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garik1379 [7]
3 years ago
14

The elemental mass percent composition of ascorbic acid (vitamin C ) is 40.92% C , 4.58% H , and 54.50% O . Determine the empiri

cal formula of ascorbic acid.
Chemistry
1 answer:
Lady bird [3.3K]3 years ago
7 0

The empirical formula of ascorbic acid is C6H8O6.

<h3>Empirical formula</h3>

To calculate the empirical mass of a compound from the mass percentage of each element, the value of the<u> molar mass </u>of each element is used, which in this case corresponds to:

                                              MM_C= 12g/mol\\MM_H=1g/mol\\MM_O=16g/mol

From this, we consider that the compound has 100 grams, so the number of moles of each element will be equal to:

                                                   C = \frac{40.92}{12} = 3.41 \\H = \frac{4.58}{1}=4.58\\ O =\frac{54.50}{16}=3.41

Now, divide all the values ​​found by the <u>smallest value</u>, to find the amount present in each element:

                                            C = \frac{3.41}{3.41} = 1\\ H =  \frac{4.58}{3.41} = 1.34\\O =  \frac{3.41}{3.41} = 1

As you can see, the value of moles of hydrogen resulted in a decimal number, so it is necessary to multiply all values ​​​​by a number in common until the three meet as integers:

                                          C = 1 \times 6 = 6\\H = 1.34 \times 6 = 8\\O = 6 \times 6 = 6

So, the empirical formula of ascorbic acid is C6H8O6.

Learn more about empirical and molecular formula in: brainly.com/question/11588623

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Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)  

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V</em>

<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s)    E°red = -0.76 V</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

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KE = 1/2*150kg*(20 m/s)^2

KE = 75kg * 400m²/s²

KE = 30,000 kg*m²/s²

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