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1 year ago

An empty vial weighs 55.32 g.

Chemistry

1 answer:

ziro4ka [17]1 year ago

7 0

The answer is: Volume of mercury = 9.63 $cm^3$ .

Density (d) is defined as mass of substance divided by its volume. Thus, if mass and density are known, then Volume can be determined.

What is the formula of volume in terms of density?

Let mass of substance be 'm' and volume be 'V'. Thus, as per the definition of density, it is expressed as-

$d =\frac{m}{V}$

Thus, volume can be expressed as-

$V = \frac{m}{d}$

Now, mass of empty vial= 55.32 g and the mass of (vial+ mercury) = 185.56 g.

Thus, mass of mercury = $185.56\ g- 55.32\g = 130.24\ g$

Thus, mass of mercury = 130.24 g and density of mercury = $13.53\ g/cm^3$ . Its volume is calculated as-

$V =\frac{130.24\ g}{13.53\ g/cm^3} = 9.63\ cm^3$

Hence, volume of mercury = 9.63 $cm^3$ .

To learn more about Density and Volume, visit:

brainly.com/question/1028144

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3 years ago

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .