300.0 mL of a 0.335 M solution of NaI is diluted to 700.0 mL. What is the new concentration of the solution?

The new concentration of the solution is 0.143 M.

Explanation:

Given: = 300.0 mL,     = 0.335 M

= 700.0 mL,         = ?

Formula used is as follows.

Substitute values into the above formula as follows.

Thus, we can conclude that the new concentration of the solution is 0.143 M.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :