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Tomtit [17]
1 year ago
9

What will be the ph of a buffer solution containing an acid of pka 7. 5, with an acid concentration exactly one fourth of that o

f the conjugate base?
Chemistry
1 answer:
Over [174]1 year ago
8 0

8.1  will be the pH of a buffer solution containing an acid of pka 7. 5, with an acid concentration exactly one-fourth of that of the conjugate base.

In chemistry, the pH scale is used to define the acidity or basicity of an aqueous solution. Historically, pH stood for the "potential of hydrogen." The pH values of acidic solutions are typically lower than those of basic or alkaline solutions.

A weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base, are combined to form the buffer solution, a water-based solvent solution. They withstand being diluted or having modest amounts of acid or alkali added to them without changing their pH.

A weak base and its salt are combined with a strong acid to create a basic buffer, which has an acidic pH. the aqueous mixture of ammonium hydroxide and ammonium chloride, both in equal concentrations.

To know more about pH refer to:  

brainly.com/question/8676275

#SPJ4

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The ratio of carbon atoms to hydrogen atoms to oxygen atoms in a molecule of dicyclohexyl maleate is 4 to 6 to 1. what is its mo
Semenov [28]
Knowing the ratio between atoms we can write an empirical formula: 

<span>C4H6O </span>

<span>we compute the molar mass of this single formula: </span>
<span>4x12 + 6 x 1 + 16 x1 = 70 g / mol </span>

<span>Now, as we know the actual molar mas being 280 g/mol, we divide this number by 70 and we get the ratio between empirical formula and molecular actual formula: </span>

<span>280 / 70 = 4 </span>

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How many valence electrons are in oxygen?
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Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2
baherus [9]

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) Cu_{(s)} →Cu^{2+}_{(aq)} +2e E°=0.337 v

(reduction) MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}→MnO_{2 (aq)}+2H_{2}O E°=1.679 v

(overall) 2MnO_{4 (aq)}+3Cu_{(s)}+8H^{+}_{(aq)}→3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346

E=1.346

5 0
2 years ago
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