Explain how you would determine the number of grams of Cu(NO3)2 that would be needed to make 1052 mL of a 2.50 M solution of Cu(NO3)2?

Explain how you would determine the mass of Na (s) needed to completely react with 25 mL of the Cu(NO3)2 solution created in #1, given the unbalanced equation below: __ Cu(NO3)2 (aq) + __ Na -> __ NaNO3 (aq) + ___ Cu (s)?

Explain how you would determine the new concentration of Cu(NO3)2 of 410 mL of distilled water was added to the solution of Cu(NO3)2 created in #1?

Explain how you would determine the freezing point of the solution created in #1, if the kբ of water is 1.853 °C/m. Assume the density of the solution is 1.00 g/mL

Explanation:

Question 1: Explain how you would determine the number of grams of Cu(NO₃)₂ that would be needed to make 1052 mL of a 2.50 M solution of Cu(NO₃)₂

1. Use the molarity (2.50M) and the volume in liters (1052mL = 1.052 liter) to calculate the number of moles of Cu(NO₃)₂, using the formula:

         ⇒ number of moles = molarity × volume in liters

2. Use the molar mass of Cu(NO₃)₂ to calculate the mass in grams using the formula:

Question 2: Explain how you would determine the mass of Na (s) needed to completely react with 25 mL of the Cu(NO₃)₂ solution created in #1, given the unbalanced equation below:

1. Balance the equation, which yields:

2. Calculate the number of moles of Cu(NO₃)₂ in 25 mL of solution, using the formula of molarity:

Subsititute with:

3. Use the theoretical mole ratio to calculate the number of moles of Na:

4. Use the atomic mass of Na to convert the number of moles into mass with the formula:

Question 3. Explain how you would determine the new concentration of Cu(NO₃)₂ if 410 mL of distilled water was added to the solution of Cu(NO₃)₂ created in #1?

1. Use the number of moles in the 1052 mL of solution, which was calculated in the first step of the question 1.

2. Add the two volumes: 410 mL + 1052 mL. This is the new volume of solution.

3. Use the formula of molarity:

Question 4. Explain how you would determine the freezing point of the solution created in #1, if the kf of water is 1.853 °C/m. Assume the density of the solution is 1.00 g/mL

1. Use the formula for the depression of the freezing point (a colligative property):

2. Kf is given: 1.853°C/m.

3. i is the van't Hoff factor.

This is used when the solute is ionic. It is the number of ions produced per mole of solute. For Cu(NO₃)₂ is its 3, because each mole of Cu(NO₃)₂ produces one mole of Cu⁺ and two moles of NO₃⁻.

4. m is the molality.

You must calculate the molality using the molarity and the density of the solution.

       m = number of moles of solute / kg of solvent.

Now that you have m, i, and Kf, just subsitute in the formula:

The freezing point of the solution created is the normal freezing point of water, 0ºC, less the depression of the freezing point of the solvent, ΔTf.

Therefore, you are done.

i have know clue i guess aluminum

i think lol

increased stability in the completion of fluorine's outer shell.