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Nastasia [14]
3 years ago
10

Write the Henderson-Hasselbalch equation for a propanoic acid solution (CH3CH2CO2H, pKa = 4.874) using the symbols HA and A–, an

d the given pKa value for propanoic acid in the expression
Chemistry
1 answer:
Luden [163]3 years ago
7 0
The Henderson-Hasselbalch approximation is for conjugate acid-base pairs in a buffered solution. We're going to call HA a weak acid, and A- its conjugate base. The equation is as follows:
pH = pKa + log([base]/[acid]), where the brackets imply concentrations
Plugging in our symbols and the pKa value, the equation becomes:
pH = 4.874 + log([A-]/[HA])
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An isotope contains 26 protons, 24 electrons, and 32 neutrons
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The correct option is 58 Fe 2+.
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How do you balance redox equations in acidic solutions?
Anuta_ua [19.1K]

Answer:

First, balance the half-reactions

Second, equalize the electrons

Third,add two reaction equations to get final answer

Explanation:

For example

H₂C₂0₄ + MnO⁻₄ ---------->CO₂+Mn²⁺

(i) Balancing the half reactions

H₂C₂O₄-------->2CO₂+2H⁺+2e⁻

5e⁻ +8H⁺+MnO₄⁻----------->Mn²⁺+4H₂O

(ii)

Equalizing the electrons

5H₂C₂O₄--------->10CO₂+10H⁺+10e⁻  ---here there is a factor of 5

10e⁻+16H⁺+2MnO₄⁻--------->2Mn²⁺+8H₂O -----here there is a factor of 2

(iii)

Add the two where electrons and some Hydrogen ions will cancel out

5H₂C₂O₄+6H⁺+2MnO₄⁻---->10CO₂+2Mn²⁺+8H₂O

7 0
3 years ago
A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo
Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>
  • Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

                                   = (2.684 - 2.156) g

                                   = 0.528 g

  • Find the number of moles of the elements in the compound

The number of moles is given by

n = \frac{m}{M}

where m = given mass and

M = Molar mass

Number of moles of Zinc = \frac{2.156}{65} = 0.033 moles

Number of moles of Oxygen =\frac{0.528}{16} = 0.033 moles

  • Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

brainly.com/question/1603500

#SPJ4

 

5 0
1 year ago
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