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lesya [120]
2 years ago
14

At 10 K Cp,m(Hg(s)) = 4.64 J K−1 mol−1. Between 10 K and the melting point of Hg(s), 234.3 K, heat capacity measurements indicat

e that the entropy increases by 57.74 J K−1 mol−1. The standard enthalpy of fusion of Hg(s) is 2322 J mol−1 at 234.3 K. Between the melting point and 298.0 K, heat capacity measurements indicate that the entropy increases by 6.85 J K−1 mol−1. Determine the Third-Law standard molar entropy of Hg(l) at 298 K.
Chemistry
1 answer:
umka21 [38]2 years ago
4 0

Answer:

S°m,298K = 85.184 J/Kmol

Explanation:

∴ T = 10 K ⇒ Cp,m(Hg(s)) = 4.64 J/Kmol

∴ 10 K to 234.3 K ⇒ ΔS = 57.74 J/Kmol

∴ T = 234.3 K ⇒ ΔHf = 2322 J/mol

∴ 234.3 K to 298.0 K ⇒ ΔS = 6.85 J/Kmol

⇒ S°m,298K = S°m,0K + ∫CpdT/T(10K) + ΔS(10-234.3) + ΔHf/T(234.3K) + ΔS(234.3-298)

⇒ S°m,298K = 0 + 10.684 J/Kmol + 57.74 J/Kmol + 9.9104 J/Kmol + 6.85 J/kmol

⇒ S°m,298K = 85.184 J/Kmol

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Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a
kvv77 [185]

The question is incomplete, the complete question is;

Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75

Answer:

A nucleus with a A nucleus with a neutron:proton ratio of 1.49

A nucleus with a mass of 187 and an atomic number of 75

Explanation:

The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1

Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.

Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.

Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.

5 0
2 years ago
0.01 M HCl solution has a pH of 2. Suppose that during the experiment, both the universal pH indicator and the cabbage indicator
NNADVOKAT [17]
It matches the universal pH indicator and is indicating the proper pH
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2 years ago
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How many molecules of carbon dioxide are in 243.6 g of carbon dioxide?
german
Hey there ! 

Molar mass carbon dioxide:

CO2 = 44.01 g/mol

1) number of moles :

1 mole CO2 ------------- 44.01 g
(moles CO2) ------------ 243.6 g

moles CO2 = 243.6 * 1 / 44.01

moles CO2 = 243.6 / 44.01

=> 5.535 moles of CO2

Therefore:

1 mole -------------------- 6.02x10²³ molecules
5.535 moles ------------ ( molecules CO2)

molecules CO2 = 5.535 * ( 6.02x10²³) / 1

=> 3.33x10²⁴ molecules of CO2
3 0
2 years ago
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K
Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S

Thus, by combining them, we obtain:

-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}

Which is related to the general line equation:

y=mx+b

Whereas:

y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

8 0
2 years ago
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