With the information given most likely in order to find the partial pressure of the gas produced you have to subtract the total air pressure in the collection flask by the atmospheric pressure since you assume that the flask started with atmospheric pressure when it was sealed and then the gas was added as the reaction took place increasing the pressure.
1.44atm-0.95 atm=0.49atm
In the crystallization process the solid compound is dissolved in the solvent at elevated temperature and the crystallize product obtained by slow cooling of the solution. Here the solubility of acetanilide at 100°C is 1g per 20mL of water. Thus to dissolve 500mg of acetanilide at high temperature that is 100°C we need 10mL of water.
Now at 25°C after the re-crystallization there will be some amount of dissolve acetanilide. Which can be calculated as- 185mL of water is needed to dissolve 1g or 1000mg of acetanilide at 25°C. Thus in 10mL of water there will be 
gmg of acetanilide.
Because of the sun reflects onto the moon and makes it brighter.
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
