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4 years ago

A surface wave is a combination of what two waves

Chemistry

2 answers:

Alika [10]4 years ago

4 0

transverse wave and a longitudinal wave

Hope it helps have a blessed day!:)

ycow [4]4 years ago

3 0

Answer:

Surface wave is a combination of transverse wave and longitudinal wave.

Explanation:

Surface wave is composed of two waves:

Longitudinal wave is the wave in which particle of the medium move in the same direction of the wave.
Transverse wave is the wave in which particle of the medium vibrates perpendicular to the direction of the wave.For example ripples on water.

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

VladimirAG [237]

Answer:

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

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

For 3: The equilibrium pressure of oxygen gas is 0.0577 atm

Explanation:

For 1:

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

For 2:

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 $1.386\times 10^{-2}$

For 3:

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

6 0

3 years ago

A sample of an unknown liquid has a volume of 30.0 mL and a mass of 6 g. What is its density? please Show your work or explain h

Leno4ka [110]

The density of the liquid is 0.2 g/mL.

The mass of the liquid is 6 g.

The volume of the liquid is 30.0 mL.

Density = mass/volume = 6 g/30.0 mL = 0.2 g/mL

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3 years ago

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artcher [175]

Answer:

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Explanation:

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notsponge [240]

Answer:

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Explanation:

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cupoosta [38]

Explanation:

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3 years ago

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