To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:
N₂ + O₂ → 2 NO
Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.
1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present
Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>
Its been almost a year since i've been in chemistry, so sorry if its not right. Cs, Ba, Rb, Sr, Xe. I think that the levels of ionozation increase from the bottom left corner of the periodic table and increase as you go up and to the right
Your weight is a measure of ibs/pounds
Density = mass / volume
= 69g / 23 ml
= 3 g / ml.
Thus, the density of the sample is 3 grams per ml or 3g/ ml
Answer:
I think it is E part electron's are lost
