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2 years ago

In the resonance form of ozone shown below, the formal charge on the central oxygen atom is ________.

1 answer:

8 0

The answer to this question would be: +1

In the resonance form of ozone, the oxygen in the middle will have 3 bonds with the other oxygen on the left and the right. One of the oxygen that gets 2 bonds will have 0 formal charges and the one with 1 bond will have -1 formal charge. The sum of the formal charge will be 0.

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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

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2

sashaice [31]

Answer:

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

(Ans)

ΔHf° of CaC2 = -59.0 kJ/mol

Explanation:

CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ

ΔHrxn = −127.2kJ

ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);

ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn

Where

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol

ΔHf°(CaC2) = -59.0 kJ/mol

7 0

2 years ago

How many electrons are transferred in the given redox reaction? Zn+2AgNO3⟶2Ag+Zn(NO3)2

Vsevolod [243]

Answer:

2 electrons are transfered in this reaction.

Explanation:

Oxidation is a reaction where an atom, ion, or molecule loses electrons, while reduction corresponds to the electron gain of an atom, ion, or molecule.

In an oxidation-reduction reaction two simultaneous processes take place, oxidation and reduction.

So, oxidation-reduction (redox) reactions involve the transfer of electrons between chemical species. They are also called electron transfer reactions since the particle that is exchanged is the electron.

In this case:

Zn(s) ⇒ Zn²⁺(aq) + 2 e⁻

2 Ag⁺ (aq) + 2 e⁻ ⇒ 2 Ag(s)

So, zinc metal loses two electrons to form the zinc(II) ions, while the two silver ions each gain one electron to form two silver metal atoms.

Then, Zn is a reducing agent (The reducing agent is the one that provides the electrons, oxidizing itself), AgNO3 is an oxidizing agent (The oxidizing agent is the one that traps the electrons, reducing itself).

Finally, you can see that 2 electrons are transfered in this reaction.

7 0

2 years ago

The density of o2 gas at 16 degrees Celsius and 1.27atm is?

velikii [3]

Answer:

The density of O₂ gas is 1.71 $\frac{g}{L}$

Explanation:

Density is a quantity that allows you to measure the amount of mass in a given volume of a substance. So density is defined as the quotient between the mass of a body and the volume it occupies:

$density=\frac{mass}{volume}$

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

So, you can get:

$\frac{n}{V} =\frac{P}{R*T}$

The relationship between number of moles and mass is:

$n=\frac{mass}{molar mass}$

Replacing:

$\frac{\frac{mass}{molar mass} }{V} =\frac{P}{R*T}$

$\frac{mass}{V*Molar mass} =\frac{P}{R*T}$

So:

$\frac{mass}{V} =\frac{P*molar mass}{R*T}$

Knowing that 1 mol of O has 16 g, the molar mass of O₂ gas is 32 $\frac{g}{mol}$ .

Then:

$\frac{mass}{V} =\frac{P*molar mass of O_{2} }{R*T}$

In this case you know:

P=1.27 atm
molar mass of O₂= 32 $\frac{g}{mol}$ .

R= 0.0821 $\frac{atm*L}{mol*K}$

T= 16 °C= 289 °K (0°C= 273°K)

Replacing:

$density=\frac{mass}{V} =\frac{1.27atm*32\frac{g}{mol} }{0.0821\frac{atm*L}{mol*K} *289 K}$

Solving:

density= 1.71 $\frac{g}{L}$

The density of O₂ gas is 1.71 $\frac{g}{L}$

3 0

2 years ago

The following chemical reaction takes place in aqueous solution: AgF (aq)+ NH4Cl (aq) →AgCl(s)+ NH4F(aq) Write the net ionic equ

Nadusha1986 [10]

Answer:

The net ionic equation is: Ag⁺ (aq) + Cl ⁻ (aq) → AgCl (s)

Explanation:

1) Start by writing the total ionic equation:

The total ionic equation shows each aqueous substance in its ionized form, while the solid or liquid substances are shown with their chemical formula.

These are the ionic species:

AgF (aq) → Ag⁺ (aq) + F⁻ (aq)

NH₄Cl (aq) → NH₄⁺ (aq) + Cl ⁻ (aq)

NH₄F(aq) → NH₄⁺ (aq) + F⁻ (aq)

Then, replace each chemical formula in the chemical equation by those ionic forms:

Ag⁺ (aq) + F⁻ (aq) + NH₄⁺ (aq) + Cl ⁻ (aq) → AgCl (s) + NH₄⁺ (aq) + F⁻ (aq)

That is the total ionic equation.

2) Spectator ions:

The ions that appear in both the reactant side and the product side are considered spectator ions (they do not change), and so they are canceled.

In our total ionic equation they are F⁻ (aq) and NH₄⁺ (aq).

After canceling them, you get the net ionic equation:

3) Net ionic equation:

Ag⁺ (aq) + Cl ⁻ (aq) → AgCl (s) ← answer

4 0

2 years ago