Answered

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It took 14.50 mL of 0.455M NaOH to fully neutralize 12.0mL of HCl. What is the concentration of the HCl?
HCl + NaOH \rightarrow→ NaCl + H2O

Sagot :

Answer:

0.550 M HCl

Explanation:

M1V1 = M2V2

M1 = 0.455 M NaOH

V1 = 14.50 mL NaOH

M2 = ?

V2 = 12.0 mL HCl

Solve for M2 --> M2 = M1V1/V2

M2 = (0.455 M)(14.50 mL) / (12.0 mL) = 0.550 M HCl

Cricetus

Answer:

The appropriate answer is "0.549 M".

Explanation:

The given values are:

N₁ = 14.50 mL

V₁ = 0.455 M

N₂ = 12 mL

Let

V₂ = C = ?

As we know,

⇒  [tex]N_1\times V_1=N_2\times V_2[/tex]

On substituting the values, we get

⇒  [tex]14.50\times 0.455 = 12\times C[/tex]

⇒            [tex]6.5975=12\times C[/tex]

⇒                   [tex]C=\frac{6.5975}{12}[/tex]

⇒                       [tex]=0.549 \ M[/tex]

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