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4 years ago

Define and explain auto-ionization of water??? Explain please??

Chemistry

2 answers:

Oduvanchick [21]4 years ago

6 0

Answer:

in pure water, H2O molecules undergo a reaction in which two molecules react in such a way that one molecule give its H+ ion to other molecule and become a hydroxyl ion(OH-) and molecule which gain H+ ion become H3O+ ion.

Explanation:

H2O + H2O ⇄ ( H3O+) +( OH-)

icang [17]4 years ago

5 0

Answer:

Hey there!

Auto-ionization of water is an ionization reaction in pure water or in another aqueous solution, in which a water molecule, H2O, loses the nucleus of one of its hydrogen atoms to become a hydroxide ion, OH−.

Let me know if this helps :)

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At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration

wlad13 [49]

Answer:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.03901 mol/liter

[Cl₂O] = 0.02351 mol/liter

Explanation:

1. Chemical reaction:

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

2. Initial concentrations:

i) 1.3 g H₂O

Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol

Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol

Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

3. ICE (Initial, Change, Equilibrium) table

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

I 0.0481 0.0326 0

C -x -x +x

E 0.0481-x 0.0326-x x

4. Equilibrium expression

$K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}$

$0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}$

5. Solve:

$x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0$

Use the quadatic formula:

$x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}$

The positive result is x = 0.00909

Thus the concentrations are:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter

[Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

3 0

3 years ago

30.00 ml of a h2so4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 44.45 ml of a 0.1489

kodGreya [7K]

The concentration of the H2SO4 solution should be 0.1101 M.

Calculation of the concentration:

Since

Volume of H2SO4 = 30.00 mL = 0.030 L

Volume of NaOH= 44.45 mL = 0.04445 L

Concentration of NaOH= 0.1489 M

Now

The balanced equation is

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Now the concentration of the H2SO4 solution is

b*Ca*Va = a*Cb*Vb

here, b = the coefficient of NaOH = 2

Ca = ?

Va = the volume of H2SO4= 30.0 mL = 0.030 L

a = the coefficient of H2SO4 = 1

Cb = the concentration of NaOH = 0.1489 M

Vb = the volume of NaOH = 44.45 mL = 0.04445 L

Now

2*Ca*0.030 = 1* 0.1489* 0.04445

0.060 * Ca = 0.00661

Ca = 0.1101 M

Hence, The concentration of the H2SO4 solution is 0.1101 M.

Learn more about concentration here:

brainly.com/question/14106631

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7 0

1 year ago

There are two main characters of matter:

Oksana_A [137]

Mass and weight are the 2 main characters of matter

3 0

4 years ago

If the temperature is kept constant, what change in volume would cause the pressure of an enclosed gas to be reduced to one-th

joja [24]

Using the ideal gas equation pV = nRT, where R is the ideal gas constant, we can show that p=nRT/V. Since n, R and T are all constants, p2/p1 = V1/V2 where p1 and p2 are the start and final pressures respectively and V1 and V2 are the start and final volumes respectively. For if p1 = 3*p2, the pressure would have fallen to one third of its original value, and it follows that V2 = 3*V1. Therefore, for the pressure to fall to a third of its original value, the volume must increase by a factor of 3.

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3 years ago

Carbon dioxide contains chemical bonds between oxygen and carbon. What particles are between the carbon and oxygen atoms? What k

natka813 [3]

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5 0

3 years ago