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romanna [79]
2 years ago
8

How long would it take the same amount of gaseous i2 to effuse from the same container under identical conditions?

Chemistry
1 answer:
77julia77 [94]2 years ago
7 0

(46 sec) x √ ((253.80894 g I2/mol) / (44.0128 g N2O/mol)) = 110 sec

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goldenfox [79]

Answer:

Solar panels and solar cells.

Explanation:

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5 0
3 years ago
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When water changes state from a liquid to a solid is energy absorbed or lost?
Brrunno [24]

Answer:

Loses

Explanation:

liquid changes into solid, heat is released. The energy released upon freezing, known as the enthalpy of fusion, is a latent heat, and is exactly the same as the energy required to melt the same amount of the solid.

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3 years ago
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What does this graph represent?<br> A. speed<br> B. velocity<br> C. acceleration<br> D. momentum?
Llana [10]
I think that it might represent Velocity
5 0
2 years ago
The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment
astraxan [27]

Answer:

The activation energy is =8.1\,kcal\,mol^{-1}

Explanation:

The gas phase reaction is as follows.

A \rightarrow B+C

The rate law of the reaction is as follows.

-r_{A}=kC_{A}

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = 10dm^{3}

Temperature "T" = 300 K

Volumetric flow rate of the reaction v_{o}=5dm^{3}s

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]

Rearrange the formula is as follows.

k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)

The feed has Pure A, mole fraction of A in feed y_{A_{o}} is 1.

\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacte.

=1(2-1)=1

Substitute the all given values in equation (1)

k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}

Therefore, the rate constant in case of the plug flow reacor at 300K is1.2s^{-1}

The rate constant in case of the CSTR can be calculated by using the formula.

\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)

The feed has 50% A and 50%  inerts.

Hence, the mole fraction of A in feed y_{A_{o}} is 0.5

\epsilon =y_{A_{o}}\delta

\delta = change in total number of moles per mole of A reacted.

=0.5(2-1)=0.5

Substitute the all values in formula (2)

\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}

Therefore, the rate constant in case of CSTR comes out to be 2.8s^{-1}

The activation energy of the reaction can be calculated by using formula

k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}

Substitute the all values.

=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}

=8.1\,kcal\,mol^{-1}

Therefore, the activation energy is =8.1\,kcal\,mol^{-1}

8 0
3 years ago
Classify..use the periodic table to classify the elements potassium, bromine a d argon according to how likely their atoms lose
iren2701 [21]

Answer:

Classification will be Potassium, Bromine, and Argon

Explanation:

  • Potassium is more likely to lose electrons and form positive ion
  • Bromine actually gain electrons and forms negative ion
  • Argon does not lose or gain electrons
6 0
3 years ago
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