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3 years ago

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The pyruvate dehydrogenase complex catalyzes the oxidative decarboxylation of pyruvate to form acetyl‑CoA. E1 , E2 , and E3 are

abbreviations for the enzymes of the complex. Classify the enzyme names, prosthetic groups, and reactions as E1 , E2 , or E3 .
a. Dihydrolipoxl (dihydrolipoamide) deydrogenase
b. Thiamine pyrophosphate (TPP) (also called TOP)
c. Lipoamide (or lipoate)
d. FAD

1 answer:

Mamont248 [21]3 years ago

4 0

It’s B Thiamine puro pyrophosphate (TPP)

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Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago

Myth: A dead organism is the same as a nonliving thing in science.<br><br> Fact:<br><br> Evidence:

Ilya [14]

I recently did this assignment!

Instructions: Read each myth (untruth). Reword it to make a factual statement. Then, give two to three reasons why the myth is untrue. Use complete sentences and support your answer with evidence, using your own words.

________________________________________

Answer:

Myth: A dead organism is the same as a nonliving thing in science.

o Fact: In science, dead is the same as nonliving.

o Evidence: Things that are nonliving never had the characteristics of life, and never will. Things that are dead once did have the characteristics of life, but when they die, they lose some of the characteristics. That is why dead and non-living are NOT the same thing.

Hope this helped!

Have an Amazing day!

~Lola

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2 years ago

Read 2 more answers

Can someone help me? <br><br>-just check my latest question.. thanks...

Harman [31]

Explanation:

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${\boxed{ \fbox{ \fbox{ \fbox{ \green{hello}}}}}}$

${\boxed{ \fbox{ \fbox{ \fbox{ \pink{hello}}}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ $\mathscr \fcolorbox{blue}{green}{BE \: BRAINLY}$ ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ $\mathscr \fcolorbox{blue}{blue}{BE \: BRAINLY}$ ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ $\mathscr \fcolorbox{blue}{red}{BE \: BRAINLY}$ ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ $\mathscr \fcolorbox{blue}{gray}{BE \: BRAINLY}$ ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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7 0

2 years ago

Calculate the value of the equilibrium constant, Kc , for the equilibrium shown below, if 0.124 moles of NO, 0.0240 mole of H2,

sdas [7]

the reaction is

2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)

Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2

Given

moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062

moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012

moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019

moles of H2O = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138

Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54

4 0

2 years ago

The bonds that link many water molecules together to form a water drop are _____ bonds.

Gelneren [198K]

The bonds found between water molecules are called:

Hydrogen bonds

which contribute to water having a high number of Van Der Waals forces.

Explanation:

A hydrogen bond is an electromagnetic attraction within polar molecules in which hydrogen is joined to a larger atom, such as oxygen or nitrogen. This is not a dividing of electrons, as in a covalent bond. Preferably, this is an attraction among the positive and negative poles of charged atoms.

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3 years ago