Which is the electronic configuration for oxygen?
1
1s 2s22p!
182282 203
o 182 252 2p4
1s 2s 2p

1s22s22p4

Explanation:

Oxygen is the eighth element with a total of 8 electrons. In writing the electron configuration for oxygen the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for O go in the 2s orbital. The remaining four electrons will go in the 2p orbital. Therefore the O electron configuration will be 1s22s22p4.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

can you attach the