If a specific element remains at charge $Q$, but there is a charge flow, contradiction with $dQ/dt = 0$? Say I have an element of a circuit with charge Q and a current flows through it. However, there is a contradiction since: dQ/dt = i = 0, so current = 0. Where's the mistake?
 A: This is a case where it's helpful to dig a little deeper into the variable names.  What you actually have is two different concepts both being referred to by Q.
Current is dQ/dt.  However, if we dig a little more, what current actually is is a flow of charge through a surface.  For example, the current going through a wire is the flux of charge going through a plane perpendicular to the wire.
When we talk about an element of a circuit having a charge, we are talking about a charge over a volume -- the volume of the element.  This is a different kind of charge measurement.  It happens to also use the variable Q when you see the equations rendered in their most typical form, but its a slightly different Q.
Let's stick to wires for an example.  Let's even make things easier and start with a charge on our wire of 0.  This is really really normal =)   We can hook this wire up to a circuit that runs current through it.
Our charge on the wire must remain 0, because we're not adding any electrostatic charge to it.  The current going into the wire is nonzero.  At any point in time, if we measure the charge in the volume of the wire, it remains basically 0 (there's some imperfections, but we can ignore them).  However, if we measure the charge flowing through any perpendicular plane, we see that there is current flowing.
What does this mean?  In the simplest terms, it means current in = current out.  As long as the same amount of charge is going into the element as is leaving the element, the charge on the element itself remains constant.
A: Use an analogy with water. Replace:


*

*circuit $\mapsto$ river system

*charge $\mapsto$ water height

*current $\mapsto$ flow of water.



If a specific element of a circuit remains at charge $Q$ but there is a
  charge flow,  is this in contradiction with $\frac{dQ}{dt} = 0$?

$\mapsto$

If a specific element of a river system remains at water height $h$ but there is a
  flow of water,  is this in contradiction with $\frac{dh}{dt} = 0$?

Obviously not! Obviously in this case there is no real relation between $\frac{dh}{dt}$ and river flow. $I=dq/dt$ is not satisfied!
Throw a capacitor/water reservoir into the mix. If $h$ is the height of a reservoir, then $dh/dt$ is the current flowing out of the reservoir. (technically height times area, but that doesn't matter here). Now $h$ might refer to the height of a reservoir or it might refer to the height of a river. If it refers to the height of a reservoir, $I=dh/dt$ is correct. If it refers to the height of a river (with unchanging height) , $I=dh/dt$ is incorrect. Here the reservoir of charge is the capacitor, and again the river is the wire.
Now what IS true, and what summarizes the idea that you're getting at, is the continuity equation. If $\vec{j}$ is a vector representing current, and $\rho$ is the charge density, then:
$$\frac{d\rho}{dt}=-\nabla \cdot \vec{j}$$
This is the statement that if more current flows into an area than out of it, the charge density must be changing.
Equivalently, using the divergence theorem, if $Q$ is the charge in a volume $V$ and $S$ is the surface enclosing $V$, then:
$$\frac{dQ}{dt}=-\oint_S \vec{j}\cdot d\vec{a}$$
The key thing here is that the rate of change of charge is the current flowing in minus the current flowing out. You only considered the current flowing in. In your circuit element, $I$ charge is flowing in to an element, and $I$ is flowing out, giving $dq/dt=0$ and no contradiction.
A: Think of the charge in the element as a volume of water in a can.
Above the can there is a tap which is open and water is entering the can.
At the bottom of the can is a hole out of which water is coming out from the can.
If the rate at which water is entering the can (flow of charge into element = electric current into element) is equal to the rate at which water is leaving the can (flow of charge out of element = electric current out of element) then the volume of water in the can (amount of charge in the element) is constant.
