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VMariaS [17]
3 years ago
12

Which of the following is false?

Chemistry
2 answers:
Alla [95]3 years ago
8 0
Protoporphyrin is bound to an iron atom in its Fe² state, not Fe³. It would need hemin to be able to bind to an iron atom in a Fe³ state.

The answer is B.
kipiarov [429]3 years ago
4 0
The accurate answer is
<span>b) Protoporphyrin is bound to a single iron atom in its Fe+3 state

Explanation:
</span>Protoporphyrin is bound to a single iron atom in its Fe+2<span> state.</span> 
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3 years ago
A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water at temperature of 288.8 K
Aloiza [94]

Answer:

a) steam used = 8440 kg/hr

b)  Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

Explanation:

Saturated steam pressure = 42KPag=1.42 bar

From steam table, Steam temperature = 110 C

Latent heat of this steam = 2230 KJ/kg

Process side pressure = 20 Kpaa = .2 bar

Water latent heat at this Pressue = 2360 KJ/kg

Water boiling Point at this Pressure =60 C

Feed Inlet temperature = 15.6 C

Total Heat required = Heat required to rise feed temperature from 15.6 to 60 degree C + Evaporation of water to concentrate the feed

Total Water evaporation required = 9072*(100-10)/100-9072*.1/.5*.5=7257.6 Kg/hr

Specific heat of feed assumed = 4.2 KJ/kg/K

=> Total heat required = 9072*4.2*(60-15.6)+7257.6*2360=18819682 KJ/hr

LMTD = ((110-60)-(110-15.6))/LN((110-60)/(110-15.6))=69.8 C

U, Overall Heat transfer coefficient = 1988 W/m2/K

Total heat required = U*A*LMTD

=> Area required of evaporator, A=18819682 *1000/3600/1988/59.8=44 m2

Steam used = 18819682/2230=8440 kg/hr

Energy required to condense vaporised feed = 7257.6*2360=17127936 KJ/hr

=> Steam efficiency = (18819682-17127936)/18819682=9%

b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

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