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Agata [3.3K]

Oxidation is when a substance gains oxygen molecules. For example when hydrogen reacts with oxygen it forms H₂O. The H₂ has been oxidised.

4 0

3 years ago

Determine the pH of a solution of 0.00278 M of HClO4. Report the answer to two digits past the decimal.

aleksandrvk [35]

Answer:

The pH of a solution of 0.00278 M of HClO₄ is 2.56

Explanation:

pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance and is calculated as:

pH= - log [H⁺]= - log [H₃O⁺]

On the other hand , a Strong Acid is that acid that in an aqueous solution dissociates completely. In other words, a strong acid completely dissociates into hydrogen ions and anions in solution.

HClO₄ is a strong acid, so in aqueous solution it will be totally dissociated. Then, the concentration of protons is equal to the initial concentration of acid and the pH will be calculated:

pH= - log 0.00278

pH= 2.56

<u><em>The pH of a solution of 0.00278 M of HClO₄ is 2.56</em></u>

8 0

3 years ago

What dissolved solid has the highest concentration in the ocean? sodium chloride

jeyben [28]

Answer:

chloride and sodium.

Explanation:

These two make up over 90% of all dissolved ions in seawater.

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3 0

3 years ago

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The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate

Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

6 0

3 years ago

Write the symbol for each of the following ions:

kap26 [50]

Answer:

a) $_{31}^{71}\textrm {P^{3+}}$

b) $_{35}^{80}\textrm {Br^{-}}$

c) $_{90}^{232}\textrm {Th^{4+}}$

d) $_{38}^{87}\textrm {Sr^{2+}}$

Explanation:

For neutral atoms:

Atomic Number (Z)= number of protons = number of electrons

Mass number (A) = number of protons + number of neutrons

For ions with positive net charge:

Number of protons = Z + net charge

For ions with negative net charge:

Number of protons= Z - net charge

a) A = 71, Charge = +3

Number of electrons = 28

Number of protons = 28 +3 =31

$_{31}^{71}\textrm {P^{3+}}$

b) A = 35, Z = 45+35=80, Charge = -1

Number of protons =35

Number of neutrons = 45

Number of electrons = 36

Charge = Number of protons- Number of electrons =35-36 = -1

$_{35}^{80}\textrm {Br^{-}}$

c) Charge = +4

Number of electrons = 86

Number of protons = Z = 86+4 = 90

mass number = A = 90+142 = 232

$_{90}^{232}\textrm {Th^{4+}}$

d) Charge = +2

Atomic number = Number of protons = Z = 38

mass number = A = 87

$_{38}^{87}\textrm {Sr^{2+}}$

3 0

3 years ago