What do units like joule * seconds imply? I can easily understand what divisive units imply, but not what multiplicative units imply.  
What I mean is, when I read "$12 \:\mathrm{eggs/carton}$", I mentally convert it to, "There are 12 eggs for each carton". I get that.
But, when I see units like joules * seconds, $\:\mathrm{J \cdot s}$, it really bothers me for some reason. I'm not sure what it's saying. It's not saying the amount of joules expended over a certain time-frame, otherwise it would be joules/second, wouldn't it? So, when I read units like joules * second, I try to relate it to my egg-carton analogy, and it comes out like this:
"There are 12 eggs for each reciprocal of a carton."  
It just doesn't make any sense for me. Is this a problem for anybody else? Should this be on math SE? Is this even a valid question or am I just asking why $2 + 2 = 4$? 
 A: This answer will make more sense if you've had some basic Calculus (derivatives and integrals).
Draw a graph of some random function $E = f(t)$, where the horizontal axis $t$ is time (in seconds), and the vertical axis $E$ is energy (in joules). It doesn't matter what shape the graph has (but let's stay in the upper-right quadrant, so $t$ and $E$ are positive quantities).


*

*At any given time, $t$, you can find the energy, $E$, at that time by tracing a vertical line from the origin to the function. The distance of that line represents the number of joules at a given instant. 

*Now look at the slope of the graph at that point. The angle of that slope represents the instantaneous change in the energy over time. It is the rate of change, in joules/second. In calculus terms, this is a derivative.

*Now pick two times along the $t$ axis ($t_1$ and $t_2$). Find the area enclosed between $t_1$ and $t_2$, and above the $t$-axis, and under the function $f(t)$. This enclosed area represent the sum of all the energy at each instant, over a period of time (from $t_1$ to $t_2$). This area is measured in joule*seconds. In calculus terms, this is an integral.
Thus, while divided units represents a rate of change in one variable relative to another, multiplied units represent an accumulation of one variable as another is changing.
Seen this way, you can view distance as merely the accumulated effect of a velocity over time (e.g. if velocity is held constant, you accumulate distance linearly). Similarly, the area of a shape can be seen as an accumulation of distance measured in one direction as distance in another direction changes.
A: To understand what Joules $\times$ seconds means, one must understand the terms by themselves.  The Joule is a unit of energy, and the second is a unit of time.  So, what 500 J $\times$ s means, is that there is a source of energy that I can use in the following manner: 500 Joules for 1 second; or 250 Joules for 2 seconds; or 100 Joules for 5 seconds; or 1 Joule for 500 seconds; or in any combination of numbers who's product is 500!
A: Seems valid to me...
I think the interpretation of your example could be that you are doing something with some joules, for some period of time, and that the product of the number of joules and the number of seconds has a certain value, and the units would be... joule-seconds!
I can picture the operator of an energy storage facility quoting a price for storing energy.  Storing $10,000\text{ joules}$ for $400\text{ seconds}$  would cost a certain amount.  Storing double the energy for half the time would use the same resources, and cost the same.
The most common example, of course, is the use of $\text { kilowatt-hour}$ as a unit of energy.  In this case, you are using the rate of energy output multiplied by the time over which the output exists, to represent the total energy.
It has been common for some time to represent the amount of work in a job in $\text{ man-hours}$.  The more people you can throw at the project, the sooner it will be done...
A less obvious unit is the measurement of the length of, say, a sailboat, in $\text{ knot-seconds}$...
A: Reciprocal cartons for storing eggs might make little sense, reciprocal seconds (Hz) do. 
Joules times seconds (J.s) is equivalent to Joules per Hertz (J/Hz). It denotes angular momentum, the amount of energy per unit of rotational frequency. 
Similarly, reciprocal meters (wave numbers) also make perfect sense. So units like Newton times meter (N.m) are also encountered in physics. 
Alternatively, one can view units like N.m (= J) as an accumulation: when I push an object with a given force, the amount of work performed increases proportional to the path length. 
A: Building on the answer that Johannes gave:
Imagine you have a broad band source of energy, transmitting on multiple frequencies. It might be the sun, or it might be a RF transmitter, or ...
Now we are going to measure the energy transmitted in each wave band for a certain time, and plot this as a function of the frequency observed. This leads to a plot with Hz along the X axis and J/Hz (the energy density per unit frequency, that is the energy measured divided by the bandwidth of my receiver) along the Y axis. 
And boom - out of nowhere I have a unit J/Hz (or J sec) which makes physical sense...
