2) what is the concentration of nacl in a solution prepared by diluting 56.98 ml of 0.5894 m stock
solution to a new volume of 1.20 l? (45 pts)

The NaCl concentration will be 0.03 M.

Explanation:

Given data:

Initial volume = V₁ = 56.98 mL (56.98/1000 = 0.05698 L)

Initial concentration = M₁= 0.5894 M

Final volume = V₂= 1.20 L

Final concentration = M₂= ?

Solution:

By diluting the solution volume of solution will increase while number of moles of solute remain the same.

Formula:

Initial concentration × Initial volume  = Final concentration × Final volume

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

M₂ = 0.5894 M × 0.05698 L / 1.20 L

M₂ = 0.0336 M /1.20  

M₂ = 0.03 M

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

wher is the rest of the question