In one experiment 20.00 mL of 5.43*10^-3 M Fe(NO3)3 and 10.0 mL of 6M HNO3 was p

Question

In one experiment 20.00 mL of 5.43*10^-3 M Fe(NO3)3 and 10.0 mL of 6M HNO3 was placed in a 100.0 mL volumetric flask and dilute to the mark with distilled water. What was the concentration of iron(III) nitrate in the final solution? (This is a dillution problem)

If you could answer both that would be amazing. I just typed #4 so people could search for it easier. Thanks for your help!

4. In one experiment 20.00 mL of 5.43x103 M Fe(NO3)3 and 10.0 mL 6 M HNO3 was placed in a 100.0 mL volumetric flask and diluted to the mark with distilled water. What was the concentration of iron(III) nitrate in the final solution? (This is a dilution problem.) 5. For the solutions that you will prepare in Step 2 of Part I below, calculate the [FeSCN2] Presume that all of the SCN ions react. In Part I of the experiment, mol of SCN mol of FeSCN2+. Thus, the calculation of [FeSCN2 ] is: mol FeSN2 L of total solution. Record these values in the table below. Beaker number[FeSCN2 2 3 4 (blank) 0.00 M AT TO DISCuss IN YOUR CONCLUSION

Explanation / Answer

1. Before dilution -

Moles of Fe(NO3)3 = Molarity*Volume in L = 5.43*10-3M * 0.02L = 0.1086*10-3moles = 0.0001086moles

Moles of HNO3 = 6M*0.01L =0.06moles

After dilution-

total number of moles after mixing = 0.0001086moles+0.06moles = 0.060moles

Volume = 100mL = 0.1L

Molarity = 0.060moles/0.1L = 0.6M

Concentration in the final solution = 0.6M

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