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jarptica
2 months ago
13

The diagram shows the different models of the atom that eventually led to the modern atomic theory. Four diagrams are shown labe

led Model A, Model B, Model C, Model D from left to right. The diagram labeled Model A has a small sphere with a positive sign on it. This small sphere at the center is surrounded by a shaded circle. The shaded circle has negative signs placed randomly inside it. The diagram labeled Model B is a large solid sphere. The diagram labeled Model C has a sphere at the center and a dotted cloud around the sphere. The dotted cloud is dense near the sphere and becomes faint as we move away from the sphere. The diagram labeled Model D has a sphere surrounded by three concentric circles. Each concentric circle has some spheres on it. Which model is based on Bohr's quantum model?
Chemistry
2 answers:
Tems11 [2.7K]2 months ago
8 0
The answer is B; I recently completed the test on Edge <3
lorasvet [2.7K]2 months ago
4 0
The answer is A on Edge 2021.
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In this lab, you will do experiments to identify types of changes. Using the question format you learned (shown above), write an
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Response:

How can you differentiate a physical change from a chemical change?

Clarification:

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2 months ago
A 250 ml flask contains 3.4 g of neon gas at 45°c. Calculate the pressure of the neon gas inside the flask.
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The solution to your inquiry yields P = 17.73 atm. Explanation: The volume V is 250 ml, equivalent to 0.25 liters (L), with a mass of 3.4 g and a temperature of 45°C, which converts to 318°K. We utilize the ideal gas law PV = nRT for the calculations.
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1 month ago
A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press
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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
1 month ago
Hydrogen has three isotopes with mass numbers of 1, 2, and 3 and has an average atomic mass of 1.00794 amu. This information ind
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The answer is actually 3, believe me.

Explanation:

7 0
3 months ago
Read 2 more answers
Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul
Anarel [2989]

Answer:

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Explanation:

Assuming all calculations occur at standard pressure and a temperature of -1.72°C :

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Where

n is the number of moles of hydrogen

n is the mass of hydrogen

\rho is the density of hydrogen

6 0
3 months ago
Read 2 more answers
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