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Anit [1.1K]
3 years ago
5

In two or more complete sentences, explain how fossil fuels play a role in your daily life.

Chemistry
1 answer:
frutty [35]3 years ago
7 0
Fossil fuels can be seen every where like when you go get gas for your car, that gas is a fossil fuel. Another thing is if you have a gas stove the propane that you use is a fossil fuel and if you use your stove every day then it plays an pretty important role.
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A solution is prepared by dissolving 38.6 g sucrose (C12H22O11) in 495 g of water. Determine the mole fraction of sucrose if the
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Answer:

4.09×10⁻³ is the mole fraction of sucrose

Explanation:

Mole fraction = Moles of solute or solvent/ Total moles

Let's convert the mass to moles (mass / molar mass)

38.6 g / 342 g/m = 0.113 moles of sucrose

495 g / 18 g/m = 27.5 moles of water

Total moles = 0.113 m + 27.5 m = 27.0613 moles

Mole fraction of sucrose = Moles of sucrose / Total moles

0.113 m / 27.0613 moles = 4.09×10⁻³

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What are two ways to describe concentration? How do they differ from one another?
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According to Hess’s law, if a series of intermediate reactions are combined, the enthalpy change of the overall reaction is
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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

3 0
4 years ago
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