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3 years ago

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Changes the environment that cause and organism to the respond called

1 answer:

marissa [1.9K]3 years ago

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<span>A change in an organisms environment is called an?</span>

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Different versions of a federal bill have successfully passed both in the House and the Senate. Which is the next step in the la

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Answer: The resulting bill returns to the House and Senate for final approval. The Government Printing Office prints the revised bill in a process called enrolling. The President has 10 days to sign or veto the enrolled bill.

Explanation:

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3 years ago

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Describe a dinner a marathon runner would eat the night before running the Boston Marathon, and give reasons for your food choic

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A dinner comprised of rice and chicken. Rice is a simple carbohydrate and will build up the ATP in his system to make his run time better and give him energy. Chicken is a long term energy source that will keep him full and his muscles working throughout the marathon.

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3 years ago

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4 years ago

Pseudomonas putida is used for fermentation of the lactose present in cheese whey. The bacteria are cultivated in a steady-state

fgiga [73]

Answer:

Explanation:

Given that:

The dilution rate D = 0.28 h⁻¹

The concentration of lactose in the feed $S_o = 2.0 \ g/L$

The effluent S = 0.10 g/L

Also;

$Y_{X/S} = 0.45 g\ X/g \ S , \\ \\ Y_{X/O2 }= 0.25 g \ X/g \ O2, \\ \\$

Saturation C* = 8 mg/l

To calculate the steady-state biomass, we use the formula:

$X = Y_{X/S}(S_o-S_e) \\ \\ X = 0.45(2.0 -0.10) \ g/L \\ \\ X= 0.45 (1.9) \ g/L \\ \\ X = 0.855\ g/L \\ \\ X = 855 \ mg/L$

The biomass is 0.855 g/L

For a steady-state condition, the oxygen uptake rate can be illustrated by using the formula:

$q_{o_2}X =\dfrac{\mu_X}{Y_{X/O_2}}$

where;

$\mu =$ dilution rate (D)

Thus, the steady-state can be expressed as:

$q_{o_2}X =\dfrac{D}{Y_{X/O_2}}$

$q_{o_2}X =\dfrac{0.28}{0.25}$

$q_{o_2}X =1.12 \ h^{-1}$

The specific rate of oxygen consumption $q_{o_2}X =1.12 \ h^{-1}$

b)

In the fermentation medium, if the desired DO concentration $C_L$ = 2 mg/L

Here, the oxygen transfer is regarded as the rate-limiting step.

As such, the oxygen transfer rate(OTR) is equivalent to the oxygen uptake rate.

In this scenario, let's determine the oxygen transfer coefficient $(K_{La})$ by using the formula:

$OTR = K_{La}(C^* - C_L)$

where;

$K_{La}$ = coefficient of oxygen transfer

C* = saturation

Since $OTR = q_{O_2}X$

$q_{o2}X = K_{La}(C^*-C_L) \\\\ (1.12 )(855) = K_{La}(8-2) \\ \\ 957.6 = K_La (6) \\ \\ K_{La}= \dfrac{957.6}{6} \\ \\ K_{La} = 159.6 \ h^{-1}$

Thus, the oxygen transfer coefficient $K_{La}$ = 159.6 h⁻¹

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3 years ago

Large organic molecules which are synthesized from multiple identical subunits are:

guapka [62]

The key word here is “synthesized”. Polymers are long chains of monomers,

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