Strain of an object is defined as:

 is the final length and  is the initial length. The equation implies that increasing the initial length of an object () would decrease the strain. Similarly, the equation also implies that increasing the final length of an object () would increase the strain; therefore, increasing the initial length decreases strain, whereas increasing the final length increases strain.

To answer this question, we need to know the equation for Young’s modulus, :

Recall the definitions of stress and strain:

Here,  is force,  is cross-sectional area,  is change in length, and  is initial length. Plugging these two equations into the definition of Young’s modulus gives us:

Rearranging this equation to solve for change in length gives us:

Since we want to increase  we look for answer choices that increase  and , or decrease  and . Among the answer choices, the greatest increase occurs when you double the elasticity of the rubber band. Recall that Young’s modulus is a measure of the stiffness of a material; therefore, a more elastic (flexible, or less stiff) material will lower Young’s modulus. Having a material twice as elastic will decrease  by one half and will give the greatest increase in the change in length of the rubber band. 

Using half the length () and using twice the cross-sectional area will decrease the change in length. Using a rubber band twice as long, but twice as elastic, will not alter the change in length.

Young’s modulus is a constant that is calculated by generating a stress vs. strain plot. The slope of the linear region of the stress vs. strain plot is defined as the Young’s modulus. Recall that the stress vs. strain plot is nonlinear past the elastic limit (stress at which the material starts to plastically deform). This means that we cannot calculate the slope past the elastic limit; Young’s modulus is invalid for stresses beyond the elastic limit.

Young’s modulus is defined as follows:

To calculate strain, you divide change in length by initial length. Since both change in length and initial length have the same units, the units cancel and strain becomes a dimensionless quantity. This means that the Young’s modulus will have the same units as stress alone. 

Young’s modulus is a material property. This means that the Young’s modulus is constant for a given type of material. It does not depend on the stress experienced by the material. Any changes to stress will be compensated for by a change in strain, which will allow Young’s modulus to be constant for the material. Similarly, Young’s modulus does not depend on strain or on the geometrical properties (length and area) of the material. It will only change if you change the type of material being used.

Since heat can be considered a reactant (endothermic), if heat is decreased, ADP is increased according to Le Chatelier’s Principle. If a reactant is removed, the equilibrium will shift toward the reactants. In this reaction, ADP and heat are both reactants.

This question ties together the reaction for oxidation of glucose with several thermodynamic principles.

The first option, stating that there is an increase of entropy, is true. When balanced, we see that there is an increase in total number of molecules when the reaction completes; thus, there will be an increase in entropy.

The second option states that enthalpy, or , is less than than zero. This is also true, as it implies that this is an exothermic (or heat generating) reaction. The oxidation of glucose is how our bodies generate heat and keep homeostasis.

The third option, stating that Gibb's free energy is zero at equilibrium, is also true. Any equation will have  at equilibrium.

The last option, stating that oxygen is the reducing agent, is false. Oxygen is actually the oxidizing agent and is itself reduced.

Since glycolysis is an exothermic reaction (ΔH is negative), enthalpy does not increase. As for entropy, the reaction begins with a large molecule and ends with smaller molecules, therefore entropy increases.

ΔG and ΔH are both negative for glycolysis, meaning that the reaction is exergonic and exothermic.

Since we know that three ATP molecules per bond are required, 1,650 ATP molecules are needed. However, we also know that 17kJ/mol of additional energy is required for each bond. This energy must come from ATP molecules. The only choice that is greater than 1,650 is 1,962.

To solve this problem, first convert molecules to moles with Avogadro’s number, then convert moles to kJ using the ΔH of ATP hydrolysis. Lastly convert kJ to J.

As water freezes, heat is released. The H value in the equation must therefore be negative.  Also as water freezes, there is a local decline in the amount of disorder. The S, or entropy, term in the Gibbs equation refers to entropy of the system. Thus, this term must be negative as well.