1s^2 = filled
2s^2 = filled
2p^6 = filled
3s^1 = partially filled (need 2)
3 orbitals are completely filled.
1. 5 ethyl, 2 methyl octane
2. 1 ethyl, 2 methyl cyclopentane
3. 3,3,5,5- tetrafluoro heptane
4. 3,4-dimethyl hexene
5. 3,4-dimethyl cyclobutene
6. 3,5 diisopropyl cyclohexene
7. 3,3,4 trimethyl pentyne
8. 2,6 dibromo phenol
keep in mind that between 4-7, there could be #1 in front of the main name. for example with #4: 3,4-dimethyl-1- hexene. this honestly depends on the professor how he/she likes it. It is not necessary because if the number is not specified, it is assumed is #1
Answer: conductive means to take charge or be in charge
metals ar like gold iron and stuff like that
and the last but not least would be Filling something up
Explanation:
Answer:
FeSO2
Explanation:
Please see attached picture for full solution.
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
