What is the formula balance, using scrips for sulfur + oxygen

Which substance has vibrating particles in regular, fixed positions

slega [8]

most likely Ca(s). calcium ( Ca) is a solid, whose particles are packed closely together

8 0

4 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

Fiesta28 [93]

This is an incomplete question, here is a complete question.

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

$M_3O_4(s)\rightleftharpoons 3M(s)+2O_2(g)$

Substance ΔG°f (kJ/mol)

M₃O₄ -9.50

M(s) 0

O₂(g) 0

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? delta G°rxn = kJ / mol.

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

What is the equilibrium pressure of O₂(g) over M(s) at 298 K?

Answer :

The Gibbs energy of reaction is, 9.50 kJ/mol

The equilibrium constant of this reaction is, 0.0216

The equilibrium pressure of O₂(g) is, 0.147 atm

Explanation :

The given chemical reaction is:

$PCl_3(l)\rightarrow PCl_3(g)$

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

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

$\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}]$

where,

$\Delta G^o$ = Gibbs energy of reaction = ?

n = number of moles

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

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)]$

$\Delta G^o=9.50kJ/mol$

The Gibbs energy of reaction is, 9.50 kJ/mol

Now we have to calculate the equilibrium constant of this reaction.

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = 9.50kJ/mol = 9500 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K = equilibrium constant = ?

$9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)$

$K=0.0216$

The equilibrium constant of this reaction is, 0.0216

Now we have to calculate the equilibrium pressure of O₂(g).

The expression of equilibrium constant is:

$K=(P_{O_2})^2$

$0.0216=(P_{O_2})^2$

$P_{O_2}=0.147atm$

The equilibrium pressure of O₂(g) is, 0.147 atm

5 0

4 years ago

The vapor pressure of water at 80 °C is 355.3 torr. Calculate the vapor pressure in mm Hg and atm. Be sure your answers have the

Darya [45]

Answer:

0.4675 atm, 355.3 mmHg

Explanation:

Given:

Pressure = 355.3 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 355.3 / 760 atm = 0.4675 atm

The conversion of P(torr) to P(mmHg) is shown below:

$P(torr)=1\times P(mmHg)$

So,

Pressure = 355.3 mmHg

8 0

3 years ago

A 186 mL sample of a 0.275 M solution is left on a hot plate overnight. The following morning, the solution is 1.10 M. What volu

Flauer [41]

Answer:

139.5 mL.

Explanation:

To solve this question we will be making use of dilution law which is given by the equation below;

C1 × V1 = C2 × V2. Where C1 = the initial concentration, V1 = Initial volume, C2 = final concentration, V2 = Initial volume.

So, from the question we are given the following parameters; V1 = 186 mL, C1 = 0.275 M, C2 = 1.10 M and V2 = unknown (??).

Therefore, slotting in the above parameters into the dilution equation we have;

0.275 × 186= 1.10 × V2.

V2 = 0.275 × 186/ 1.10.

V2 = 46.5 mL.

The volume of solvent that has evaporated from the 0.275 M solution = 186 mL - 46.5 mL= 139.5 mL.

3 0

4 years ago

Read 2 more answers

Balancing equations <br><br> H2SO4 +<br> NaOH<br> Na2SO4 +<br> H20

notsponge [240]

Explanation:

<h3>2 in sodium hydroxide </h3><h3>And 2 in water </h3>

6 0

3 years ago

What is the formula balance, using scrips for sulfur + oxygen > sulfur dioxide