Chemistry: 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 added to a sample of Mg metal. The reaction goes to completion and all of the Mg metal is gone. 27.08 mL of 0.30 M NaOH solution is required to reach the end point when titrating the remaining HCl.

Determine the moles of H+ reacting with the metal

Ответ:

kmafromhannah2905

19.01.2022

Chemistry

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

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

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Ответ:

carligreen090

03.04.2021

Chemistry

The answer is option C

Explanation: The law of conservation of momentum states that when two bodies collide, in a closed or an isolated system, the total momentum before the collision is equal to the total momentum after the collision provided that no external force is acting on them.

That is, the system momentum remain the same.

Momentum, p = mv (measured in Kg. m/s)

Where

V = velocity of the body

m = mass of the body

For conservation of momentum

mv = constant

That is,

m1v1 = m2v2

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