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1 year ago

9

You connect an ammeter to a circuit. The ammeter shows you that there are 5 amperes of __________ in the circuit. What one word

completes the sentence?

2 answers:

pickupchik [31]1 year ago

5 0

Answer:

Current

Explanation:

Ampere is the unit for current.

balu736 [363]1 year ago

5 0

Answer: current

Explanation:

Ampere is a unit of current

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Pseudomonas sp. has a mass doubling time of 2.4 h when grown on acetate. The saturation constant using this substrate is 1.3 g/l

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Answer:

a) Cell concentration when the dilution rate is one-half of the maximum is 0.598g cell/L

b) the substrate concentration when the dilution rate is 0.8 $D_{max}$ is 5.2g/l

c) the maximum dilution rate is : 0.41 h⁻¹

d) the cell productivity at 0.8 $D_{max}$ is 2.40g cell/L

Explanation:

Given data :

Mass doubling time of Pseudomonas sp. = 2.4 hr

Saturation constant = 1.3 g/L

Cell yield on acetate = 0.46g cell/g acetate

We are to find;

a. Cell concentration when the dilution rate is one-half of the maximum.

Here, cell yield =amount of cell produced / amount of substrate consumed.

[S] at 0.5D max is determined using the Monod's equation.

Using the formula :

$D = \frac {D_{max}[S] }{ks+[S]}$

, where D is the dilution rate,

[S] is substrate concentration; &

Ks is the saturation constant.

By replacing the values, we get :

$0.5 = \frac{S}{1.3+[S]}$

$\\\0.65=0.5[S]$

[S]=1.3g/L

The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.

=0.46×1.3

= 0.598g cell/L

b)

Substrate concentration when the dilution rate is 0.8 $D_{max}$ is calculated as:

$D = \frac {D_{max}[S] }{ks+[S]}$

0.8=[S]/1.3+[S]

1.04+0.8[S]=[S]

[S]= 5.2g/L

Therefore , the substrate concentration when the dilution rate is 0.8 $D_{max}$ is 5.2g/l

c)

Maximum dilution rate is calculated using the expression $D_{max} = \frac{1}{time}$

=1/2.4

=0.41 h⁻¹

So, the maximum dilution rate is : 0.41 h⁻¹

d)

The cell productivity at 0.8 $D_{max}$ can be calculated by multiplying the amount of the cell yield with the amount of the substrate consumed at 0.8 $D_{max}$

Cell yield = $\frac {cell \ productivity \ at \ 0.8Dmax}{amount \ of \ substrate\ consumed \at\ 0.8 \D_{max}}$

Cell productivity at 0.8 $D_{max}$ = 0.46 × 5.2

=2.40g cell/L

Therefore, the cell productivity at 0.8 $D_{max}$ is 2.40g cell/L

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