8.00 mL of 6.00 x 10-3 M Fe(NO3)3 (6.00 times 10 to the minus 3rd power M Fe (NO3)3) is added to 6.00 mL of 5.00 x 103 M KSCN (5.00 times 10 to the minus 3rd power M KSCN) along with 3.00 mL of water. The concentration of [F e(SCN)2+] was found to be 1.00 x 1o-4 M (1.00 times 10 to the minus 4th power M) at equilibrium.How many moles of [F e(SCN)2+] are present in the solution? Express your answer as a decimal number (no exponents).

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8.00 mL of 6.00 x 10-3 M
Fe(NO3)3 (6.00 times 10 to the minus 3rd
X
power M Fe (NO3)3) is added to 6.00 mL
of 5.00 x 10-3 M KSCN (5.00 times 10 to
the minus 3rd power M KSCN) along with
3.00 mL of water. The concentration of [F
e(SCN)2+] was found to be 1.00 x 10-4 M
(1.00 times 10 to the minus 4th power M)
at equilibrium.How many moles of [F
e(SCN)2*] are present in the solution?
Express your answer as a decimal number
(no exponents).
Your Answer:
Answer
units
Transcribed Image Text:8.00 mL of 6.00 x 10-3 M Fe(NO3)3 (6.00 times 10 to the minus 3rd X power M Fe (NO3)3) is added to 6.00 mL of 5.00 x 10-3 M KSCN (5.00 times 10 to the minus 3rd power M KSCN) along with 3.00 mL of water. The concentration of [F e(SCN)2+] was found to be 1.00 x 10-4 M (1.00 times 10 to the minus 4th power M) at equilibrium.How many moles of [F e(SCN)2*] are present in the solution? Express your answer as a decimal number (no exponents). Your Answer: Answer units
Expert Solution
Step 1

Given, 

Volume of Fe(NO3)3 = 8.00 mL 

Molarity of Fe(NO3)3 = 6.00 x 10-3

Volume of KSCN = 6.00 mL 

Molarity of KSCN = 5.00 x 10-3

Volume of water = 3.00 mL 

Molarity of [Fe(SCN)2+] at equilibrium stage = 1.00 x 10-4 M

Number of moles of [Fe(SCN)2+] in the solution = ?

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