield is given 23.1. Now the percentage yield is calculated as follows:

Percentage yield = (Actual yield/ heretical yield)×100

= (23.1/30.81)×100

= 74.9 %.

Hence, the percentage yield of the reaction is 74.9 %.

To learn more about percentage yield, refer the link:

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Similar question just through and use it to answer your. Thanks. Hope it helps.

Could someone pls help me :)

Answers

liquid 1 and 2 have the same color and mass so the answer would be liquid 1 and 2

Explanation:

hope this is helpful

A 0.1153-gram sample of a pure hydrocarbon was burned in a C-H combustion train to produce 0.3986 gram of CO2and 0.0578 gram of H2O. Determine the masses of C and H in the sample and the percentages of these elements in this hydrocarbon.

Answers

Answer: The mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=0.3986g$

Mass of $H_2O=0.0578g$

Mass of sample = 0.1153 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.3986 g of carbon dioxide, $(12)/(44)* 0.3986=0.1087g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.0578 g of water, $(2)/(18)* 0.0578=0.0066g$ of hydrogen will be contained.

To calculate the percentage composition of a substance in sample, we use the equation:

$\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}* 100$ ......(1)

For Carbon:

Mass of sample = 0.1153 g

Mass of carbon = 0.1087 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=(0.1087g)/(0.1153g)* 100=94.27\%$

For Hydrogen:

Mass of sample = 0.1153 g

Mass of hydrogen = 0.0066 g

Putting values in equation 1, we get:

$\%\text{ composition of hydrogen}=(0.0066g)/(0.1153g)* 100=5.72\%$

Hence, the mass of carbon and hydrogen in the sample is 0.1087 g and 0.0066 g respectively and the percentage composition of carbon and hydrogen in the sample is 94.27 % and 5.72 % respectively.

What is the molarity of the potassium hydroxide if 27.20 mL of KOH is required to neutralize 0.604 g of oxalic acid, H2C2O4?H2C2O4(aq)+2KOH(aq)→K2C2O4(aq)+2H2O(l)

Answers

Answer:

Concentration of KOH = 1.154 M

Explanation:

$H_2C_2O_4(aq) + 2KOH(aq) \rightarrow K_2C_2O_4(aq) + 2H_2O(l)$

In the above reaction, 1 mole of oxalic acid reacts with 2 moles of KOH.

Mass of oxalic acid = 0.604 g

$Mole = (Mass\; in\;g)/(Molecular\;mass)$

Molecular mass of oxalic acid = 90.03 g/mol

$Mole = (0.604)/(90.03)=0.0067\;mol$

1 mol of oxalic acid reacts with 2 moles of KOH

0.0067 mol of oxalic acid reacts with $0.0067* 2 = 0.0134 mol\; of\;KOH$

Volume of the solution = 27.02 mL = 0.0272 L

$Molarity=(Mole)/(Volume\;in\;L)$

No. of mole of KOH = 0.0134 mol

$Molarity=(0.0134)/(0.0272)=1.154\;M$

Concentration of KOH = 1.154 M

Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine produces 18.13 g of CO₂, 4.639 g of H₂O, and 2.885 g of N₂. Determine the molar mass of the compound if it is between 150 and 210 g/mol.

Answers

The molar mass of the compound is found by finding the empirical and

molecular formula of the compound.

The molar mass of the compound, MM = 2 × 97 g/mol = 194 g/mol

Reasons:

Molar mass of CO₂ = 44.01 g/mol

Number of moles of CO₂ produced = $(18.13)/(44.01)$ ≈ 0.412 moles

Number of moles of produced C = 0.412 moles

Mass of C = 12 × 0.412 = 4.944 g

Molar mass of H₂O = 18.015 g/mol

Moles of H₂O produced = $(4.639)/(18.015 )$ = 0.2575 moles

Moles of produced H = 2 × 0.2575 = 0.515 moles

Mass of H = 1.00784 × 0.515 ≈ 0.519 g

Molar mass of N₂ = 28.0134 g/mol

Moles of N produced = 2 × $(2.885)/(28.0134 )$ = 2 × 0.103 = 0.206 moles

Mass of oxygen = 10 - 4.944 - 2.885 - 0.519 = 1.652

Moles of oxygen, O = $(1.652 \, g)/(16 \, g/mol)$ ≈ 0.103 moles

Therefore, we get;

Number of moles of produced C = 0.412 moles

Number of moles of produced H = 0.515 moles

Number of moles of oxygen, O ≈ 0.103 moles

Number of moles of N produced = 0.206 moles

Dividing by 0.103 gives;

Mole ratio of C = $(0.412)/(0.103) = 4$

Mole ratio of H = $(0.515)/(0.103) = 5$

Mole ratio of O = 1

Mole ratio of N = $(0.206)/(0.103) = 2$

The empirical formula of the compound is therefore; C₄H₅N₂O

The general molecular formula is of the form (C₄H₅N₂O)ₙ

Molar mass of the compound is between 150 g/mol and 210 g/mol (given)

The molar mass of C₄H₅N₂O = 4×12 + 5×1.00784 + 2×14 + 16 ≈ 97

The molar mass of C₄H₅N₂O ≈ 97 g/mol

Molar mass of the compound is between 150 and 210 g/mol, therefore, n in

(C₄H₅N₂O)ₙ = 2, which gives;

The molar mass of the compound, MM = 2 × 97 g/mol = 194 g/mol

Learn more here:

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Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

$C=18.13gCO_2*(12gC)/(44gCO_2) =4.945gC\nH=4.639gH_2O*(2.016gH)/(18.0152gH_2O)=0.519gH\nN=2.885gN_2\nO=10.0g-4.945g-0.519g-2.885g=1.651gO$

Next, the corresponding moles:

$C=4.945gC*(1molC)/(12gC)=0.412mol\nH=0.519gH*(1molH)/(1gH)=0.519mol\nN=2.885gN*(1molN)/(14gN)=0.206molN\nO=1.648gO*(1molO)/(16gO) =0.103molO$

Then, each element's subscripts is found to be:

$C=(0.412)/(0.103)=4\nH=(0.519)/(0.103)=5\nN=(0.206)/(0.103) =2\nO=(0.103)/(0.103)=1$

Therefore, the empirical formula is:

$C_4H_5N_2O$

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

$C_8H_10N_4O_2$

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

Need help asap please! This is for a very important test. Giving brainliest to best answerTwo different atoms have four protons each and the same mass. However, one has a negative charge while the other is neutral. Describe what each atomic structure could be, listing the possible number and location of all subatomic particles.

Answers

Answer:

The three subatomic particles are the particles contained in the iota. They are protons, neutrons, and electrons. Protons and electrons convey a positive and negative charge, individually, while neutrons don't convey any change

Explanation:

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