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svlad2 [7]
3 years ago
10

Identify all correct statements about the ionization of water.check all that apply.check all that apply.water ionizes to form pe

roxide and hydronium ions.dissociation of water produces equal numbers of oh- and h+.dissociation of water is reversible.water ionizes to form hydroxide and hydronium ions.dissociation of water produces equal masses of oh- and h+.dissociation of water is not reversible.
Chemistry
1 answer:
DerKrebs [107]3 years ago
6 0
The ionization of water occurs in pure water or in an aqueous solution where the water molecule loses one nucleus from one of its hydrogen atoms forming OH- and H+. This H+ immediately combines with another water molecule forming H3O+.

This reaction has the following postulates (these are the correct choices):
1- <span>Water ionizes to form peroxide and hydronium ions
</span>2- <span>Dissociation of water produces equal numbers of OH- and H+.
3- </span><span>Dissociation of water is reversible</span>
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1.02 mg of a compound that is known to absorb at 340 nm is dissolved and diluted to 5.00 mL in a volumetric flask. A 1.00 mL ali
Vinvika [58]

Answer:

6.97 X 10^{-4} M

Explanation:

The concentration of the analyte in the 5-mL flask would be 6.97 X 10^{-4} M

This is a problem of simple dilution that can be solved using the dilution equation;

C1V1 = C2V2,

where C1 = initial concentration, V1 = initial volume, C2 = final concentration, and V2 = final volume.

<em>In this case, the initial concentration (C1) is not known, the initial volume (V1) is 1.00 mL, the final concentration is 6.97 x 10-5 M, and the final volume is 10.00 mL.</em>

Now, let us make the initial concentration the subject of the formula from the equation above;

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C1 = 6.97 x 10-5 x 10/1 = 6.97 X 10^{-4} M

7 0
3 years ago
The 1995 Nobel Prize in chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th
Nikolay [14]

Answer:

\Delta H_{rxn3}=-162.5 kJ/mol

Explanation:

The reaction we need to calculate:

O_3 (g) + Cl (g) \longrightarrow ClO (g) + O_2 (g)

1) O_3 (g) + ClO (g) \longrightarrow Cl (g) +2 O_2 (g)

\Delta H_{rxn}=-122.8 kJ/mol

We need the ClO in the products side, so we use the inverse of this reaction:

Cl (g) +2 O_2 (g) \longrightarrow O_3 (g) + ClO (g)

\Delta H_{rxn1}=122.8 kJ/mol

2) 2 O_3 (g) \longrightarrow 3 O_2 (g)

\Delta H_{rxn2}=-285.3 kJ/mol

Now we need to combine this two:

Cl (g) +2 O_2 (g) + 2 O_3 (g)\longrightarrow O_3 (g) + ClO (g) + 3 O_2 (g)

Cl (g) + O_3 (g)\longrightarrow ClO (g) + O_2 (g)

The enthalpy of reaction:

\Delta H_{rxn3}=\Delta H_{rxn1}+ \Delta H_{rxn2}=122.8 kJ/mol-285.3 kJ/mol

\Delta H_{rxn3}=-162.5 kJ/mol

4 0
3 years ago
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