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3 years ago

Photosynthesis produces

Chemistry

1 answer:

dmitriy555 [2]3 years ago

8 0

Photosynthesis produces glucose and oxygen.... Respiration on the other hand would produce the other options.

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What will happen if you put a metal object to the beaker??

miss Akunina [59]

Answer:

umm i guess if there is a liquid inside it would rise to the top and the metal would sink to the bottom

3 0

3 years ago

If a man has brown eye with the recessive gene for blue eyes (Bb), each of his sex cells will have

iren [92.7K]

B - brown eyes
b - blue eyes

If we were to fill out a punette square each of his sex cells would contain at least one brown eyed gene.

8 0

3 years ago

Read 2 more answers

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

3 years ago

A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained. How many

beks73 [17]

<u>Answer: </u>

A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed

<u>Explanation:</u>

Given, the initial value of the sample, $A_0$ = 150mg

Final value of the sample or the quantity left, A = 18.75mg

Time = 11.4 days

The amount left after first half life will be ½.

The number of half-life is calculated by the formula

$\frac{A}{A_{0}}=\left(\frac{1}{2}\right)^{N}$

where N is the no. of half life

Substituting the values,

$\frac{18.75}{150}=\left(\frac{1}{2}\right)^{N}$

$\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{N}$

On equating, we get, N = 3

Therefore, 3 half-lives have passed.

7 0

3 years ago

What is the name of the structures that help move substances across a tract surface?

dimaraw [331]

Answer:

cilia and flagella

Explanation:

In prokaryotic species , cilia are present , and in eukaryotic species , flagella is present .

Cilia and flagella both have same function , i.e. , to enable the movement of the cell , along with the movement of some substance and direct the flow of these substance along the tracts.

Cilia and flagella are composed of basal bodies.

Hence , from the given statement of the question,

The correct term is cilia and flagella .

4 0

4 years ago