I would say when an atom has its valence electron shell filled like a noble gas has, it is not easily changed.
I’m not entirely sure of what you’re asking, but if you’re talking about bonding then it would be an ionic bond that is not easily changed.
First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:
-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
x = 177.75*100/44.4 = 400.33
The boiling point of water in ∘a would be 400.33∘a.
Answer:
Mole fraction O₂= 0.43
Explanation:
Mole fraction is the moles of gas/ total moles.
Let's determine the moles of each:
Moles O₂ → 15.1 g / 16 g/mol = 0.94
Moles N₂ → 8.19 g / 14 g/mol = 0.013
Moles H₂ → 2.46 / 2 g/mol = 1.23
Total moles = 2.183
Mole fraction O₂= 0.94 / 2.183 → 0.43
Answer:
Explanation:
To calculate the cell potential we use the relation:
Eº cell = Eº oxidation + Eº reduction
Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .
So the species that is going to be oxidized is the Aluminium, and therefore:
Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V
Equally valid is to write the equation as:
Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species
These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.
The activity series goes top to bottom, most active to least active elements, going: Li, K, Ba, Sr, Ca, Na, Mg, Mn, Zn, Fe, Cd, Co, Ni, Sn, Pb, H, Cu, Ag, Hg, Au.
Thus, your list of metals would go from most reactive to least reactive: Li, K, Mg, Zn, Fe, Cu, Au
