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gayaneshka [121]
3 years ago
8

Consider the reaction:

" id="TexFormula1" title="NiO(s)+CO(g) \rightleftharpoons Ni(s)+CO_2(g)" alt="NiO(s)+CO(g) \rightleftharpoons Ni(s)+CO_2(g)" align="absmiddle" class="latex-formula">; Kc=4.0×10³ at 1500 K.
If a mixture of solid nickel(II) oxide and 0.18000 MM carbon monoxide is allowed to come to equilibrium at 1500 K, what will be the equilibrium concentration of CO₂?
Chemistry
1 answer:
sasho [114]3 years ago
5 0

Answer:

Molar concentration of CO₂ in equilibrium is 0.17996M

Explanation:

Based on the reaction:

NiO(s) + CO(g) ⇆ Ni(s) + CO₂(g)

kc is defined as:

kc = [CO₂] / [CO] = 4.0x10³ <em>(1)</em>

As initial concentration of CO is 0.18M, the concentrations in equilibrium are:

[CO] = 0.18000M - x

[CO₂] = x

Replacing in (1):

4.0x10³ = x / (0.18000-x)

720 - 4000x = x

720 = 4001x

x = 0.17996

Thus, concentrations in equilibrium are:

[CO] = 0.18000M - 0.17996 = 4.0x10⁻⁵

[CO₂] = x = <em>0.17996M</em>

<em></em>

Thus, <em>molar concentration of CO₂ in equilibrium is 0.17996M</em>

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I hope it helps!

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3 years ago
Suppose you have been given the task of distilling a mixture of hexane + toluene. Pure hexane has a refractive index of 1.375 an
miv72 [106K]

Explanation:

The given data is as follows.

         Refractive index of mixture = 1.456

        Refractive index of hexane = 1.375

        Refractive index of toulene = 1.497

Let mole fraction of hexane = x_{1}

and, mole fraction of toulene = x_{2}

Also,        x_{1} + x_{2} = 1

or,            x_{1} = 1 - x_{2}

Hence, calculate the mole fraction of hexane as follows.

        refractive index mixture= mole fraction hexane × ref index hexane + mole fraction toluene × ref index toluene.

              1.456 = (1 - x_{2}) \times 1.375 + x_{2} \times 1.497

               1.456 = 1.375 - 1.375x_{2} + 1.497x_{2}

               0.081 = 0.122x_{2}

                   x_{2} = \frac{0.081}{0.122}

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Since,     x_{1} = 1 - x_{2}

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Thus, we can conclude that mole fraction of hexane in your sample is 0.34.        

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3 years ago
What is the mass in grams of 394 mL of ethylene glycol? Express your answer with the appropriate units.
AleksandrR [38]

Answer : The mass of ethylene glycol is, 437.34 grams

Solution : Given,

Volume of ethylene glycol = 394 ml

Density of ethylene glycol = 1.11 g/ml    (Standard value)

Formula used :

Density=\frac{Mass}{Volume}

Now put all the given values in this formula, we get the mass of ethylene glycol.

1.11g/ml=\frac{Mass}{394ml}

Mass=437.34g

Therefore, the mass of ethylene glycol is, 437.34 grams

3 0
3 years ago
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