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Zinaida [17]
3 years ago
13

Explain why an empirical formula alone cannot be used to derive the actual molecular formula for a given compound. Provide one o

r more examples.
Chemistry
1 answer:
dsp733 years ago
7 0

Answer:

Molecular formulas indicate how many atoms there are of each element in a compound, while empirical formulas indicate the reduced proportion of elements in a given compound. The molecular formula can be achieved through the empirical formula if the molecular weight of the compound is known, therefore the empirical formula alone cannot be used to obtain the molecular formula.

Explanation:

An empirical formula expresses the relative relationships of the different atoms in a given compound. For example, H2O is made up of two hydrogen atoms and 1 oxygen atom. In the same way, 1.0 mol of H2O is made up of 2.0 mol of hydrogen and 1.0 mol of oxygen. .

The molecular formula for a compound is always the empirical formula. The molecular formula can be obtained from the empirical formula if the molecular weight of the compound is known.

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Elden [556K]

Answer:

5 mol H₂. this is the answer bro all the best for your test/exam

7 0
3 years ago
For the hypothetical compound AX2, which of the following statements is true?. a.) A is a nonmetal from Group 3A, and X is a met
Ierofanga [76]

The cations has positive charges that are metals while the anions have negative charges that are non-metals. Upon reaction, there is an exchange in charges that are reflected in the subscripts of the atoms. In this case, compound AX2 must have a cation, A belonging to group 2 A with +2 charge and anion, X  belonging to Group 7A with -1 charge. Answer is D.
4 0
3 years ago
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Consider the following reaction: COCl2(g) ⇌ CO(g) + Cl2(g) A reaction mixture initially contains 1.6 M COCl2. Determine the equi
professor190 [17]

Answer:

The equilibrium concentration of CO is 0.0361 M

Explanation:

Step 1: Data given

Kc = 8.33 *10^-4

Molarity of COCl2 = 1.6 M

Step 2: The balanced equation:

COCl2(g) ⇌ CO(g) + Cl2(g)

Step 3: Calculate final concentrations

The initial concentration of COCl2 = 1.6M

The initial concentration of CO and Cl2 = 0M

There will react xM of COCl2

Since the mole ratio is 1:1

The final concentration of CO and Cl2 will be X M

The final concentration of COCl2 will be (1.6 -X)M

Step 4: Define Kc

Kc=  [CO] *[Cl2] /  [COCl2]  = 8.33*10^-4

Kc = X*X / 1.6-X = 8.33 * 10^-4

8.33 * 10^-4  = X² /(1.6-X)

8.33 * 10^-4 *(1.6 -X) = X²

0.0013328 - 8.33*10^-4 X = X²

X² + 8.33*10^-4 X  - 0.0013328= 0

X = 0.0361 M = [CO] = [Cl2]

[COCl2] = 1.6 - 0.0361 = 1.5639 M

To control this we can calculate the Kc

(0.0361*0.0361)/1.5639 = 0.000833

5 0
3 years ago
what is the half life of a radioactive sample if 100.0 grams of it decays to 12.5 grams in 24.3 hours
frutty [35]

The Half-Life of the radioactive sample is 8.1 hours.

<h3>What is half life of a radioactive element?</h3>

The half-life of a radioactive substance is the time taken for half the amount of atoms in the given substance to decay.

12.5 grams remains out of 100 grams of the sample.

This is 1/8 of the original samples.

1/8 of the original sample reperesent three half-lives undergone.

Half-Life of the radioactive sample = 24.3/3 = 8.1 hours

Therefore, the Half-Life of the radioactive sample is 8.1 hours.

Learn more about Half-Life at: brainly.com/question/25750315

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7 0
2 years ago
What is the basic buildings blocks of all living things and are in constant motion
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Matter. The answer is matter.
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3 years ago
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