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evablogger [386]
4 years ago
12

1.00 mL of a 250.0 µM solution of KCl is diluted to 50.0 mL. What is concentration of this solution?

Chemistry
1 answer:
Sever21 [200]4 years ago
3 0

Answer:

5.00 µM

Explanation:

Given data

  • Initial concentration (C₁): 250.0 µM
  • Initial volume (V₁): 1.00 mL
  • Final concentration (C₂): ?
  • Final volume (V₂): 50.0 mL

We can find the final concentration using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 250.0 µM × 1.00 mL / 50.0 mL

C₂ = 5.00 µM

The concentration of the diluted solution is 5.00 µM.

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[Ba^2+] = 0.160 M

Explanation:

First, let's calculate the moles of each reactant with the following expression:

n = M * V

moles of K2CO3 = 0.02 x 0.200 = 0.004 moles

moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles

Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.

Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3

As you can see, 0.04 moles of  K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of

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