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The enthalpy change involved in the oxid...

The enthalpy change involved in the oxidation of glucose is `-2880kJ mol^(-1)`. Twenty five per cent of this energy is available for muscular work . If `100kJ` of muscular work is needed to walk one kilometre, what is the maximum distance that a person will be able to walk after eating `120g` of glucose ?

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To solve the problem, we will follow these steps: ### Step 1: Calculate the number of moles of glucose in 120 g. The molecular weight of glucose (C₆H₁₂O₆) is 180 g/mol. \[ \text{Number of moles of glucose} = \frac{\text{mass}}{\text{molar mass}} = \frac{120 \text{ g}}{180 \text{ g/mol}} = 0.6667 \text{ moles} \] ...
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The enthalpy change involved in the oxidation of glucose is -2880 kJ mol^(-1) . Twenty five per cent of this energy is available for muscular work. If 100 kJ of muscular work is needed to walk one kilometre, what is the maximum distance that a person will be able to walk eating 120 g of glucose ?

Standard molar enthalpy of combustion of glucose is -2880 kJ. If only 25% of energy is available for muscular work and 1.0 km walk consumes 90 kJ of energy, what maximum distance (in km) a person can walk after eating 90 g of glucose.

If 150 kJ of energy is needed for muscular work to walk a distance of one km, than how much of gulcose one has to consume to walk a distance of five km, provided only 30% energy is available for muscular work.The enthalpy of combustion of glucose is 3000 kJ mol^(-1)

If 150 kJ of energy is needed for muscular work to walk a distance of one km, than how much of gulcose one has to consume to walk a distance of five km, provided only 30% energy is available for muscular work.The enthalpy of combustion of glucose is 3000 kJ mol^(-1)

An average person eats carbohydrates equivalent to 0.350 kg of glucose and 0.200 kg of fats every day. The person needs about 10000 kJ per day. If the body consumes carbohydrates preferentially, how much will his weight increase per year assuming that only 50% of the excess fats are excreted. Give that the heat of combustion of glucose is 2900 kJ and combustion of fats produces energy equivalent to 39000 kJ kg^(-1) ?

Animals operate under conditons of constant pressure and most of the process tht maintain life are isothermal ( in a broad sense) . How much energy is available for sustaining this type of muscular and nervous activity from the combustion of 1mol of glucose molecules under standard conditons at 37^(@) C (blood temperature) ? The entropy change is +182.4 JK^(-1) for the reaction stated above DeltaH_("combustion") [glucose]=-2808 KJ

The enthalpy of combustion of mol. Wt. 180 glucose is -2808 KJ "mol"^(-1) at 25^(@) C . X and Y grams of glucose do you need to consume respectively cases [Assume wt=62.5 Kg]. (a) to climb a flight of stairs rising through 3M. to climb a mountain of altitude 3000 M? Assume that 25% of enthalpy can be converted to useful work. X and Y are related as X=mY, then find m.

Photosynthesis is a bio process by which plants make energy rich molecules from low energy molecules with the help of energy from sunlight . The photosynthesis of glucose can be represented as: 6CO_(2(g))+ 6H_(2)O_(g) + hv rarr C_(5)H_(12)O_(6(s)) + 6O_(2)(g)..........(i) The energy of one mole of a photon of wave lenght is known as one Einstein. A weight lifter lifts a weight of 160Kg through of 2.4 m . Assuming all the energy required for this task is obtained by the combustion of glucose , calculate the change in the current produced by a sample of blood of the weight lifter . Same volume of blood (5mL) is tested both before and after lifting the weight and the total volume of blood in his body is 5L. (1 g of glucose releases 15.58 KJ of energy )

An athelete is given 100g of glucose (C_(6)H_(12)O_(6)) of energy equivalent to 1560 kJ . He utilises 50% of this gained enegry in the event. In order to avoid storage of enegry in the body, calculate the weight of water he would need to perspire. The enthalpy of evaporation of water is 441kJ//mol .

An athlete is given 100 g of glucose ( C_6H_12O_6 ) of energy equivalent to 1560 kJ. He utilises 50% of this gained energy in the event. In order to avoid the storage of energy in the body, calculate the weight of water he would need to perspire. The enthalpy of evaporation of water is 44 kJ mol^(-1) .

CENGAGE CHEMISTRY ENGLISH-THERMODYNAMICS-Archives (Subjective)
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  2. The enthalpies for the following reactions (DeltaH^(Theta)) at 25^(@)C...

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  3. The standared enthalpies of formation at 298K for C C1(g), H(2)O(g), C...

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  4. Given that: i. C(s) + O(2)(g) rarr CO(2)(g) , DeltaH =- 94.05 kcal ...

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