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3 years ago

8 Fe + Sg → 8 Fes How many grams of FeS is produced from 0.3 mol Sg?

Chemistry

1 answer:

frozen [14]3 years ago

6 0

Answer:

Explanation:

Molar ratio for Sg : FeS = 1:8

If there are 0.3 moles for Sg

Therefore, 0.3 × 8 =2.4 moles of FeS

Mass = Moles/ Mr

Mr of FeS = 56+32=88

So mass = 2.4/88

Mass= 0.027g

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Calculate δg∘rxn and e∘cell for a redox reaction with n = 3 that has an equilibrium constant of k = 4.4×10−2. you may want to re

lapo4ka [179]

a) First, to get ΔG°rxn we have to use this formula when:

ΔG° = - RT ㏑ K

when ΔG° is Gibbs free energy

and R is the constant = 8.314 J/mol K

and T is the temperature in Kelvin = 25 °C+ 273 = 298 K

and when K = 4.4 x 10^-2

so, by substitution:

ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)

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b) then, to get E° cell for a redox reaction we have to use this formula:

ΔE° Cell = (RT / nF) ㏑K

when R is a constant = 8.314 J/molK

and T is the temperature in Kelvin = 25°C + 273 = 298 K

and n = no.of moles of e- from the balanced redox reaction= 3

and F is Faraday constant = 96485 C/mol

and K = 4.4 x 10^-2

so, by substitution:

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3 years ago

Which of the following statements is true? A) This reaction will be spontaneous only at high temperatures. B) This reaction will

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Answer:

D) This reaction will be nonspontaneous only at high temperatures.

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature

∆S = is the change in entropy .

And , the sign of the ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

ΔG < 0 , the reaction is Spontaneous

ΔG > 0 , the reaction is non Spontaneous

ΔG = 0 , the reaction is at equilibrium

The reaction has the value for ∆H = negative , and ∆S = negative ,

Now ,

∆G = ∆H -T∆S

= ( - ∆H ) - T( - ∆S )

= ( - ∆H ) +T( ∆S )

Now, for making the reaction Spontaneous ΔG = negative ,

Hence ,

The temperature is low, then the value for ΔG will be negative , i.e. , Spontaneous reaction .

And , vice versa , at higher temperature , the reaction will have ΔG positive , and the reaction will be non -Spontaneous reaction .

The standard free energy of formation will be zero , only for the compounds that are in their pure form ,

Hence , Al(s) will have ΔG = 0 .

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Beginning with commercial grade hydrochloric acid, 1.00 • 10^2 mL of a 12.4 M HCl is added to water to bring the total volume if

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How does the presence of a catalyst affect the enthalpy of a reaction? Group of answer choices 1. It depends on whether you are

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Answer:

Option 3. The catalyst does not affect the enthalpy change ( $\Delta H_\text{rxn}$ ) of a reaction.

Explanation:

As its name suggests, the enthalpy change of a reaction ( $\Delta H_\text{rxn}$ ) is the difference between the enthalpy of the products and the reactants.

On the other hand, a catalyst speeds up a reaction because it provides an alternative reaction pathway from the reactants to the products.

In effect, a catalyst reduces the activation energy of the reaction in both directions. The reactants and products of the reaction won't change. As a result, the difference in their enthalpies won't change, either. That's the same as saying that the enthalpy change $\Delta H_\text{rxn}$ of the reaction would stay the same.

Refer to an energy profile diagram. Enthalpy change of the reaction $\Delta H_\text{rxn}$ measures the difference between the two horizontal sections. Indeed, the catalyst lowered the height of the peak. However, that did not change the height of each horizontal section or the difference between them. Hence, the enthalpy change of the reaction stayed the same.

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