How fast would a bullet travel through liquid air? According to Wikipedia, a bullet from a .22 Long Rifle travels at a speed of 320 m/s and the density of liquid air is 870 kg/m3.
What equation would calculate the speed of a bullet fired into liquid air at a certain distance, say 10 meters?
The reason for the rather odd question is to see if a potential answer to a brain teaser is feasible.
Here's the teaser:

Alan fires a bullet from his gun and his friend Wade catches the bullet with his bare hands. The bullet does not touch anything but air after it leaves the gun and until it reaches Wade’s hand. Wade is uninjured. How does he do it?

 A: The speed of the bullet is not a constant.  The value given of 320 m/s most likely refers to the barrel exit velocity of the bullet.  For a quick and dirty approximation to how the bullet speed varies with distance you can simply use Newton's famous equation $$F = ma$$ With $F$ being the total force acting on the bullet, $m$ the bullet mass, and $a$ the acceleration.  In this case, the force acting on the bullet is primarily due to drag from the fluid.  Assuming supersonic flow, to accurately calculate the drag you would need to know something about the shape and equation of state of the fluid.  From there you could estimate the flow over the bullet and come up with an approximation to the force exerted on the bullet by the fluid.  However, how to do that is probably too much detail for the answer that you are looking for.  So, lets assume that the drag force acting on the bullet is given by $$F = \frac{1}{2}C_d \rho u^2A$$ Where $C_d$ is the drag coefficient measured from experimental data, $\rho$ is the fluid density, $u$ is the bullet's speed and $A$ is the frontal area of the bullet in the direction of travel.  Unfortunately, in reality, $C_d$ is not a constant value across changes in density and velocity.  However, lets assume that it is, then we get $$-\frac{1}{2}C_d \rho u^2A = ma$$ or with $$ a =  \frac{du}{dt} $$ we get the system of ordinary differential equations for position and speed $(x,u)$ $$-\frac{1}{2}C_d \rho u^2A = m\frac{du}{dt}$$ with $x$ the position of the bullet $$u = \frac{dx}{dt}$$  Or, since we are interested in the speed as a function of distance, using the fact that $$u\frac{du}{dx} = a$$ we have $$mu\frac{du}{dx} = -\frac{1}{2}C_d \rho u^2A $$ giving $$\frac{m}{u}\frac{du}{dx} = -\frac{1}{2}C_d \rho A $$  Integrating gives $$ Log(\frac{u}{u_0})= -\frac{C_d \rho A }{2m} x$$ or $$u(x) = u_0 e^{-\frac{C_d \rho A }{2m} x}$$  From that you can estimate how much difference a change in density makes.  A very rough estimate, but an estimate none the less.
