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Deffense [45]
2 years ago
8

In a different experiment, a student places a piece of pure Ba(s) in a beaker containing 250.mL of 6.44MHCl(aq) and observes tha

t the rate of appearance of H2(g) bubbles when Ba(s) was added is much faster than the rate of appearance of bubbles when the Mg(s) was added. The student claims that the greater reactivity is due to the difference in ionization energy between Mg and Ba. Explain the difference in ionization energy between Mg and Ba in terms of atomic structure.
Chemistry
1 answer:
kkurt [141]2 years ago
6 0

Answer: The Ionization energy  of Barium is lower and hence it can lose its electrons easily.

Explanation:

Ionization energy is defined as the energy required to remove an electron from an isolated gaseous atom. It is represented as E_i

This energy will be higher for fully filled and half-filled electronic configuration than partially filled electronic configuration. This is so because half filled and fully filled configurations are stable.

Magnesium is the 12th element of the periodic table having electronic configuration of = [Ne]3s^2

Barium is the 56th element of the periodic table having electronic configuration of [Xe]6s^2

Both magnesium and barium belong to the same group as they have similar valence electrons. But as Barium is larger in size, the valence electrons lie farther from the nucleus and lesser energy is required to remove the electron and thus ionization energy will be lower for barium and thus will be more reactive.

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The chemical reaction that causes chromium to corrode in air is given by 4Cr+3O2→2Cr2O3 in which at 298 K ΔH∘rxn = −2256 kJ ΔS∘r
MAVERICK [17]

Answer:

-2092 kJ

Explanation:

Let's consider the chemical reaction that causes chromium to corrode in air.

4 Cr + 3 O₂ → 2 Cr₂O₃

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ΔH° - T × ΔS°

where,

  • ΔH°: standard enthalpy of the reaction
  • T: absolute temperature
  • ΔS°: standard entropy of the reaction

ΔG° = -2256 kJ - 298 K × (-0.5491 kJ/K)

ΔG° = -2092 kJ

5 0
2 years ago
Read 2 more answers
A chemist is working on a reaction represented by this chemical equation:
erastova [34]

Answer is: A. 1.81 mol.

Balanced chemical reaction: FeCl₂ + 2KOH → Fe(OH)₂ + 2KCl.

n(FeCl₂) = 4.15 mol; amount of iron(II) chloride.

n(KOH) = 3.62 mol; amount of potassium hydroxide, limiting reactant.

From chemical reaction: n(KOH) : n(Fe(OH)₂) = 2 : 1.

n(Fe(OH)₂) = n(KOH) ÷ 2.

n(Fe(OH)₂) = 3.62 mol ÷ 2.

n(Fe(OH)₂) = 1.81 mol; amount of iron(II) hydroxide.

6 0
2 years ago
Read 2 more answers
The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T
kifflom [539]

Answer : The equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

Explanation :

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

First we have to calculate the standard free energy of reaction (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]

where,

\Delta G^o = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]

\Delta G^o=-90.72kJ/mol

Now we have to calculate the value of K_p

\Delta G^o=-RT\ln K_p

where,

\Delta G_^o =  standard Gibbs free energy  = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

K_p = equilibrium constant  = ?

Now put all the given values in this expression, we get:

-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p

K_p=7.98\times 10^{15}

Now we have to calculate the value of K_p.

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

The expression for equilibrium constant will be :

K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium SO_2 pressure be, x

Pressure of SO_2 = Pressure of H_2S = x

Now put all the given values in this expression, we get

7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}

x=3.93\times 10^{-5}torr

Thus, the equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

4 0
2 years ago
Equal numbers of moles of He(g), Ar(g), and Ne(g) are placed in a glass vessel at room temperature. If the vessel has a pinhole-
mrs_skeptik [129]

Answer:

VP as function of time => VP(Ar) > VP(Ne) > VP(He).

Explanation:

Effusion rate of the lighter particles will be higher than the heavier particles. That is, the lighter particles will leave the container faster than the heavier particles. Over time, the vapor pressure of the greater number of heavier particles will be higher than the vapor pressure of the lighter particles.

=> VP as function of time => VP(Ar) > VP(Ne) > VP(He).

Review Graham's Law => Effusion Rate ∝ 1/√formula mass.

4 0
2 years ago
A chemist reacted 12.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equation to d
Nataly_w [17]

Answer: 91.73g of NaCl

Explanation:

First, we solve for the number of moles of F2 using the ideal gas equation

V = 12L

P = 1.5 atm

T = 280K

R = 0.082atm.L/mol/K

n =?

PV = nRT

n = PV /RT

n = (1.5x12)/(0.082x280)

n = 0.784mol

Next, we convert this mole ( i.e 0.784mol) of F2 to mass

MM of F2 = 19x2 = 38g/mol

Mass conc of F2 = n x MM

= 0.784 x 38 = 29.792g

Equation for the reaction is given below

F2 + 2NaCl —> 2NaF + Cl2

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass conc. of NaCl from the equation = 2 x 58.5 = 117g

Next, we find the mass of NaCl that reacted with 29.792g of F2.

From the equation,

38g of F2 redacted with 117g of NaCl.

Therefore, 29.792g of F2 will react with Xg of NaCl i.e

Xg of NaCl = (29.792 x 117)/38

= 91.73g

Therefore, 91.73g of NaCl reacted with f2

3 0
2 years ago
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