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1 month ago

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On the graph, which shows the potential energy curve of two N atoms, carefully sketch a curve that corresponds to the potential

energy of two O atoms versus the distance between their nuclei.

1 answer:

Anarel [2.9K]1 month ago

6 0

Answer:

Explanation:

We need to carefully draw a graph depicting the potential energy associated with two O atoms concerning the separation of their nuclei.

According to the diagram, O2 exhibits a greater potential energy compared to the N2 molecule. This is due to the fact that on the periodic table, atomic size increases moving from left to right within a period, making the O2 atoms larger in size than N2 atoms.

Thus, the bond length between two O atoms will be larger when compared to that of two N atoms.

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A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press

Tems11 [2777]

Answer: The vapor pressure of naphthalene within the flask remains at $2.906\times 10^{-4}$ atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

$C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)$

The formula employed to determine the enthalpy change for the reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})$

The provided information includes:

$\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol$

Substituting the values into the previous equation produces:

$\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol$

The formula utilized to compute Gibbs free energy change is of a reaction:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]$

The equation for the enthalpy change for the reaction is:

$\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})$

The given factors include:

$\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol$

By inserting values from the above equation, we arrive at:

$\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol$

For the calculation of $K_1$ (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

$\Delta G^o=-RT\ln K_1$

where,

$\Delta G^o$ = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_1$ = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

$22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}$

To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

$\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$K_2$ = Equilibrium constant at 35°C =?

$K_1$ = Equilibrium constant at 25°C = $1.14\times 10^{-4}$

$\Delta H$ = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = $8.314J/K mol$

$T_1$ = Initial temperature = $25^oC=[273+25]K=298K$

$T_2$ = Final temperature = $35^oC=[273+35]K=308K$

By plugging values into the equation above, we obtain:

$\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}$

In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for $K_p$ , which is:

$K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)$

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for $K_2$ will equal $K_p$

$p_{C_{10}H_8}=2.906\times 10^{-4}$

Consequently, the partial pressure of naphthalene at 35°C is $2.906\times 10^{-4}$ atm.

3 0

18 days ago

Write a balanced equation depicting the formation of one mole of NaBr(s) from its elements in their standard states.

alisha [2963]

Answer:

Refer to the explanation.

Explanation:

Formation reactions involve the creation of one mole of a compound from its elements in their standard states.

NaBr (s)

The equation for the standard formation is

Na (s) + (1/2)Br₂ (g) → NaBr (s)

As per appendix C, the standard heat of formation for NaBr(s) is

ΔH∘f = -359.8 kJ/mol.

SO₃ (g)

The equation for the standard formation is

S (s) + (3/2) O₂ (g) → SO₃ (g)

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ΔH∘f = -395.2 kJ/mol.

Pb(NO₃)₂ (s)

The equation for the standard formation is

Pb (s) + N₂ (g) + 3O₂ (g) → Pb(NO₃)₂ (s)

According to appendix C, the standard heat of formation for Pb(NO₃)₂(s) is

ΔH∘f = -451.9 kJ/mol.

I hope this is helpful!

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6 0

1 month ago

In KCI how are the valence electrons distributed

eduard [2782]

Answer:

Explanation:

In KCl, the two elements that combine to create KCl are potassium (K) and chlorine (Cl).

Potassium, as a Group 1 element, possesses one valence electron in its outermost shell which it readily donates during bonding. Every element aims to achieve a stable electron configuration, typically with 2 or 8 electrons in its outer shell. Potassium is characterized by its lower electronegativity and higher ionization energy, making it more likely to donate its electron than to accept one. On the other hand, chlorine belongs to Group 17 and has 7 electrons in its outer shell, requiring just one additional electron to complete its octet. Chlorine’s higher electronegativity and lower ionization energy facilitate its tendency to accept an electron rather than donate it.

The bond between potassium and chlorine that results in KCl is termed an electrovalent bond.

Reaction equation:

K + Cl → KCl

3 0

1 month ago

Determine how many grams of silver would be produced, if 12.83 x 10^23 atoms of copper react with an excess of silver nitrate. G

Anarel [2989]

1) The chemical equation is

Cu + 2AgNO3 ---> Cu (NO3)2 + 2Ag

2) Molar ratios are as follows:

1 mol Cu: 2 moles AgNO3: 1 mol Cu (NO3)2: 2 mol Ag

3) Converting 12.83 * 10^23 atoms of Cu to moles gives:

12.83 * 10^23 atoms / (6.02 * 10^23 atoms / mol) = 2.131 mol Cu

4) Using the ratios:

2.131 mol Cu * 2 mol Ag / 1 mol Cu = 4.262 mol Ag

5) To convert 4.262 mol of silver to grams, use the atomic weight of silver:

mass = moles × atomic mass = 4.262 mol * 107.9 g / mol = 459.9 grams

Answer: 459.9 g

5 0

1 month ago

A 20.0–milliliter sample of 0.200–molar K2CO3 solution is added to 30.0 milliliters of 0.400–molar Ba(NO3)2 solution. Barium c

KiRa [2933]

Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

O sea,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Dado que:

Para $K_2CO_3$ :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de $K_2CO_3$ son:

$Moles=0.200 \times {20.0\times 10^{-3}}\ moles$

Moles de $K_2CO_3$ = 0.004 moles

Para $Ba(NO_3)_2$ :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de $Ba(NO_3)_2$ son:

$Moles=0.400 \times {30.0\times 10^{-3}}\ moles$

Moles de $Ba(NO_3)_2$ = 0.012 moles

Según la reacción:

$Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3$

1 mol de $Ba(NO_3)_2$ reacciona con 1 mol de $K_2CO_3$

Por lo tanto,

0.012 mol de $Ba(NO_3)_2$ reacciona con 0.012 mol de $K_2CO_3$

Moles disponibles de $K_2CO_3$ = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces $K_2CO_3$ es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de $K_2CO_3$ reacciona con 1 mol de $Ba(NO_3)_2$ y produce 1 mol de $BaCO_3$

0.004 mol de $K_2CO_3$ reacciona con 0.004 mol de $Ba(NO_3)_2$ y genera 0.004 mol de $BaCO_3$

Los moles restantes de $Ba(NO_3)_2$ son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, $Ba^{2+}$ , después de la reacción es:

$Molarity=\frac{0.008}{0.050}\ M = 0.16\ M$

3 0

1 month ago