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lara [203]
2 years ago
15

The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles

in liquid water according to the equation 2HgO(s)+H2O(l)+2Cl2(g)⇌2HOCl(aq)+HgO⋅HgCl2(s) What is the equilibrium-constant expression for this reaction? View Available Hint(s) The acid (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles in liquid water according to the equation What is the equilibrium-constant expression for this reaction? K=[HOCl]2[HgO⋅HgCl2][Cl2]2[H2O][HgO]2 K=[HOCl]2[Cl2]2[H2O] K=[Cl2]2[HOCl]2 K=[HOCl]2[Cl2]2
Chemistry
1 answer:
skelet666 [1.2K]2 years ago
4 0

Answer:

k = \frac{[HOCl]^2}{[Cl]^2}

Explanation:

The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.

Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium

The global reaction is:

2 HgO (s) + H₂O (l) +2 Cl₂ (g) ⇌ 2 HOCl (aq) + HgO⋅HgCl₂ (s)

Thus, equilibrium-constant expression is:

<em>k = \frac{[HOCl]^2}{[Cl]^2}</em>

You don't include HgO nor HgO⋅HgCl₂ because are pure solids nor water because is pure liquid.

I hope it helps!

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A student runs an experiment in the lab and then uses the data to prepare an Arrhenius plot of the natural log of the rate const
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21.86582KJ

Explanation:

The graphical form of the Arrhenius equation is shown on the image attached. Remember that in the Arrhenius equation, we plot the rate constant against the inverse of temperature. The slope of this graph is the activation energy and its y intercept is the frequency factor.

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2 years ago
The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
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9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

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[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained
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Answer:

A. Yes, there is more than enough sodium carbonate.

Explanation:

Hello,

In this case, based on the given reaction which is:

H_2SO_4(aq) + Na_2CO_3(s) \rightarrow  Na_2SO_4(aq) + CO_2(g) + H_2O(l)

By stoichiometry, one computes the grams of sodium carbonate that will neutralize 1,665 g of sulfuric acid as shown below:

1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3

Thus, the available mass is 2.0 kg so 0.2 kg are in excess, therefore: A. Yes, there is more than enough sodium carbonate.

Best regards.

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Answer:

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1 year ago
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