Answer: The ion formed after the reduction of bromine is 
Explanation:
The electronic configuration of Sodium (Na) = ![[Ne]3s^1](https://tex.z-dn.net/?f=%5BNe%5D3s%5E1)
The electronic configuration of Bromine (Br) = ![[Ar]3d^{10}4s^24p^5](https://tex.z-dn.net/?f=%5BAr%5D3d%5E%7B10%7D4s%5E24p%5E5)
From the above configurations, Sodium ion will loose 1 electron in order to gain stable electronic configuration and that electron is accepted by the Bromine atom because it is 1 electron short of the stable electronic configuration.
(oxidation reaction)
(Reduction reaction)
Bromine atom is reduced to form
Reduction reactions are the reactions in which the element gain electrons.
Oxidation reactions are the reactions in which the element looses its electrons.
Answer: N2(g) + 3H2-> 2NH3(g) This is the balanced equation
Note the mole ratio between N2, H2 and NH3. It is 1 : 3 : 2 This will be important.
moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present
moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present
Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.
moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced
grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced
NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.
Explanation: here you go!! good luck! hope this helped
Answer:
11.29Kj
Explanation:
1. find moles of 33.8g of water
Molar mass of H2O: 18.02g/Mol
33.8/18.02= 1.88mols
2. find energy
1.88 x 6.02= 11.29Kj
The correct answer is B)wave mechanical. Hope that helps! If you have any questions about the answer i gave you please let me know!
