Find the enthalpy of hydration for an iodide ion in the process Rbl - Rb+ +|-. Use Hsol= -Hlat+Hhydr. Hsol = 31.0 kJ/mol
Hlat = -630.0 kJ/mol
Hhydr of Rb+ = -315.0 kJ/mol

-661.0 kJ/mol
-661 kJ/mol
-284.0 kJ/mol
-284 kJ/mol

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2