Let $\mathcal{A}$ be a $C^*$-algebra and $p\in\mathcal{A}^{**}$ be an open projection, that is, $p=p^*=p^2$ and $p\in\overline{(p\mathcal{A}^{**}p\cap\hat{\mathcal{A}})}^{\operatorname{w}^*}$, where $\hat{\mathcal{A}}$ is the canonical copy of $\mathcal{A}$ in $\mathcal{A}^{**}$ and the closure is taken in the weak$^*$ topology on $\mathcal{A}^{**}$.

Question: Is any orthogonal projection in $\mathcal{A}^{**}$ which is Murray-von Neumann equivalent to $p$ open?

Let $\mathcal{A}$ be a $C^*$-algebra and $p\in\mathcal{A}^{**}$ be an open projection, that is, $p=p^*=p^2$ and $p\in\overline{(p\mathcal{A}^{**}p\cap\hat{\mathcal{A}})}^{\operatorname{w}^*}$, where $\hat{\mathcal{A}}$ is the canonical copy of $\mathcal{A}$ in $\mathcal{A}^{**}$ and the closure is taken in the weak$^*$ topology on $\mathcal{A}^{**}$.

Question: Is every orthogonal projection in $\mathcal{A}^{**}$ which is Murray-von Neumann equivalent to $p$ open?