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Please answer the following question with explanation
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Asked by deepakudgiri29 | 04 Jan, 2019, 05:14: PM
answered-by-expert Expert Answer
The combustion of starch takes place as:
 
  xO subscript 2 space plus space left parenthesis straight C subscript 6 straight H subscript 10 straight O subscript 5 right parenthesis subscript straight x space rightwards arrow with blank on top space 6 straight x space CO subscript 2 space plus space 5 straight x space straight H subscript 2 straight O
This problem may be solved using the empirical formula of starch i.e., C6H5O5.
 
increment straight H subscript straight f superscript straight o space left parenthesis starch right parenthesis space equals space 6 straight x increment straight H subscript straight f superscript straight o space left parenthesis CO subscript 2 right parenthesis space plus space 5 straight x space increment straight H subscript straight f superscript straight o space left parenthesis straight H subscript 2 straight O right parenthesis space minus space increment straight H to the power of straight o space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 6 straight x space open parentheses fraction numerator negative 94.05 space kcal over denominator mol end fraction close parentheses space plus space 5 straight x open parentheses fraction numerator negative 68.32 space kcal over denominator mol end fraction close parentheses space minus space open parentheses fraction numerator negative 4.18 space kcal over denominator gm end fraction close parentheses open parentheses fraction numerator 162 straight x space straight g over denominator mol end fraction close parentheses space equals space minus space 229 straight x space kcal For space per space gm space of space starch comma space space space open parentheses fraction numerator negative 229 straight x space kcal over denominator mol end fraction close parentheses open parentheses fraction numerator 1 space mol over denominator 162 straight x space gm end fraction close parentheses space equals space minus 1.41 space kcal divided by straight g
Answered by Ramandeep | 07 Jan, 2019, 02:35: PM
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NEET neet - Chemistry
for a gaseous reaction at 300k, ∆H-∆U=-4.98kj assuming that r=8.3JK^-1mol^-1 ∆n(g) is
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Asked by vasantagomasi23 | 05 Apr, 2024, 08:35: AM
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Thermodynamics  
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Asked by prakriti12oct | 29 Oct, 2019, 12:54: AM
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NEET neet - Chemistry
Please answer the following question with explanation
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Asked by deepakudgiri29 | 04 Jan, 2019, 05:14: PM
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