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?
Answer:
Electrons
Explanation:
In an atom there would be three subatomic particles: Neutrons, electrons, protons. The smallest and lightest in terms of mass is electrons. This is because the nucleus is comprised of the protons and the neutrons, these have a greater mass than electrons as electrons has very little mass that can considered to be 0.
Okay so the answer will be.
The total amount of heat required is the sum of all the sensible heat and latent heats involved in bringing the ice to a desired temperature and state. The latent heat of fusion and vaporization of water 333.55 J/g and 2260 J/g, respectively. Solving for the total amount of heat,
total amount of heat = 13.0 g (2.09 J/gC)(12) + 13(333.55 J/g) + 13.0 g (4.18 J/gC)(100 - 0) + (13.0 g)(2260 J/g) + (13 g)(2.01 J/g)(113-100)
= 39815.88 J
= 39.82 kJ
