If 1 kg of mass is converted to energy and used to accelerate a second 1 kg mass, what would be the final velocity? 
A bit rusty, tried using $E=mc^2$ to figure out how much kinetic energy would 1kg of mass convert to, and then work backwards to figure out what would be the final velocity of a 1kg mass having kinetic energy equivalent to 1kg mass as kinetic energy.

Not sure how correct this is but here it goes:
$E = 1kg\times c^2$ is the (kinetic) energy equivalent of 1kg mass.
I am hazy about the next step, 1kg mass having kinetic energy of $1kg\times c^2$ w.r.t. to a stationary point would appear to be moving at what (relativistic) speed?  (not sure even if the first step is correct)
 A: If you take a mass $m=1\:\rm kg$ and turn it straight into energy, by whatever means, then indeed you will get an energy $E=mc^2$ (a.k.a. its rest energy) out of it. 
If you then use all of that energy to accelerate a second $1\:\rm kg$ mass, that second mass will definitely be accelerated to relativistic domains. This means that you cannot use the old relationship $E_\mathrm{kin}=\frac12 mv^2$ between the particle's velocity $v$ relative to its initial rest frame and its kinetic energy $E_\mathrm{kin}$. Instead, you need to use the relativistic version for the total energy, which reads
$$
E=\gamma mc^2 = \frac{mc^2}{\sqrt{1-v^2/c^2}},
$$
where $\gamma=1/\sqrt{1-v^2/c^2}$ is known as the Lorentz factor and reduces to $1$ at zero velocity (thereby giving the rest energy $E=mc^2$).
In your case, the second mass already has its rest energy, and you're doubling this, so
$$
E=\gamma mc^2 = 2mc^2,
$$
which then gives you the equation
$$\frac{1}{\sqrt{1-v^2/c^2}}=2$$
that you can solve for $v$ to give $v=\frac{\sqrt{3}}{2}c\approx 0.8660\,c$.
