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gulaghasi [49]
3 years ago
6

Glucose, C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the equation C

6H12O6(aq)+6O2(g)⟶6CO2(g)+6H2O(l) Calculate the number of grams of oxygen required to convert 48.0 g of glucose to CO2 and H2O. mass of O2: g Calculate the number of grams of CO2 produced. mass of CO2: g
Chemistry
1 answer:
Crank3 years ago
7 0

Answer:

70.3824 grams of carbondioxide is produced.

Explanation:

C_6H_{12}O_6(aq)+6O_2(g)⟶6CO_2(g)+6H_2O(l)

Moles of glucose =  \frac{48.0 g}{180 g/mol}=0.2666 mol

According to reaction 1 mole of glucose reacts with 6 mole of oxygen gas.

Then 0.2666 mol of gluocse will reactwith :

\frac{6}{1}\times 0.2666 mol =1.5996 mol of oxygen

Mass of 1.5996 moles of oxygen gas:

1.5996 mol\times 32 g/mol = 51.1872 g

51.1872 grams of oxygen are required to convert 48.0 grams of glucose to CO_2 and H_2O.

According to reaction, 1 mol, of glucose gives 6 moles of carbon dioxide.

Then 0.2666 moles of glucose will give:

\frac{6}{1}\times 0.2666 mol =1.5996 mol of carbon dioxide

Mass of 1.5996 moles of carbon dioxide gas:

1.5996 mol\times 44 g/mol = 70.3824 g

70.3824 grams of carbondioxide is produced.

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Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.

<h3>What is the Ki for the inhibitor?</h3>

The Ki of an inhibitor is known as the inhibition constant.

The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.

Using the Michaelis-Menten equation for inhibition:

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Making Ki subject of the formula:

  • Ki = [I]/{(Kma/Km) - 1}

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Solving for Ki:

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Ki = 26.7 μM/{(2.5/1) - 1)

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2 years ago
Why does potassium permanganate need to be standardized right before a titration?.
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If the volume of wet gas collected over water is 85.0 mL at 20°C and 760 mm Hg , what is the volume of dry gas at STP conditions
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Explanation:

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The combined gas equation is:

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P_1 = initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg

P_2 = final pressure of dry gas at STP =  760 mm Hg

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V_2 = final volume of dry gas at STP = ?

T_1 = initial temperature of dry gas = 20^oC=273+20=293K

T_2 = final temperature of dry gas at STP = 0^oC=273+0=273K

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V_2=77.4mL

Volume of dry gas at STP is 77.4 mL.

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