Hadamard gate on entangled qubit
If you mean result in terms of state, then you can compute this by applying the matrix $I \otimes H$ to the Bell state.

If you mean what happens operationally, then observe that applying a hadamard gate on each qubit of the Bell state results in no change -- the resulting state is again the Bell state. So, if you apply two hadamard gates in parallel or none at all and then you measure both qubits, you will get perfect correlations between the measurement outcomes (measurement outcomes on the qubits are guaranteed to be equal). However, if you apply a hadamard gate to either the second qubit or the first qubit and then measure both qubits, then the measurement outcomes  will be completely independent of each other, that is, the measurement result of the first qubit can be $0$ or $1$ with $50\%$ probability and the measurement result of the second qubit can be $0$ or $1$ with $50\%$ probability, regardless of what was seen in the other measurement.



Algebra:

Given the state $|00\rangle + |11\rangle$, you can not compute the Hadamard gate by factoring the state and applying a Hadamard to only one qubit, as the state is entangled and cannot be factored into two independent qubits. You can, however, apply the result of $I \otimes H$, which is the operation of applying Identity gate on the first qubit and the Hadamard gate on the second. The resulting operation is, with scaling factor $s$,

$$ I \otimes H = \left[\begin{matrix}
1 &amp; 0 \\
0 &amp; 1
\end{matrix}\right]
\otimes
\frac{1}{\sqrt{2}}
\left[\begin{matrix}
1 &amp; 1 \\
1 &amp; -1
\end{matrix}\right]
= s\left[\begin{matrix}
1 &amp; 1 &amp; 0 &amp; 0 \\
1 &amp; -1 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 1 &amp; 1 \\
0 &amp; 0 &amp; 1 &amp; -1 \\
\end{matrix}\right]$$

For more details on this, see the question How to apply a 1-qubit gate to a single qubit from an entangled pair?.

Now, you can pass your entangled state, $
\left[\begin{matrix}
\frac{1}{\sqrt{2}} &amp; 0 &amp; 0 &amp; \frac{1}{\sqrt{2}}\\
\end{matrix}\right]^{T}
$ for $\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle$, through the gate and get
$$
s\left[\begin{matrix}
1 &amp; 1 &amp; 0 &amp; 0 \\
1 &amp; -1 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 1 &amp; 1 \\
0 &amp; 0 &amp; 1 &amp; -1 \\
\end{matrix}\right]
\left[\begin{matrix}
\frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}}\\
\end{matrix}\right]
= 
\frac{1}{2}\left[\begin{matrix}
1 \\ 1 \\ 1 \\ -1\\
\end{matrix}\right]
$$
Which is the state 
$$|00\rangle + |01\rangle + |10\rangle - |11\rangle.$$
And measurement of the first qubit or the second qubit would be 0 or 1 with equal probability, and give no information on the state of the other qubit.