Suppose of sodium iodide is dissolved in of a aqueous solution of potassium carbonate. Calculate the final molarity of sodium cation in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Round your answer to significant digits.

0.2762M Na+ in the solution

Explanation:

2.07g of sodium iodide Is Dissolved In 50.ML Of A 0.30M...

To solve this question we need to find the moles of sodium iodide, NaI, that are the same than the moles of sodium cation, Na+. The volume in liters of the solution is 0.050L. The molarity is:

Moles NaI = Moles Na+ -Molar mass NaI: 149.89g/mol-

2.07g NaI * (1mol / 149.89g) = 0.01381 moles NaI = Moles Na+

Molarity:

0.01381 moles Na+ / 0.0500L =

0.2762M Na+ in the solution

i'm assuming that the correct answer is indeed the first one/ "a"..

the formula for the conversion of temperature from degree celsius to kelvin is:

the sample of water at is converted to kelvin temperature by using the above formula as:

on comparing the two samples of water at or we can say that at and at respectively. the difference between the temperatures of two samples of water is:

hence, one can say that the sample of water at is greater than the sample of water at by .