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11 days ago

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An ab4 molecule has two lone pairs of electrons on the a atom (in addition to the four b atoms). what is the electron-domain geo

metry around the a atom?

2 answers:

eduard [2.6K]11 days ago

7 0

In the described structure, A is the central atom with four surrounding B atoms creating four covalent bonds. Ideally, this configuration could imply a square planar molecular geometry. However, due to the presence of two lone pairs of electrons on atom A, it is essential to acknowledge this when interpreting the electron geometry.

KiRa [2.8K]11 days ago

4 0

The electron-domain geometry surrounding the 'A' atom is octahedral.

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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 [2624]

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

11 days ago

Next we need to determine the mass of oleic acid in the monolayer. The concentration of the oleic acid/benzene solution is 0.02g

Alekssandra [2891]

Response:

$m=1x10^{-6}g$

Clarification:

Hello,

In this scenario, since a single drop equates to 0.05 mL of the solution provided, with a concentration of 0.02 g/mL, the mass of oleic acid in one drop calculates to:

$m=0.02\frac{g}{L}*0.05mL*\frac{1L}{1000mL}\\ \\m=1x10^{-6}g$

Best wishes.

3 0

1 month ago

Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When 1.00 g of X is dissolved in water and allowed to react w

eduard [2645]

Let's assume that the compound formula is as follows: Experiment 1: 1.00 g of the compound yields 1.95 g of AgCl. The molar mass of AgCl is 143.32 g/mol. Thus, the moles of AgCl for 1.95g are: The moles of Cl also equal 0.0136, considering that 1 mole of AgCl corresponds to 1 mole of Cl. Experiment 2: 1.00 g of the compound results in 0.900 g of CO2 and 0.735 g of H2O. The molar mass of CO2 is 44 g/mol, and for H2O, it's 18 g/mol. Therefore, the moles of C come to 0.0205 and the moles of H stand at 0.0816 (which is 2 times the moles of H2O). Now, from the provided details, it's derived that in 1.00 g of the compound, there are 0.0136 moles of Cl, 0.0205 moles of C, and 0.0816 moles of H. In terms of mass: Mass of Cl = 0.0136 * 35.5 = 0.4828 g. Mass of C = 0.0205 * 12 = 0.246 g. Mass of H = 0.0816 * 1 = 0.0816 g. Total mass = 0.4828 + 0.246 + 0.0816 + mass of N. Given that 1.00 = 0.8104 + Mass of N, it follows that Mass of N = 0.1896. Thus, upon dividing all moles by the smallest value, we find Cl = 0.0136 / 0.0135 = 1.0007; C = 0.0205 / 0.0135 = 1.52; H = 0.0816 / 0.0135 = 6.04; N = 0.0135 / 0.0135 = 1. Multiplying by 2 allows us to reach integer values: Cl = 2, C = 3, H = 12, N = 2.

7 0

19 days ago

What fact do scientists believe provides evidence for the big bang theory?a. all comets follow elliptical orbits.b. most extra-s

alisha [2865]

Answer:

D

Reasoning:

This is due to the belief that at one time, the galaxy was condensed into a minute spot and the Big Bang event initiated the universe's expansion, propelling galaxies outward into space, distancing them from one another. Dark energy is theorized to exert the force from the Big Bang that drives this expansion.

8 0

19 days ago

Read 2 more answers

Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr

eduard [2645]

Answer:

The integer value of x in the hydrate is 10.

Explanation:

$Molarity=\frac{Moles}{Volume(L)}$

Molar concentration of the solution = 0.0366 M

Volume of the solution = 5.00 L

Moles of hydrated sodium carbonate = n

$0.0366 M=\frac{n}{5.00 L}$

$n=0.0366 M\times 5 mol=0.183 mol$

Weight of hydrated sodium carbonate = n = 52.2 g

Molar mass of hydrated sodium carbonate = 106 g/mol + x * 18 g/mol

$n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}$

$0.183 mol=\frac{52.2 g}{106 g/mol+x\times 18 g/mol}$

$106 g/mol+x\times 18 g/mol=\frac{52.2 g}{0.183 mol}$

By solving for x, we arrive at:

x = 9.95, approximating to 10

The integer x in the hydrate equals 10.

6 0

1 month ago