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3 years ago

Can a molecular formula ever be the same as an empirical formula? Explain your answer.

Chemistry

2 answers:

Marina CMI [18]3 years ago

8 0

Answer:

The molecular formula for a compound can be the same as or a multiple of the compound's empirical formula. Molecular formulas are compact and easy to communicate; however, they lack the information about bonding and atomic arrangement that is provided in a structural formula.

Explanation:

Artyom0805 [142]3 years ago

8 0

Answer: Hello Luv......

Hope this helps. Please mark me brainest... Anna ♥

Explanation:

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What is the difference between ores and fosssil fuels

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An ore contains minerals (metals etc) that have economic value.
Fossil fuels are organic material, which means that they are made from dead organisms.

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Which describes a step in the process of forming an ionic bond?

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Answer:

Explanation:

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What is the total number of hydrogen atoms contained in one molecule of (NH4)2C204?

Ber [7]

Answer:

=8 atoms

Explanation:

In (NH4)2C2O4 there are four moles of Hydrogen in the compound (NH4), but there two molecules of (NH4) in this compound. That's what the 2 in (NH4)2 means, so multiply 4 x 2 = 8.

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3 years ago

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The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst

OLEGan [10]

Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product?

A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol.

Not feasible. ΔG°rxn = ΔG°f(product) > 0.

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol

Feasible. ΔG°rxn = ΔG°f(product) < 0.

C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol

Not feasible. ΔG°rxn = ΔG°f(product) > 0.

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Feasible. ΔG°rxn = ΔG°f(product) < 0.

3 0

4 years ago

Determine the oxidation number of Cl in each of the following species.Cl2O7AlCl4-Ba(ClO2)2CIF4+

DIA [1.3K]

These are four questons and four answers:

Answers:

1) 7⁺

2) 1⁻

3) 3⁺

4) 5⁺

Explanation:

Question 1) Cl₂O₇:

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

2×X + 7 (-2) = 0

2X - 14 = 0

2X = +14

X = +14 /2 = + 7

Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.

Question 2) AlCl₄⁻

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

1 (+3) + 4X = - 1

+3 + 4X = - 1

4X = - 1 - 3

4X = - 4

X = - 1

Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.

Question 3) Ba(ClO₂)₂

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

+2 + 2X + 4 (-2) = 0

2X +2 - 8 = 0

2X - 6 = 0

2X = +6

X = + 3

Conclusion: the oxidation number of Cl in Ba(ClO₂)₂ is 3⁺.

Question 4) CIF₄⁺

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

X + 4(-1) = +1

X - 4 = +1

X = +1 + 4

X = + 5

Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.

6 0

3 years ago