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?
Heat= latent heat of fusion+sensible heat+ latent heat of vapourization
=(79.7*5)+(5*100*1)+(540*5)
=3598.5 cal
The answer is D: Products combine to produce new reactants... Hope this helps! :)
Answer:In determining the energy of activation, why was it prudent to run the slowest trial done at room temperature in the hot water bath and the fastest trial done at room temperature in the cold water bath?
Explanation:
Answer:
The solution is 50 %wt
Explanation:
50% wt is a sort of concentration and means, that 50 g of solute (in this case, the potassium bromide) dissolved in 100 g of water.
It is the same to say, that there are 50g of KBr for every 100g of H₂O
