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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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Calculate the pH of each of following buffered solutions.?a. 0.10 M acetic acid/0.25 M sodium acetate b. 0.25 M acetic acid/0.10
Masteriza [31]

Answer:

a. 5.10.

b. 4.35.

c. 5.10.

d. 4.35.

Explanation:

<u><em>a. 0.10 M acetic acid/0.25 M sodium acetate </em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.25)/(0.10)

∴ pH = 4.744 + 0.34 = 5.084 ≅ 5.10.

<u><em>b. 0.25 M acetic acid/0.10 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.10)/(0.25)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

<u><em>c. 0.080 M acetic acid/0.20 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.20)/(0.08)

∴ pH =  4.744 + 0.34 = 5.084 ≅ 5.10.

<u><em></em></u>

<u><em>d. 0.20 M acetic acid/0.080 M sodium acetate</em></u>

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.08)/(0.20)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

6 0
2 years ago
How many mg does an 830 kg sample contain?
Aleksandr-060686 [28]

Answer:

A sample of 830 kg would contain 830000000 mg

Explanation:

3 0
1 year ago
The stock concentration of dye is 3.4E-5M. The stock concentration of bleach is 0.36M. Assuming that the reaction goes to comple
Aleks04 [339]

Answer:

0.036 M

Explanation:

To do this, let's mark the dye as D and bleach as B.

We have the concentrations of both, and we already know that they react in a 1:1 mole ratio. The total volume of reaction is 9 + 1 = 10 mL or 0.010 L, and we hava both concentrations.

The problem already states that the dye reacts completely, so this is the limiting reagent, while bleach is the excess.

To know the remaining amount of bleach, we need to do this with the moles. First, let's calculate the initial moles of D and B:

moles D = 3.4x10⁻⁵ * 0.009 = 3.06x10⁻⁷ moles

moles B = 0.36 * 0.001 = 3.6x10⁻⁴ moles

Now that we have the moles, and that we know that all the dye reacts completely, let's see how many moles of bleach are left:

moles of B remaining = 3.6x10⁻⁴ - 3.06x10⁻⁷ = 3.597x10⁻⁴ moles

These are the moles presents of B after the reaction has been made. The concentration of the same will be:

[B] = 3.597x10⁻⁴ / 0.010

[B] = 0.0357

With 2 SF it would be:

[B] = 3.6x10⁻² M

5 0
1 year ago
1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theore
Alina [70]

Answer :

(1) The theoretical yield of product, MgO is, 1.257 grams.

(2) The percent yield of MgO is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2Mg(s)+O_2(g)\rightarrow 2MgO(s)

From the reaction, we conclude that

As, 2 mole of Mg react to give 2 mole of MgO

So, 0.03142 moles of Mg react to give 0.03142 moles of MgO

Now we have to calculate the mass of MgO

\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO

\text{ Mass of }MgO=(0.03142moles)\times (40g/mole)=1.257g

Theoretical yield of MgO = 1.257 g

Experimental yield of MgO = 0.8922 g

Now we have to calculate the percent yield of MgO

\% \text{ yield of }MgO=\frac{\text{ Experimental yield of }MgO}{\text{ Theretical yield of }MgO}\times 100

\% \text{ yield of }MgO=\frac{0.8922g}{1.257g}\times 100=70.97\%

Therefore, the percent yield of MgO is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

4 0
2 years ago
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Stolb23 [73]

Answer:

Explanation:

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4 0
1 year ago
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