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1 year ago

14

When a 0.764 g almond is burned, it heats 225 mL of water from 14.8oC to 18.4oC. Calculate the heat of combustion of the almond

in kcal/g

1 answer:

Soloha48 [4]1 year ago

8 0

It’s 1.071 BWIWNWOWLQQMKWBWJWNWKWNQKKQJWKWKKQKWKAMQNMWNWNWKWWKWKWNWKW

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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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Answer:

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so the moles of HCl reacted with NaOH

1.5 × 10^ -3

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