Answer:
33.8 m
Explanation:
(a) We want to know the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO₃ by mass).
Step 1: Determine the mass of HNO₃ and water in 100 grams of solution.
Step 2: Convert the mass of HNO₃ to moles.
Step 3: Convert the mass of water to kilograms.
Step 4: Calculate the molality.
(b)
Step 1
In 100 g of solution, there are 68.0 g of HNO₃ and 100 g - 68.0 g = 32.0 g of water.
Step 2
The molar mass of HNO₃ is 63.01 g/mol. The moles corresponding to 68.0 g are:
68.0 g × (1 mol/63.01 g) = 1.08 mol
Step 3
The mass of water is 32.0 g = 0.0320 kg
Step 4
The molality of HNO₃ is:
m = moles of solute / kilograms of solvent
m = 1.08 mol / 0.0320 kg
m = 33.8 m
Answer:
When an experiment is replicanle it means it is able to be done again.
Explanation:
It is important because You need to see if you get different results or if you messed up the first time
Answer: The E for Silver-silver Chloride electrode = 0.287 V
Explanation:
Silver/Silver Chloride (Ag/AgCl) with a value for E° that is actually +0.222 V or approximately 0.23 V has the actual potential of the half-cell prepared in this way as +0.197 V vs SHE, (Standard Hydrogen Electrode) which arises because in addition to KCl, there is the contribuion of AgCl to the chloride activity, which isn't exactly unity.
Therefore, the E for the Ag/AgCl electrode would approximately equal 0.287 V
