Calculate the mass of glucose metabolized by a 60.0 −kg person in climbing a mountain with an elevation gain of 1550 m . Assume that the work performed in the climb is about four times that required to simply lift 60.0 kg by 1550 m . (ΔfH∘ of C6H12O6(s) is -1273.3 kJ/mol.) w = mgh (mass in kg, g = 9.8 m/s2, h in meters)

Mass of glucose = 515.34 g

Explanation:

We are given;

Mass; m = 60 kg

Elevation; h = 1550 m

Acceleration due to gravity; 9.8 m/s²

Now, work performed to lift 60kg by 1550m is given by the formula;

W = mgh

W = 60 × 9.8 × 1550

W = 911400 J

We are told the actual work is 4 times the one above.

Thus;

Actual work = 4W = 4 × 911400 = 3,645,600 J

Now,

Molar mass of Glucose(C6H12O6) = 180 g/mol

We are given standard enthalpy of combustion = -1273.3 KJ/mol = -1273300

Moles of glucose = 3645600/1273300 = 2.863mol

Mass of glucose = 2.863 mol × 180 g/mol

Mass of glucose = 515.34 g

i can ! i just need to know what the paragraph is! : p

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.