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Sedaia [141]
3 years ago
15

which number should be placed before F2 on the reactants side equation to make equation balanced? Xe + ___F2 > XeF4

Chemistry
2 answers:
jolli1 [7]3 years ago
5 0

Answer:

The number that should be placed before F₂ is 2

Explanation:

For a chemical reaction to be balanced we need to have the same amount of mass in the reactants and in the products because by the<em> law of conservation of mass</em>, mass cannot be created or destroyed.

We can acomplish this by having the same amount of atoms of each element in the reactant side that in the product side.

For example:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂0

In this case in the reactants side we have 3 C, 8 H and 10 O, so we have  the same amount in the product side, so the equation is balanced.

In the equation Xe + F₂ → XeF₄ we have 1 Xe in each side, 2 F in the reactant side and 4 F in the product side, so we multiply the 2 F in reactan side to balance the equation.

Xe + 2Fe₂ → XeF₄

Ivahew [28]3 years ago
3 0

There are four F atoms on the products side.

Since two more F atoms are required on the reactant side, you multiply the number of F2 molecules by two.

So 2 should be placed in front of F2

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Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol

b) To calculate the constant we have the following expression:

lnK=-\frac{deltaG_{rxn} }{RT}

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066

c) The equilibrium pressure of O₂ over M is:

K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm

3 0
3 years ago
Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

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