Can someone help me with these two questions?

Thank you!

a.) Describe (give instructions) how you would prepare 7.00 L of 0.750 M K2CrO4 from a 2.25 M K2Cro4 stock solution

b.) A stock solution containing Ba2+ ions was prepared by dissolving 23.0 grams of pure barium metal in nitric acid and diluting to a final volume of 2.000 L. Calculate the concentrations of Ba2+ ions in the stock solution and the diluted solution if 125 mL of the stock solution was diluted to 500.0 mL

Again, thank you for your help!

you want to dilute it 2.25/.750 times, or 10/3 times. Which means, one part stock, 7/3 parts water.

What is one part? 7L/(10/3)=21/10=2.10liter

So, take 2.10 l of stock, add to 7/3 * 210 l water, stir and label. BTW, 7/3*2.10 l is 4.90 l

check: 4.90 + 2.1 l= 7liters

a.

The dilution formula is
L1 x M1 = L2 x M2
7 x 0.750 = L x 2.25
Solve for L of the 2.25M solution. Add that volume of a 2.25M solution to a flask and add enough water to make the final volume 7.00 L. This may be exactly what your prof wants and it COULD be a harder problem than that. WHY? Because you want to take exactly 2.333333333 L of the 2.25 M solution and add (7.000L-2.33333L) of water; BUT there are no EASY ways to measure out that exact volume to begin with nor to add exactly 4.6667 L H2O.

b.
mols Ba = grams/atomic mass = ?
M final solution = mols/2.000 L = ?
Then use the dilution formula for the M of the diluted solution.

a.)

To prepare 7.00 L of 0.750 M K2CrO4 from a 2.25 M K2CrO4 stock solution, you can use the dilution formula:

C1V1 = C2V2

where C1 is the concentration of the stock solution, V1 is the volume of the stock solution, C2 is the desired concentration, and V2 is the final volume.

In this case, we have:
C1 = 2.25 M (concentration of the stock solution)
V1 = ?
C2 = 0.750 M (desired concentration)
V2 = 7.00 L (final volume)

Rearranging the formula, we get:
V1 = (C2 * V2) / C1

Plugging in the values:
V1 = (0.750 M * 7.00 L) / 2.25 M
V1 = 2.333 L

So, you would need to measure 2.333 L of the 2.25 M K2CrO4 stock solution and then add enough solvent (like water) to make the final volume 7.00 L.

b.)

To calculate the concentrations of Ba2+ ions in the stock solution and the diluted solution, we need to determine the number of moles of Ba2+ ions present.

First, let's calculate the number of moles of Ba2+ in the stock solution:
From the given mass of pure barium metal (23.0 g), we can use the molar mass of barium (Ba) to find the number of moles.
Molar mass of Ba = 137.33 g/mol
Number of moles of Ba = mass / molar mass = 23.0 g / 137.33 g/mol ≈ 0.1677 mol

Since Ba2+ is a divalent ion, the number of moles of Ba2+ ions is equal to twice the moles of barium:
Number of moles of Ba2+ ions = 0.1677 mol × 2 = 0.3354 mol

Next, let's calculate the concentration of Ba2+ ions in the stock solution:
Concentration = number of moles / volume
Given that the final volume is 2.000 L, the concentration of Ba2+ ions in the stock solution is:
Concentration = 0.3354 mol / 2.000 L ≈ 0.1677 M

For the diluted solution, we can use the dilution formula as mentioned in the previous response:
C1V1 = C2V2

In this case:
C1 = 0.1677 M (concentration of the stock solution)
V1 = 125 mL = 0.125 L (volume of the stock solution)
C2 = ?
V2 = 500.0 mL = 0.500 L (final volume)

Rearranging the formula, we get:
C2 = (C1 * V1) / V2

Plugging in the values:
C2 = (0.1677 M * 0.125 L) / 0.500 L
C2 = 0.0419 M

Therefore, the concentration of Ba2+ ions in the diluted solution is approximately 0.0419 M.