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Phantasy [73]
3 years ago
7

Glucose, C 6 H 12 O 6 , is used as an energy source by the human body. The overall reaction in the body is described by the equa

tion C 6 H 12 O 6 ( aq ) + 6 O 2 ( g ) ⟶ 6 CO 2 ( g ) + 6 H 2 O ( l ) Calculate the number of grams of oxygen required to convert 58.0 g of glucose to CO 2 and H 2 O . mass of O 2 : 61.76 g Calculate the number of grams of CO 2 produced.
Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
6 0

Answer:

m_{O_2}=61.87gO_2

m_{CO_2}=85.07gCO_2

Explanation:

Hello,

Considering the given reaction's stoichiometry, grams of oxygen result:

m_{O_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molO_2}{1molC_6H_{12}O_6}*\frac{32gO_2}{1molO_2}\\m_{O_2}=61.87gO_2

Moreover, the mass of produced carbon dioxide turns out:

m_{CO_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molCO_2}{1molC_6H_{12}O_6}*\frac{44gCO_2}{1molCO_2}\\m_{O_2}=85.07gCO_2

Best regards.

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Given the standard entropy values in the table, what is the value of º for the following reaction?
Mandarinka [93]
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The answer is <span>+313.766 J/mol·K
</span>
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The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
Cloud [144]

Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

7 0
3 years ago
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Is there any more info lol
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