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Natasha2012 [34]

2 years ago

3

Biological oxidation of phenylacetic acid is inhibited by propionic acid present in wastewater. The following data were obtained

for enzymatic oxidation of phenylacetic acid at different concentrations of propionic acid:S (mM) 0.25 0.40 0.50 0.60 0.75 1.00V (mM/l-h), I = 0 1.0 1.7 1.9 2.1 2.4 2.5V (mM/l-h), I = 1 mM 0.65 1.1 1.3 1.4 1.8 2.2V (mM/l-h), I = 2 mM 0.55 0.9 1.0 1.3 1.4 1.8(a) What kind of inhibition is this?(b) Determine Vm, Km, and KI.

1 answer:

musickatia [10]2 years ago

7 0

Answer:

The solution to this problem is given in the attachments.

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The molecular weight of a 10g rubber band

Archy [21]

Answer:

The answer to this question is 1 * 10g /mole = 10

Explanation:

From the question given, the first step to take is to find the weight of a 10g rubber band.

Now,

The molecular weight of rubber band , tires the primary chains may have molecular weights is of order of 1 x 10 g/

Therefore it is 1 * 10g/mole =10

8 0

2 years ago

Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure

dezoksy [38]

Answer:

(a). Entropy change of refrigerant is = 0.7077 $\frac{KJ}{K}$

(b). Entropy change of cooled space $dS_{space} = - 0.6844$ $\frac{KJ}{K}$

(c). Total entropy change is dS = 0.0232 $\frac{KJ}{K}$

Explanation:

Given data

Saturation pressure = 140 K pa

Saturation temperature from property table

$T_{sat}$ = - 18.77 °c = - 18.77 + 273 = 254.23 K

(a). Entropy change of refrigerant is given by

$dS_{ref} = \frac{Q}{T_{sat}}$

Since heat absorbed by refrigerant Q = 180 KJ

$dS = \frac{180}{254.23}$

dS = 0.7077 $\frac{KJ}{K}$

(b). Entropy change of cooled space

$dS_{space} = - \frac{Q}{T_{space}}$

$T_{space}$ = - 10 °c = 263 K

$dS_{space} = - \frac{180}{263}$

$dS_{space} = - 0.6844$ $\frac{KJ}{K}$

(c). Total entropy change is given by

$dS = dS_{ref} + dS_{space}$

dS = 0.7077 - 0.6844

dS = 0.0232 $\frac{KJ}{K}$

This is the value of total entropy change.

4 0

2 years ago

A heat recovery device involves transferring energy from the hot flue gases passing through an annular region to pressurized wat

Elina [12.6K]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

4 0

2 years ago

The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to

skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 $kN/m^{2}$

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = $0.02\ m^{3}/s$

(Since, 1 litre = $10^{-3} m^{3}$ )

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, $H_{loss} = 45 cm = 0.45\ m$

Height, $h_{roof} = 2.5\ m$

Now,

The velocity in the bore is given by:

$v = \frac{Q}{\pi (\frac{d}{2})^{2}}$

$v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s$

Now, using Bernoulli's eqn:

$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k$ (1)

The velocity head is given by:

$\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553$

Now, by using energy conservation on the surface of water on the roof and that in the tank :

$\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}$

$\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45$

$P_{tank} = 5.5\times \rho \times g$

$P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}$

4 0

2 years ago

A steel rotating-beam test specimen has an ultimate strength Sut of 1600 MPa. Estimate the life (N) of the specimen if it is tes

ziro4ka [17]

Answer:

the life (N) of the specimen is 46400 cycles

Explanation:

given data

ultimate strength Su = 1600 MPa

stress amplitude σa = 900 MPa

to find out

life (N) of the specimen

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 1600

Se = 800 Mpa

and we know

Se for steel is 700 Mpa for Su ≥ 1400 Mpa

so we take endurance limit Se is = 700 Mpa

and strength of friction f = 0.77 for 232 ksi

because for Se 0.5 Su at $10^{6}$ cycle = (1600 × 0.145 ksi ) = 232

so here coefficient value (a) will be

a = $\frac{(f*Su)^2}{Se}$

a = $\frac{(0.77*1600)^2}{700}$

a = 2168.3 Mpa

so

coefficient value (b) will be

a = - $\frac{1}{3}$ log $\frac{(f*Su)}{Se}$

b = - $\frac{1}{3}$ log $\frac{(0.77*1600)}{700}$

b = -0.0818

so no of cycle N is

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value

N = $(\frac{ 900}{2168.3})^{1/-0.0818}$

N = 46400

the life (N) of the specimen is 46400 cycles

5 0

2 years ago