Assuming the volumes are additive, what is the [Cl-]solution obtained by mixing 210mL of 0.600M KCl and 630ml of 0.385M MgCl2 ?

You have to assume each chloride is soluble. How many Cl in each?

Add the Cl, you have the volume, so molar concentration is moles Cl per liter.

i don't understand

How many millicuries should be given to a 59.0g- patient

To find the concentration of chloride ions ([Cl-]) in the final solution obtained by mixing KCl and MgCl2, we need to use the concept of stoichiometry and the principle of volumes being additive. Here's how you can calculate it:

Step 1: Calculate the number of moles of chloride ions in each solution.
First, let's start by calculating the number of moles of chloride ions in the KCl solution.
- Concentration of KCl solution (C1) = 0.600 M
- Volume of KCl solution (V1) = 210 mL = 0.210 L (converted to liters)
We can use the formula:

moles = concentration (M) x volume (L)

moles of chloride ions in KCl solution = C1 x V1
= 0.600 M x 0.210 L
= 0.126 moles

Next, let's calculate the number of moles of chloride ions in the MgCl2 solution.
- Concentration of MgCl2 solution (C2) = 0.385 M
- Volume of MgCl2 solution (V2) = 630 mL = 0.630 L (converted to liters)

moles of chloride ions in MgCl2 solution = C2 x V2
= 0.385 M x 0.630 L
= 0.24255 moles

Step 2: Determine the total moles of chloride ions in the final solution.
Since volumes are additive, we can sum up the moles of chloride ions from both solutions:

total moles of chloride ions = moles from KCl + moles from MgCl2
= 0.126 moles + 0.24255 moles
= 0.36855 moles

Step 3: Calculate the final volume of the solution.
To calculate the final volume, we need to add the volumes of KCl and MgCl2 solutions:

final volume of solution = V1 + V2
= 0.210 L + 0.630 L
= 0.840 L

Step 4: Calculate the concentration of chloride ions in the final solution.
Using the formula:

concentration (M) = moles / volume (L)

concentration of chloride ions in the final solution = total moles / final volume
= 0.36855 moles / 0.840 L
= 0.4383 M

Therefore, the [Cl-]solution obtained by mixing 210 mL of 0.600 M KCl and 630 mL of 0.385 M MgCl2 is approximately 0.4383 M.