The answer is: (5696 J) / (155 g) / (40.0 - 25.0)°C = 2.45 J/g·°C
We can use a variety of formulas to determine our answers here.
Our formula for pOH is -log(mol), and we can plug it in as -log(0.010). Take note that OH- is a base, not an acid.
So, the pOH of OH- is 2.
To find pH we can set up this simple equation:
pH + pOH = 14
All we need to do is subtract 2 from 14, therefore the pH is 12.
This makes sense since acids range in the pH of 1-6, and we are dealing with a base. Hope I could help!
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
