Probability and expected value of rolling two dice with a twist

Let this be a game with two phases.

In phase 1, roll two 6-sided dice and compute the difference between the rolls. Call this difference x.
In phase 2, roll x dice, and add up the total of the rolls. This is the payout in dollars of the game.

In the first phase of the game, x can take on any value 0 to 5. Compute the probability (as a fraction)
Now for the second phase of the game, for each value of x, what is the average payout (expected value) of the game?

Now I believe I got the right probabilities for 0 - 5, but I'm questioning if i'm doing the second part correctly.

When $x = 0$ I said $E(X) = 0$

And for $x = 1$ I said that $E(X) = \frac{10}{36} * 3.5$

Is this correct? My reasoning was the expected value for rolling a 6 sided die, would be 3.5 and since we have a $\frac{10}{36}$ of getting a value of 1 in the first part we need to multiply 3.5 by it.

Is this the correct logic? Thanks!!