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2 years ago

What fraction of the carbon dioxide exhaled by animals is generated by the reactions of the citric acid cycle

Chemistry

1 answer:

puteri [66]2 years ago

4 0

Answer:

1/3

Explanation:

Pyruvate is produced by the glycolysis in cytoplasm. The oxidation of pyruvate takes place in mitochondrial matrix.

Pyruvate is converted to acetyl-CoA in the reaction given below:

Pyruvate + NAD⁺ + CoA-SH ⇒ acetyl-CoA + NADH + CO₂

1 molecule of carbon dioxide is eliminated from 1 molecule of pyruvate.

Also,

2 molecules of carbon dioxide is eliminated from 2 molecules of pyruvate (as glucose on glycolysis yields 2 molecules of pyruvate).

Also, acetyl-CoA further goes into the citric acid cycle and produces 2 molecules of carbon dioxide.

Thus pyruvate produces total 3 molecules of CO₂ and hence glucose produces 6 molecules of CO₂ (as glucose on glycolysis yields 2 molecules of pyruvate)

Thus,

<u>Fraction = 2/6 = 1/3</u>

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A sample of gas contains 0.1800 mol of CO(g) and 0.1800 mol of NO(g) and occupies a volume of 23.2 L. The following reaction tak

baherus [9]

Answer:

The volume of the sample is 17.4L

Explanation:

The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:

0.1800mol + 0.1800mol reactants =

0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.

Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:

V1n2 = V2n1

<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>

Replacing:

V1 = 23.2L

n2 = 0.2700 moles

V2 = ??

n1 = 0.3600 moles

23.2L*0.2700mol = V2*0.3600moles

17.4L = V2

<h3>The volume of the sample is 17.4L</h3>

8 0

1 year ago

If an aqueous solution of urea n2h4co is 26% by mass and has density of 1.07 g/ml, calculate the molality of urea in the soln

Ksju [112]

Answer is: molality of urea is 5.84 m.

If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.

5 0

2 years ago

Read 2 more answers

The empirical formula of a gaseous fluorocarbon is CF2. At a certain temperature and pressure, a 1-L volume holds 8.93 g of this

dimaraw [331]

Answer:

C₄F₈

Explanation:

Using their mole ratio to compute their mass

molar mass of carbon = 12.0107 g/mol

molar mass of fluorine gas = 37.99681

let x = mass of carbon

given mass of fluorine = 1.70 g

x / 12.01067 = 1.70 / 37.99687

cross multiply

x = ( 1.70 × 12) / 37.99687 = 20.4 / 37.99687 = 0.53688 g

mass of one mole of CF₂ = 0.53688 + 1.70 = 2.23688 g

number of mole of CF₂ = 8.93 g / 2.23688 = 3.992 approx 4

molecular formula of CF₂ = 4 (CF₂) = C₄F₈

3 0

2 years ago

trans-2-Butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm-1); however, IR spectroscopy is

spayn [35]

Answer:

The other signal that would indicate the presence of a C= C bond appears close to 3100 $cm^{-1}$ .

Explanation:

Bands that appear above 3000 $cm^{-1}$ are often unsaturation diagnoses suggest. The band at 3000- 3100 $cm^{-1}$ is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.

4 0

2 years ago

A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me

DiKsa [7]

Problem One

You will use both m * c * deltaT and H = m * heat of fusion.

Givens

m = 12.4 grams

c = 0.1291

t1 = 26oC

t2 = 1204

heat of fusion (H_f) = 63.5 J/grams.

Equation

H = m * c * deltaT + m * H_f

Solution

H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5

H = 1660.1 + 787.4

H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.

Problem Two

Formula and Givens

t1 = 14.5

t2 = 50.0

E = 5680

c = 4.186

m = ??

E = m c * deltaT

Solution

5680 = m * 4.186 * (50 - 14.5)

5680 = m * 4.186 * (35.5)

5680 = m * 148.603 * m

m = 5680 / 148.603

m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.

m = 38.2 to 3 sig digs.

8 0

2 years ago

Read 2 more answers