What property holds for $\mathbb{R}$, but not for $\mathbb{Q}$ when we can use just multiplication and equal sign?

Question

Formally, let $L = \{\cdot \}$ be language with equality, let $Q$ and $R$ be models with universe $\mathbb{Q}$ and $\mathbb{R}$ and let for both models be $\cdot$ interpreted as multiplication. For which sentence $\phi$ holds $Q \not\models \phi$, but $R \models \phi$?

Note: this question was taken from my logic course textbook. There was no answer, thus I'm asking here.

Answer 1

Every real number has a cube root.

Answer 2

$$\exists x\colon \forall y\colon \exists z\colon (z\cdot z=y\lor (z\cdot z)\cdot x=y)$$

Namely, in $\Bbb R$, we can pick $x=-1$ and observe that one of $\pm y$ is non-negative and thereby a square.

In $\Bbb Q$ no such $x$ exists: Clearly picking $x=0$ won't work for $y=2$, and for any other $x$, pick a prime $p$ not occurring in either numerator or denominator of $x$; then neither $p$ nor $\frac px$ is a square.