Calculate the volume of dry co2 produced at body temperature (37 ∘c) and 0.970 atm when 25.5 g of glucose is consumed in this reaction.

I believe the balanced chemical equation is:

C6H12O6 (aq) + 6O2(g) > 6CO2(g) + 6H2O(l) 

First calculate the moles of CO2 produced:

moles CO2 = 25.5 g C6H12O6 * (1 mol C6H12O6 / 180.15 g) * (6 mol CO2 / 1 mol C6H12O6)

moles CO2 = 0.8493 mol

Using PV = nRT from the ideal gas law:

V = nRT  / P

V = 0.8493 mol * 0.08205746 L atm / mol K * (37 + 273.15 K) / 0.970 atm

V = 22.28 L

The volume of the dry CO₂ produced is 22.30 liters.

The given parameters;

The balanced chemical reaction of the given compounds;

Molecular mass of glucose ( = 180 g/mol)

180 g of glucose ---------------- 6 moles of CO₂

25.5 g of glucose ---------------- ? moles of  CO₂

The volume of the  dry CO₂ produced is calculated form Ideal gas law;

PV = nRT

where;

R is the ideal gas constant = 0.082057 L. atm /mol K

Thus, the volume of the dry CO₂ produced is 22.30 liters.

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the answer is learned

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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