Answer:
Moles=2.04×10^(-6)
Explanation:
No. Of moles=no. Of particles/ Avogadro's no
(Where no. Of particles may be atoms molecules or compounds)
Moles=1.23×10^18/6.022×10^23
Moles=0.204×10^(-5)
Moles=2.04×10^(-6)
He used a tube of mercury and marked the height of the mercury when placed in an ice bath as 0 degrees celsius, when he placed the tube in a boiling, he marked the height of mercury and called that 100 degrees celsius, he marked it linearly between 0-100 degrees celsius
Delta H of solution = -Lattice Energy + Hydration
<span>Delta H of solution=- (-730)+(-793) </span>
<span>Delta H of solution= -63kJ/mol </span>
<span>Now we find moles of LiI: </span>
<span>10gLiI/133.85g=.075moles </span>
<span>multiply moles to the delta H of solution to cross cancel moles. .75moles x -64kJ/mol =4.7</span>
Ammonia is the base out of the 4