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Please answer ASAP and I'll give a thumbs up. Please show step by step solution A two-stage chemostat system is used to produce a secondary

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A two-stage chemostat system is used to produce a secondary metabolite. In the first chemostat (V1=500L), the microorganism grows with max=0.3h1 and Ks=0.1g/L. The biomass yield from substrate YXs=0.4g/g and all substrate is consumed for growth only. The feed, which contains 5.0g/L of substrate, enters this chemostat at a volumetric flow rate of 100L/h. Secondary metabolite production is achieved in a second chemostat (V2=300L), where cell growth is negligible and substrate consumption is for product formation only. The feed to this chemostat is the effluent from the first chemostat. The specific rate of product formation is qp=0.02g product/g cell-h, and YPs=0.6g/g a. What is the steady-state cell concentration in the first chemostat? b. What is the steady-state substrate concentration in the first chemostat? c. What is the steady-state product concentration in the second chemostat? d. What is the steady-state substrate concentration in the second chemostat

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