I need help with 3 and 4

Reil [10]

i think it is the cell wall.

6 0

3 years ago

Analyses of enzymes found in the blood are used as indicators that tissue damage (heart, liver, muscle) has occurred and resulte

iren [92.7K]

Answer:

Lactate dehydrogenase (LDH)

Explanation:

LDH is tetrameric enzyme found in the muscles (M-type) and the heart (H-type) of living cells, responsible for the conversion of pyruvate to lactate and back, to promote generation of nicotinamide adenine dinucleotide (NAD).

3 0

4 years ago

BRAINLIST!!

Natali5045456 [20]

B erosion is the answer

6 0

3 years ago

Read 2 more answers

Electrons are mapped beginning with the ____ electrons.

kifflom [539]

D the lowest amount of energy

7 0

3 years ago

Pulverized coal pellets, which may be approximated as carbon spheres of radius ro= 1 mm, are burned in a pure oxygen atmosphere

Ganezh [65]

Answer:

Explanation:

SO; If we assume that:

P should be the diffusion of oxygen towards the surface ; &

Q should be the diffusion of carbondioxide away from the surface.

Then the total molar flux of oxygen is illustrated by :

$Na,x = - cD_{PQ}\frac{dy_P}{dr} +y_P(NP,x + N_Q,x)$

where;

r is the radial distance from the center of the carbon particle.

Since ;

$N_P,x = - N_Q, x$ ; we have:

$Na,x = - cD_{PQ}\frac{dy_P}{dr}$

The system is not steady state and the molar flux is not independent of r because the area of mass transfer $4\pi r^{2}$ is not a constant term.

Therefore, using quasi steady state assumption, the mass transfer rate $4\pi r^{2}N_{P,x}$ is assumed to be independent of r at any instant of time.

$W_{P}=4\pi r^{2}N_{P,x}$

$W_{P}=-4\pi r^{2}cD_{PQ}\frac{dy_{P}}{dr}$

= constant

The oxygen concentration at the surface of the coal particle $yP,R$ will be calculated from the reaction at the surface.

The mole fraction of oxygen at a location far from pellet is 1.

Thus, separating the variables and integrating result into the following:

$W_{P}\int_{R}^{\infty} \frac{dr}{r^{2}}=-4\pi$

$r^{2}cD_{PQ}\int_{y_{P,R}}^{y_{P,\infty }}dy_{P}$

$-W_{P}\frac{1}{r}\mid ^{\infty }_{R}= -4\pi cD_{PQ}(y_{P,\infty }-y_{P,R})$

$=> W_{P}= - 4\pi cD_{PQ}(1-y_{P,R})R$

The mole of oxygen arrived at the carbon surface is equal to the mole of oxygen consumed by the chemical reaction.

$W_{P} = 4 \pi R^2R"$

$W_{P}= 4\pi R^{2}k_{1}"C_{O_{2}}\mid _{R}$

$W_{P}= 4\pi R^{2}k_{1}"c y _{P,R}$

$-4\pi cD_{PQ}(1-y_{P,R})R= - 4\pi R^{2}k_{1}"c y _{P,R}$

$y_{P,R}=\frac{D_{PQ}}{D_{PQ}+Rk_{1}}$

$y_{P,R}=\frac{1.7 \times 10^{-4}}{1.7\times 10^{-4}+10^{-3}\times 0.1}$

$\mathbf{= 0.631}$

Obtaining the total gas concentration from the ideal gas law; we have the following:

where;

R= $0.082m^3atm/kmolK$

$c=\frac{P}{RT} \\ \\ c=\frac{1}{0.082\times 1450} \\ \\ = 0.008405kmol/m^3$

The steady state $O_2$ molar consumption rate is:

$W_{P}= -4\pi cD_{PQ}(1-y_{P,R})R$

$W_{P}= -4\pi (0.008405)(1.7\times 10^{-4})(1-0.631)(10^{-3})$

$W_{P}= - 6.66\times 10^{-9}kmol/s$

5 0

3 years ago