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Oksana_A [137]
4 years ago
6

Significant digits of 1.0 x 10^-2

Chemistry
1 answer:
son4ous [18]4 years ago
8 0

There's two sig figs: the 1 and the zero. The 10^-2 contributes no sig figs.

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Visible light, which has a wavelength of between 380 nm and 760 nm, is usually
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Here is the reaction of carbamic acid and ammonia to form an amide and water. There is a scheme of a reversible reaction where c
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Iron(III) oxide and hydrogen react to form iron and water, like this: (s)(g)(s)(g) At a certain temperature, a chemist finds tha
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The question is incomplete, here is the complete question:

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 certain temperature, a chemist finds that a 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition:

Compound        Amount

   Fe_2O_3         3.54 g

      H_2             3.63 g

      Fe             2.37 g

     H_2O           2.13 g

Calculate the value of the equilibrium constant for this reaction. Round your answer to 2 significant digits

<u>Answer:</u> The value of equilibrium constant for the given reaction is 2.8\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 3.63 g

Molar mass of hydrogen gas = 2 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

\text{Molarity of hydrogen gas}=\frac{3.63}{2\times 5.4}\\\\\text{Molarity of hydrogen gas}=0.336M

  • <u>For water:</u>

Given mass of water = 2.13 g

Molar mass of water = 18 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

\text{Molarity of water}=\frac{2.13}{18\times 5.4}\\\\\text{Molarity of water}=0.0219M

The given chemical equation follows:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of K_c for above equation follows:

K_c=\frac{[H_2O]^3}{[H_2]^3}

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression. So, the concentration of iron and iron (III) oxide is not present in equilibrium constant expression.

Putting values in above equation, we get:

K_c=\frac{(0.0219)^3}{(0.336)^3}\\\\K_c=2.77\times 10^{-4}

Hence, the value of equilibrium constant for the given reaction is 2.8\times 10^{-4}

7 0
3 years ago
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