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uranmaximum [27]

3 years ago

7

How can you distinguish A from B?

2 answers:

mel-nik [20]3 years ago

8 0

* Lab question: How can you distinguish a physical change from a chemical change?

> Format: How can you distinguish A from B?

Answer! Comparison: physical changes vs. chemical changes

Paul [167]3 years ago

4 0

I think we can distinguish A from B because they are totally different letters but bro what do you mean in what form of letters, is there supposed to be like an equation or something please tell me if there is so i can help you or whatever it is hope i can help you :)

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Using the equations

Anna [14]

Considering the Hess's Law, the enthalpy change for the reaction is 221.8 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol

Equation 2: C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol

Equation 3: 2 C(s) + 2 H₂ (g) → C₂H₄ (g) ∆H° = -97.6 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

<h3 /><h3>FIRST STEP</h3>

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C₂H₄ (g) on reactant side and it is present in first equation. Since this equation has one mole of C₂H₄ (g) on the product side, it is necessary to locate it on the reactant side (invert it).

When an equation is inverted, the sign of ΔH° also changes.

<h3>SECOND STEP</h3>

Now, you need 2 moles of CF₄ (g) on the product side. The second equation has 1 mole of CF₄ (g) on the product side, so it is necessary to multiply it by 2 to obtain 2 moles of CF₄ (g).

Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

<h3>THIRD STEP</h3>

Finally, you need 4 moles of HF (g) on the product side. The first equation has 2 moles of HF (g) on the product side, so it is necessary to multiply it by 2 to obtain 4 moles of the compound.

Since the equation is multiply by 2, the variation of enthalpy also is multiplied by 2.

<h3>SUMMARY</h3>

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 H₂ (g) + 2 F₂ (g) → 4 HF (g) ∆H° = -158.4 kJ/mol

Equation 2: 2 C (s) + 4 F₂ (g) → 2 CF₄ (g) ∆H° = 282.6 kJ/mol

Equation 3: C₂H₄ (g) → 2 C(s) + 2 H₂ (g) ∆H° = 97.6 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g) ΔH°= 221.8 kJ/mol

Finally, the enthalpy change for the reaction is 221.8 kJ/mol.

Learn more about molar enthalpy:

7 0

2 years ago

Is rubbing alcohol polar, non-polar, or ionic?

Karolina [17]

Answer: Rubbing alcohol molecules have a polar and nonpolar part, which means they are able to form hydrogen bonds with water and therefore able to mix with it.

Explanation:

7 0

3 years ago

Describe the displacement method for measuring the volume of an object

jeyben [28]

<span>Volume is a measure of the amount of space an object takes up. When a cylinder is submerged in the water it pushes water out of the way. If you measure the amount the water level increases, you can find the volume of the water pushed out of the way. </span>

6 0

3 years ago

2, 3-diethyl–4-octene what is structure of this pls help

pantera1 [17]

Answer:

(CH2CH3).

!

CH-CH-CH-CH=CH-CH2-CH2-CH3

!

(CH2CH3)

2, 3-diethyl–4-octene

i am not sure.

4 0

3 years ago

What is the law of multiple proportions

dmitriy555 [2]

When two element combine to form more than one compound i hope this helps you with work have a nice day :)

7 0

3 years ago