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3 years ago

A container is filled with a gas. The container has a wall controlled by a moveable piston. The

Chemistry

1 answer:

ioda3 years ago

6 0

Answer:

The new volume will be 3.67 L.

Explanation:

As the volume increases, the gas particles (atoms or molecules) take longer to reach the walls of the container and therefore collide with them fewer times per unit of time. This means that the pressure will be lower because it represents the frequency of collisions of the gas against the walls. In this way pressure and volume are related, determining Boyle's law which says:

"The volume occupied by a certain gaseous mass at constant temperature is inversely proportional to pressure"

Boyle's law is expressed mathematically as:

P*V=k

Now it is possible to assume that you have a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment. If you vary the volume of gas to a new value V2, then the pressure will change to P2, and it will be fulfilled:

P1 * V1 = P2 * V2

In this case:

P1= 1.85 atm
V1= 4.64 L
P2= 2.34 atm
V2= ?

Replacing:

1.85 atm* 4.64 L= 2.34 atm* V2

Solving:

$V2= \frac{1.85 atm* 4.64 L}{2.34 atm}$

V2= 3.67 L

The new volume will be 3.67 L.

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Consider the decomposition of hydrogen peroxide reaction. If the rate of decomposition of hydrogen peroxide at a particular temp

Natasha2012 [34]

Answer:

0.049 mol/L.s

Explanation:

The decomposition of hydrogen peroxide is:

$H_2O_2 \to H_2O + \dfrac{1}{2}O_2$

$Rate = -\dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{ 2 \Delta [H_2O_2]}{\Delta t}$

The rate of decomposition reaction = the rate of formation of $H_2O$ = 0.098 mol/L.s

∴

Rate of formation of $O_2$ $= \dfrac{ rate \ of \ reaction }{2}$

$= \dfrac{ 0.098 }{2}$

= 0.049 mol/L.s

8 0

3 years ago

Which best describes the relationship between air temperature and high and low pressure air

Vsevolod [243]

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5 0

3 years ago

How many moles of water, H2O contain 2.0x10^22 molecules of water

Licemer1 [7]

I mole of water has an Avogadro number of molecules.

1 mole = 6.02 * 10^ 23 molecules.

6.02 * 10^ 23 molecules = 1 mole of water

1 molecule = 1/(6.02 * 10^23) mole of water

2.0 * 10^22 molecules would have = (2*10^22) * 1/(6.02*10^23)

= 0.033

2* 10 ^22 molecules of water would have 0.033 moles of water.

5 0

3 years ago

The entropy of a system at 337 K increases by 221.7 J/mol•K. The free energy value is found to be –717.5 kJ/mol. Calculate the c

OlgaM077 [116]

Answer: The change in enthalpy for the given system is -642.8 kJ/mol

Explanation:

To calculate the change in enthalpy for given Gibbs free energy, we use the equation:

$\Delta G=\Delta H-T\Delta S$

where,

$\Delta G$ = Gibbs free energy = -717.5 kJ/mol = -717500 J/mol (Conversion factor: 1 kJ = 1000 J)

$\Delta H$ = change in enthalpy = ?

T = temperature = 337 K

$\Delta S$ = change in entropy = 221.7 J/mol.K

Putting values in above equation, we get:

$-717500J/mol=\Delta H-(337K\times 221.7J/mol.K)\\\\\Delta H=-642787J/mol=-642.8kJ/mol$

Hence, the change in enthalpy for the given system is -642.8 kJ/mol

4 0

4 years ago

Calculate ΔS° for the reaction: 4Cr(s) + 3O2(g) → 2Cr2O3(s), Substance: Cr(s) O2(g) Cr2O3(s), S°(J/K⋅mol): 23.77 205.138 81.2

Mariana [72]

Answer: The value of $\Delta S^o$ for the reaction is 1051.93 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

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

For the given chemical reaction:

$4Cr(s)+3O_2(g)\rightarrow 2Cr_2O_3(s)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cr_2O_3(s))})]-[(4\times \Delta S^o_{(Cr(s))})+(3\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(Cr_2O_3(s))}=881.2J/K.mol\\\Delta S^o_{(O_2(g))}=205.13J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (881.2))]-[(4\times (23.77))+(3\times (205.13))]\\\\\Delta S^o_{rxn}=1051.93J/K$

Hence, the value of $\Delta S^o$ for the reaction is 1051.93 J/K

5 0

4 years ago