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1 year ago

Science please help me with this

Chemistry

2 answers:

nataly862011 [7]1 year ago

7 0

Answer:

a lack of genetic variety

first one is the correct one

kvv77 [185]1 year ago

6 0

A lack of genetic variety

The first one

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Fire extinguishers that spray carbon dioxide on the fire, work very effectively because it forms a blanket around the burning ma

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I think it will option B it will retain enough heat

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2 years ago

List two uses of H2SO4<br>

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2 years ago

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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 is $1.2s^{-1}$

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

2 years ago

Okay #PlatoFam. I need help!!!

Vikki [24]

-2 for O
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+1 for Li
+3 for Fe

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3 years ago

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1. Based on the observed performance of the air bag models and the amounts of sodium bicarbonate and acetic acid (vinegar) neede

Ira Lisetskai [31]

Sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags since the require more mass and produce less gas.

<h3>Which is the better chemical for an airbag?</h3>

The chemical equation for the production of nitrogen gas from sodium azide is given below:

NaN₃ → 2 Na + 3 N₂

1 mole or 66 go of sodium azide produces 3 moles or 67.2 L of nitrogen gas.

The equation for the production of carbon dioxide from sodium bicarbonate and acetic acid is given below:

Na₂CO₃ + CH₃COOH → CH₃COONa + CO₂ + H₂O

1 mole, 106 g of Na₂CO₃ and 1 mole, 82 g of CH₃COOH are required to produce 1 mole or 22.4 L of CO₂.

The mass of sodium azide required is less than that of sodium bicarbonate and acetic acid required. Also, sodium azide produces a greater volume of gas. Therefore, sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags.

In conclusion, sodium azide is a better choice in airbags.

Learn more about airbags at: brainly.com/question/14954949

#SPJ1

7 0

1 year ago