Bohr-haber cycles - a level chemistry

could someone check my answers, please? Thank you
I got a decimal for (ii) so not sure if I have gone wrong anywhere. Also, for (iii) it asked to calculate the lattice enthalpy but I calculated it in (ii) which confused me.

(i) The lower dotted line is the elements in their standard states, i.e. Mg_(s) + Br₂(l)

The upper dotted line is Mg²⁺(g) + 2Br(g)

(ii)
You need the enthalpy of hydration of bromide ions, i.e. Br^-(g) ==> Br^-(aq)
There is no lattice enthalpy step in the cycle.

You do need to add up (magnitude) all of the LHS and it should equal all of the RHS (just one way to do it.
The other way is to go around the cycle in the other direction to the step that you are trying to determine (bottom right)

186 + 525 + 2(112) + 148 + 736 + 1450 = (2 x dH(hydration) Br-) + 1926 + (2 x 325)

Rearrange

(2 x dH(hydration) Br-) = 693

Hydration enthalpy of bromide ions = -346.5 kJ (hydration is always exothermic)

is (iii) right? I got 1383

MgBr₂(s) ==> Mg²⁺(g) + 2Br^-(g)

As you are working with solutions in parts (i) and (ii) you should get from MgBr₂(s) to Mg²⁺(g) + 2Br^-(g) via MgBr₂(aq)

MgBr₂(s) ==> MgBr₂(aq) ............... ΔH = -186

Then add in the reverse of the hydration enthalpies of the two ions

Mg²⁺(aq) ==> Mg²⁺(g) ...........ΔH = +1926 kJ

2Br-(aq) ==> 2Br-(g) ........... ΔH = +693kJ

Add all three together = +2433 kJ

Isn't lattice enthalpy always negative?