What are the formulas for these elements

1. Arsenic (V) chloride
2. Antimony (V) Iodide
3. Bismuth (lll) bromide
4. Uranium (Vl) fluoride
5. Manganese (ll) oxide
6. Manganese (lll) oxide
7. Manganese (IV) oxide
8. Manganese (V) oxide
9. Manganese (Vll) oxide
10. Copper (l) sulfide
11. Copper (ll) cyanide
12. Iron (lll) hydroxide

Thank you (^_^)

As(Cl)5
Sb(I)5
Bi(Br)3
U(F)6
MnO
Mn2(O)3
Mn(O)2
Mn2(O)5
Mn2(O)7
Cu2S
Cu(CN)2
Fe(OH)3

the AP chemistry guy is completely wrong, he has wrong symbols for manganese and uranium

First off, the listed substances are not elements, but compounds, or, more specifically, metal salts. Second, here's how to solve this kind of problem:

In the given substance names, the term in front of the brackets names a metal; the roman number in parentheses indicates the oxidation state of that metal, and behind the brackets names the anion in the salt. Now, the anions have fixed charges: halogen ions have a charge of (-), as do hydroxide ions, nitrate ions and pseudohalogenides such as cyanide. Oxygen as an anion has a charge of (2-), as does sulfur, if you ever encounter that, of sulfate anions SO4(2-).
The oxidation state in brackets is equal to the number of electrons removed from the metal in its elemental form, which leaves a corresponding positive charge on the metal; e.g. iron (III) is Fe(3+), copper(I) is Cu(+), copper(II) is Cu(2+), etc.

What you have to do to find the stoichiometry of the compounds is simply match these numbers - charge of the metal cation and charge of the anion - so you get a neutral compound with no overall charge. I'll do one of the exercises as an example:
In manganese(III) oxide, the manganese cation obviously has a charge of +3, so it's Mn(3+). Each oxide anion has a charge of (-2), so you need 1.5 oxide anions to compensate for the charge of the metal. Now, MnO(1.5) isn't very beautiful, because we prefer integer numbers for the indices, so we just multiply the whole thing by 2 and get Mn2O3, which is the formula of your compound. All the other exercises can be solved in the same manner.

What you do is write the chem symbol for each element. Then know the charge of the second. For example Cloride has a -1 charge. For Arsenic to have a charge of 5, the charges of Cl have to add up to 5 (you need the overall charge of the compound to be zero). That means you need 5 Cl and it becomes ArCl5 Ar = +5 Cl = -1 times 5 +5 -5= 0

1. AsCl5
2. (whatever the chem sybol for Antimony is) I5
3. (W/e the chem symbol for Bismuth is) Br3
4. UrF6
5. MgO2
6. MgO3
7.MgO4
8. MgO5
9.MgO7
10. CuS
11. Cu(CN)2
12. Fe(OH)3