Difference between stiffness and damping? I understand stiffness as the extent to which an object (e.g. a mass spring) resists deformation from an applied force, or the rigidity of an object. 
And I understand damping as the energy dissipative properties of an object/system (e.g. a mass spring) under cyclic stress. In the context of a spring/oscillator, damping is what causes a spring to eventually stop oscillating.
But I don't understand the real differences between the two. Couldn't a spring's stiffness explain it's damping? Why not?
 A: So given a spring with spring constant $k$ can one predict what dissipative force the spring will exert when extended?
The answer is "No" because they depend on different things.
The stiffness depends on the elasticity of the bonds between the atoms/molecules which make up the spring and the damping depends on the permanent distortion of the bonds which are not (simply) related.

The equation of motion of a mass at the end of a spring undergoing damped harmonic motion can be written as
$$m\ddot x + c \dot x + k x=0$$
where $x$ is the displacement at a time, $c$ is a constant relating the frictional force opposing the motion to the velocity $\dot x$ of the mass of the object $m$ and $k$ is the spring constant.
$c \dot x$ is often called the damping term and $kx$ the stiffness term.
As far as I know there is no simple (or even complex) relationship between $c$ and $k$ and indeed the $c$ in this context is to do with the air rather than the spring.
$c$ tells you something about the frictional force which intimately means that it is to do with the reduction of mechanical energy of the system and $k$ tells you something about how the spring resists changes to its length.
In the context of your question the parameter $c^2 - 4mk$ is important because it controls the type of damping which the spring-mass system undergoes and that parameter contains both $c$ and $k$.
So you could say that the spring constant $k$ does influence the type of damping that exists which in turn means that the spring constant also influences the rate at which mechanical energy of the spring-mass system is reduced.
A: Spring stiffness is not responsible for energy loss. Consider a spring with stiffness $k$ but no damping. The work done in compressing the spring by a distance $x$ will be stored as the potential energy(PE) of the spring. For a spring compressed by a distance $x$, the PE is given by $\frac12 k x^2$. This energy is not lost and can be used by letting the spring expand (Old mechanical pendulum clocks work on this principle, the energy of a compressed spring drives the clock  while the spring slowly expands). This is similar to how you have to do work to move a mass up against gravity, but the work you do is not lost. It is stored as the potential energy of the mass and can be reused. 
