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Alexxx [7]
4 years ago
14

4. Is it possible for two different compounds to be made from the exact same two elements? Why or why not? With a limited number

of elements (less than 120 are known), does this mean we also have a small number of compounds or do we have a large number of compounds in this world?
Chemistry
1 answer:
densk [106]4 years ago
3 0

no it is not possible, because they both have the same number of valence electrons in each element. in a compound you are supposed to have two or more elements that have different numbers of valence electrons so when put together they for a compound.

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A 0.15 M NaOH solution has a volume of 0.125 L but is then diluted to 0.15 L. What is the concentration of the new solution
tresset_1 [31]

Answer:

.125 M

Explanation:

.15 M/L   * .125 L = .01875 moles

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2 years ago
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.
VladimirAG [237]

<u>Answer:</u>

<u>For 1:</u> The standard Gibbs free energy change of the reaction is 10.60 kJ/mol

<u>For 2:</u> The equilibrium constant for the given reaction at 298 K is 1.386\times 10^{-2}

<u>For 3:</u> The equilibrium pressure of oxygen gas is 0.0577 atm

<u>Explanation:</u>

  • <u>For 1:</u>

The equation used to calculate standard Gibbs free energy change of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]

For the given chemical reaction:

M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)

The equation for the standard Gibbs free energy change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})]

We are given:

\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol

Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol

  • <u>For 2:</u>

Relation between standard Gibbs free energy and equilibrium constant follows:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol    (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 298 K

K_{eq} = equilibrium constant = ?

Putting values in above equation, we get:

10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}

Hence, the equilibrium constant for the given reaction at 298 K is 1.386\times 10^{-2}

  • <u>For 3:</u>

The expression of K_{eq} for above equation follows:

K_{eq}=p_{O_2}^{3/2}

The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.

Putting values in above expression, we get:

1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm

Hence, the equilibrium pressure of oxygen gas is 0.0577 atm

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Answer:

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