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NARA [144]
2 years ago
14

Calculate the hight of a column of liquid glycerol in meters required to exert the same pressure as 3.02 m if CCl4?

Chemistry
1 answer:
katovenus [111]2 years ago
4 0

Answer:

Approximately 3.81\; \rm m.

Explanation:

Look up the density \rho of carbon tetrachloride, \rm CCl_4, and glycerol:

  • Density of carbon tetrachloride: approximately 1.59\times 10^{3}\; \rm kg \cdot m^{-3}.
  • Density of glycerol: approximately 1.26\times 10^{3}\; \rm kg \cdot m^{-3}.

Let g denote the gravitational field strength. (Typically g \approx 9.81\; \rm N \cdot kg^{-1} near the surface of the earth.) For a column of liquid with a height of h, if the density of the liquid is \rho, the pressure at the bottom of the column would be:

P = \rho\cdot g \cdot h.

The pressure at the bottom of this carbon tetrachloride column would be:

\begin{aligned} P &= \rho \cdot g \cdot h \\ & \approx 1.59\times 10^{3} \; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1} \times 3.02 \; \rm m \approx 4.71 \times 10^{4} \; \rm N \cdot m^{-2} \end{aligned}.

Rearrange the equation P = \rho\cdot g \cdot h for h:

\displaystyle h = \frac{P}{\rho \cdot g}.

Apply this equation to calculate the height of the liquid glycerol column:

\begin{aligned}h &= \frac{P}{\rho \cdot g} \\ &\approx \frac{4.71 \times 10^{4}\; \rm N \cdot m^{-2}}{1.26 \times 10^{3}\; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1}} \approx 3.81\; \rm m\end{aligned}.

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<h3>What is a reducing agent?</h3>

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