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ser-zykov [4K]
3 years ago
15

Identify the correct order of electron transfers in the electron transport chain starting from FADH2.

Chemistry
1 answer:
Damm [24]3 years ago
6 0

Answer:

FADH₂ → Q coenzyme → Complex III → c cytochrome → Complex IV → O₂

Explanation:

During oxidative phosphorylation, the electrons from NADH and FADH₂ are combined with O₂ and the energy released in the process is used to synthesize ATP from ADP.

The components of the electron transport chain are located in the internal part of the mitochondrial membrane in eukaryotic cells, and in the cell membrane in bacteria. The transporters in the electron transport chain are organized into four complexes in the inner mitochondrial membrane. A fifth complex then couples these reactions to the ATP synthesis.

Complex II receives the electrons from the succinate, which is an intermediary in the Krebs cycle. These electrons are transferred to the FADH₂ and then to the Q coenzyme. This liposoluble molecule will transport the electrons from Complex II to Complex III. In this complex, the electrons are transferred from the <em>b</em> cytochrome to the <em>c</em> cytochrome. This <em>c </em>cytochrome, which is a peripheric membrane protein located in the external part of the inner membrane, then transports the electrons to Complex IV where finally they are transferred to the oxygen.

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3 years ago
Be sure to answer all parts. Dimercaprol (HSCH2CHSHCH2OH) was developed during World War I as an antidote to arsenic-based poiso
Sauron [17]

<u>Answer:</u>

<u>For A:</u> The number of arsenic atoms are 3.4\times 10^{21}

<u>For B:</u> The percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

<u>Explanation:</u>

  • <u>For A:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of dimercaprol = 696 mg = 0.696 g    (Conversion factor:  1 g = 1000 mg)

Molar mass of dimercaprol = 124.21 g/mol

Putting values in above equation, we get:

\text{Moles of dimercaprol}=\frac{0.696g}{124.21g/mol}=0.0056mol

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.0056 moles of dimercaprol will contain 0.0056\times 6.022\times 10^{23}=3.4\times 10^{21} number of molecules.

As, 1 molecule of dimercaprol binds with 1 atom of Arsenic

So, 3.4\times 10^{21} number of dimercaprol molecules will bind with = 1\times 3.4\times 10^{21}=3.4\times 10^{21} number of arsenic atoms

Hence, the number of arsenic atoms are 3.4\times 10^{21}

  • <u>For B:</u>

We know that:

Molar mass of dimercaprol = 124.21 g/mol

Molar mass of mercury = 200.59 g/mol

Molar mass of thallium = 204.38 g/mol

Molar mass of chromium = 51.99 g/mol

Also, 1 molecule of dimercaprol binds with 1 metal atom.

To calculate the percentage composition of metal in a complex, we use the equation:

\%\text{ composition of metal}=\frac{\text{Mass of metal}}{\text{Mass of complex}}\times 100     ......(1)

  • <u>For mercury:</u>

Mass of Hg-complex = (200.59 + 124.21) = 324.8 g

Mass of mercury = 200.59 g

Putting values in equation 1, we get:

\%\text{ composition of mercury}=\frac{200.59g}{324.8g}\times 100=61.76\%

  • <u>For thallium:</u>

Mass of Tl-complex = (204.38 + 124.21) = 328.59 g

Mass of thallium = 204.38 g

Putting values in equation 1, we get:

\%\text{ composition of thallium}=\frac{204.38g}{328.59g}\times 100=62.2\%

  • <u>For chromium:</u>

Mass of Cr-complex = (51.99 + 124.21) = 176.2 g

Mass of chromium = 51.99 g

Putting values in equation 1, we get:

\%\text{ composition of chromium}=\frac{51.99g}{176.2g}\times 100=29.51\%

Hence, the percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

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Soloha48 [4]

Explanation:

It is known that equation for ideal gas is as follows.

               PV = nRT

The given data is as follows.

     Pressure, P = 1500 psia,     Temperature, T = 104^{o}F = 104 + 460 = 564 R

     Volume, V = 2.4 cubic ft,      R = 10.73 psia ft^{3}/lb mol R

Also, we know that number of moles is equal to mass divided by molar mass of the gas.

                n = \frac{mass}{\text{molar mass}}

            m = n \times W

                = 0.594 \times 16.04

                = 9.54 lb

Hence, molecular weight of the gas is 9.54 lb.

  • We will calculate the density as follows.

                d = \frac{PM}{RT}

                    = \frac{1500 \times 16.04}{10.73 \times 564}

                    = 3.975 lb/ft^{3}

  • Now, calculate the specific gravity of the gas as follows.

  Specific gravity relative to air = \frac{\text{density of methane}}{\text{density of air}}

                         = \frac{3.975 lb/ft^{3}}{0.0765 lb/ft^{3}}

                         = 51.96

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Answer:

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