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Chemistry

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nevsk [136]3 years ago

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I think it’s c hope it helps.

Andreyy893 years ago

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What else will happen.

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Determine the moles of H+ reacting with the metal based on the following experimental data. 10.00 mL of 1.00 M HCl solution is a

musickatia [10]

Answer:

Approximately $1.876 \times 10^{-3}\; \rm mol$ .

Explanation:

Convert both volumes to standard units (that is: liters.)

$10.00 \; \rm mL = 10.00 \times 10^{-3}\; \rm L = 1.000 \times 10^{-2}\; \rm L$ .
$27.08 \; \rm mL = 27.08 \times 10^{-3}\; \rm L = 2.708 \times 10^{-2}\; \rm L$ .

Number of moles of $\rm HCl$ initially present (in the $10.00\; \rm mL$ solution at $1.00\; \rm M$ .)

$n(\mathrm{HCl}, \, \text{initial}) = \displaystyle c \cdot V = 1.00\times 1.000 \times 10^{-2}= 1.000\times 10^{-2}\; \rm mol$ .

Number of moles of $\rm NaOH$ from the titration:

$n(\mathrm{NaOH}) = c \cdot V = 0.30 \times 2.708 \times 10^{-2} = 8.124 \times 10^{-3}\; \rm mol$ .

$\rm NaOH$ neutralizes $\rm HCl$ at a $1:1$ ratio:

$\rm HCl + NaOH \to NaCl + H_2O$ .

Hence, $n(\mathrm{HCl},\, \text{leftover}) = n(\mathrm{NaOH}) = 8.124 \times 10^{-3}\; \rm mol$ .

$\begin{aligned}&n(\mathrm{HCl},\, \text{consumed}) \\ =& n(\mathrm{HCl},\, \text{initial}) - n(\mathrm{HCl},\, \text{leftover}) \\ =& 1.000\times 10^{-2} - 8.124\times 10^{-3} \\ =& 1.876 \times 10^{-3}\; \rm mol\end{aligned}$ .