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3 years ago

15

in the laboratory, a group of students was assigned to determine the density of an unknown liquid. They used a buret to find a v

olume of 2.04 mL. The mass was determined on an analytical balance to be 2.260 g. How should they report the density of the liquid?

1 answer:

Daniel [21]3 years ago

4 0

They should report the density as 1.11 g/L .

Density = mass/volume = 2.260g/2.04 mL

My calculator says the density <em>1.107 843 137 g/mL</em>

However, the answer can have <em>no more</em> significant figures than are in the number with the <em>fewest </em>significant figures.

The volume measurement has only three significant figures, so we must round off the density to three significant figures.

We drop all the digits after the zero.

The digit to be dropped is 7, so we <em>round up </em>the last significant figure of the answer.

1.10<u>7 843 137</u> → 1.11

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The intermediate balanced chemical reaction will be,

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(1) $6CO_2(g)+3H_2O(l)\rightarrow C_6H_6(g)+\frac{15}{2}O_2(g)$

$\Delta H_1=-\frac{-6271kJ/mole}{2}=3135.5kJ/mol$

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$\Delta H_2=6\times (-393.5kJ/mole)=-2361kJ/mol$

(3) $3H_2(g)+\frac{3}{2}O_2(g)\rightarrow 3H_2O(l)$

$\Delta H_3=\farc{3}{2}\times (-483.6kJ/mole)=-725.4kJ/mol$

The expression for enthalpy of formation of $C_6H_6$ will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(3135.5kJ/mole)+(-2361kJ/mole)+(-725.4kJ/mole)$

$\Delta H=49.1kJ/mole$

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Here equal moles of $NH_{3}$ and $H_{2}S$ is formed.

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