Answer:
Final Temperature = 36.54 ⁰C
Explanation:
Lets suppose the gas is acting ideally, then according to Charle's Law, "<em>The volume of a fixed mass of gas at constant pressure is directly proportional to the absolute temperature</em>". Mathematically for initial and final states the relation is as follow,
V₁ / T₁ = V₂ / T₂
Data Given;
V₁ = 32 L
T₁ = 10 °C = 283.15 K ∴ K = °C + 273.15
V₂ = 35 L
T₂ = ??
Solving equation for T₂,
T₂ = V₂ × T₁ / V₁
Putting values,
T₂ = (35 L × 283.15 K) ÷ 32 L
T₂ = 309.69 K ∴ ( 36.54 °C )
Result:
As the volume is increased from 32 L to 35 L, therefore, the temperature must have increased from 10 °C to 36.54 °C.
<span>Avogadro's number
represents the number of units in one mole of any substance. This has the value
of 6.022 x 10^23 units / mole. This number can be used to convert the number of
atoms or molecules into number of moles. We calculate as follows:
0.340 mol Br2 ( </span>6.022 x 10^23 molecules / mol ) = 2.05 x 10^23 molecules
Answer:
Option C. Energy Profile D
Explanation:
Data obtained from the question include:
Enthalpy change ΔH = 89.4 KJ/mol.
Enthalpy change (ΔH) is simply defined as the difference between the heat of product (Hp) and the heat of reactant (Hr). Mathematically, it is expressed as:
Enthalpy change (ΔH) = Heat of product (Hp) – Heat of reactant (Hr)
ΔH = Hp – Hr
Note: If the enthalpy change (ΔH) is positive, it means that the product has a higher heat content than the reactant.
If the enthalpy change (ΔH) is negative, it means that the reactant has a higher heat content than the product.
Now, considering the question given, the enthalpy change (ΔH) is 89.4 KJ/mol and it is a positive number indicating that the heat content of the product is higher than the heat content of the reactant.
Therefore, Energy Profile D satisfy the enthalpy change (ΔH) for the formation of CS2 as it indicates that the heat content of product is higher than the heat content of the reactant.
Answer: 1.59atm
Explanation:
We have that for the Question "Calculate the final pressure of the gas mixture, assuming that the container volume does not change."
it can be said that
The final pressure of the gas mixture, assuming that the container volume does not change =
From the question we are told
A container of N2O3(g) has a pressure of 0.265 atm. When the absolute temperature of the N2O3(g) is tripled, the gas completely decomposes, producing NO2(g) and NO(g).
Answer:
Explanation:
Hello!
In this case, since the definition of entropy in a random mixture is:
![\Delta S=-n_TR\Sigma[x_i*ln(x_i)]](https://tex.z-dn.net/?f=%5CDelta%20S%3D-n_TR%5CSigma%5Bx_i%2Aln%28x_i%29%5D)
For this silver-gold mixture we write:
![\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]](https://tex.z-dn.net/?f=%5CDelta%20S%3D-%28n_%7BAu%7D%2Bn_%7BAg%7D%29R%5CSigma%5B%5Cfrac%7Bn_%7BAu%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%2Aln%28%5Cfrac%7Bn_%7BAu%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%29%2B%5Cfrac%7Bn_%7BAg%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%2Aln%28%5Cfrac%7Bn_%7BAg%7D%7D%7Bn_%7BAu%7D%2Bn_%7BAg%7D%7D%20%29%5D)
By knowing the moles of gold:
It is possible to write the aforementioned formula in terms of the variable 
representing the moles of silver:
![20\frac{J}{mol}=-(0.508+x)8.314\frac{J}{mol*K} \Sigma[\frac{0.508}{0.508+x} *ln(\frac{0.508}{0.508+x} )+\frac{x}{0.508+x} *ln(\frac{x}{0.508+x} )]](https://tex.z-dn.net/?f=20%5Cfrac%7BJ%7D%7Bmol%7D%3D-%280.508%2Bx%298.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%5CSigma%5B%5Cfrac%7B0.508%7D%7B0.508%2Bx%7D%20%2Aln%28%5Cfrac%7B0.508%7D%7B0.508%2Bx%7D%20%29%2B%5Cfrac%7Bx%7D%7B0.508%2Bx%7D%20%2Aln%28%5Cfrac%7Bx%7D%7B0.508%2Bx%7D%20%29%5D)
Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:
So the mass is:
Best regards!
