All categories

2 years ago

Are used them since made of two or more elements that have been chemically combined is a(n)

1 answer:

5 0

Elements can be chemically combined into compounds, therefore, a compound consists of two or more elements combined, in definite proportions, by chemical means. Compounds may be formed by combining atoms of their constituent elements by ionic bonds or by covalent bonds.

hope this helped

You might be interested in

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

Explain the law of conservation of energy.

Ulleksa [173]

<h2>Energy is conserved. It never appears or disappears. It just changes from one kind of type to another. The Universe contains the same amount of energy it started with. This amount can never change.</h2><h2></h2>

6 0

3 years ago

Read 2 more answers

When a small amount of 12 M HNO3(aq) is added to a buffer solution made by mixing CH3NH2(aq) and CH3NH3Cl(aq) , the pH of the bu

fredd [130]

Answer:

a. CH3NH2(aq) + H⁺ → CH3NH3⁺

Explanation:

The mixture of a weak base as CH3NH2 with its conjugate acid CH3NH3Cl produce a buffer. As the weak acid is in equilibrium with water, the mixture of the weak base and its conjugate base produce that the acid or base released react avoiding the change in pH.

For example, when a strong acid as HNO3 reacts, the weak base will react producing the conjugate base, that is:

CH3NH2(aq) + H⁺ → CH3NH3⁺

Right answer is:

5 0

2 years ago

A redox reaction can also be a?

AveGali [126]

Losing eltron is the answer

7 0

3 years ago

What mass of oxygen contains the same number of molecules as 42g of nitrogen?

lakkis [162]

Answer:

Given: 42 g of N2

Solve for O2 mass that contains the same number of molecules to 42 g of N2.

Solve for the number of moles in 42 g of N2

1 mole of N2 = (14 * 2) g = 28 g so the number of moles in 42 g of N2 is equal to 42 g / 28 g per mole = 1.5 moles

Solve for mass of 1 mole of oxygen

1 mole of O2 = 16 g * 2 = 32 g per mole

Solve for the mass of 1.5 moles of oxygen

mass of 1.5 moles of O2 = 32 g per mole * 1.5 moles

mass of 1.5 moles of O2 = 48 g

So 48 g of O2 contains the same number of molecules as 42 g of N2

3 0

3 years ago