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3 years ago

7

You are determining the density of an unknown solid. You figure out the volume by water displacement. Then you weigh the solid,

but forget to dry it off first. The calculated density will be larger, smaller, or the correct value? Why?

1 answer:

WINSTONCH [101]3 years ago

8 0

Answer:

The calculated density will be larger

Explanation:

The calculated density will be <u>larger</u>. Because, the volume is taken accurately, by the water displacement method. But, when we the took the mass, the water was present on the unknown solid. So, the mass of that water was added to the original mass of the solid. Hence, the mass measured was larger than the original mass. We, know from the formula of density that density is directly proportional to the mass of the object.

Density = Mass/Volume

Hence, the larger measured mass means the larger value of density.

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light energy is converted to chemical energy during photosynthesis.

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Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a cer

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is $K_c= 2.8*10^{-4}$

Explanation:

From the question we are told that

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The molar mass of $H_{2}_{(g)}$ is a constant with value of $H_2 = 2g/mol$

The molar mass of $H_{2}O$ is a constant with value of $H_2O = 18g/mol$

Generally the number of moles is mathematically given as

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$No \ of\ moles = \frac{3.63}{2}$

$= 1.815 \ mols$

For $H_{2}O$

$No \ of\ moles = \frac{2.13}{18}$

$= 0.12 \ mols$

Generally the concentration of a compound is mathematicallyrepresented as

$Concentration = \frac{No \ of \ moles }{Volume }$

For $Fe_{2} O_{3}_{(s)}$

$Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}$

$= 4.10*10^{-3}M$

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$Concentration[H_2] = \frac{1.815}{5.4}$

$= 0.336M$

For $H_{2}O$

$Concentration [H_2O] = \frac{0.12}{5.4}$

$= 0.022M$

The equilibrium constant is mathematically represented as

$K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}$

Considering $H_2O \ for \ product$

And $H_2 \ for \ reactant$

At equilibrium the

$K_c = \frac{0.022}{0.336}$

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