A balloon occupies a volume of 44 L at 200K, what will the temperature of the balloon become if it is changed to 500K?

Chemistry

1 answer:

natta225 [31]3 years ago

8 0

Answer:

110L

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 44L

Initial temperature (T1) = 200K

Final temperature (T2) = 500K

Final volume (V2) =..?

The new volume of the balloon can be obtained by using Charles' law equation as shown below:

V1/T1 = V2/T2

44/200 = V2/500

Cross multiply

200 x V2 = 44 x 500

Divide both side by 200

V2 = 44 x 500/200

V2 = 110L

Therefore, the new volume of the balloon is 110L

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what is the ratio of the rate of effusion of helium (atomic mass 4.00 amu) to that of oxygen gas (molecular mass 32.0 amu)?

nignag [31]

Answer:

3 : 1

Explanation:

Let the rate of He be R1

Molar Mass of He (M1) = 4g/mol

Let the rate of O2 be R2

Molar Mass of O2 (M2) = 32g/mol

Recall:

R1/R2 = √(M2/M1)

R1/R2 = √(32/4)

R1/R2 = √8

R1/R2 = 3

The ratio of rate of effusion of Helium to oxygen is 3 : 1

8 0

3 years ago

1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction

nikklg [1K]

Answer:

$H_{comb}=-4406kJ/mol$

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

$H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}$

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

$H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol$