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Sergio039 [100]
3 years ago
9

Which of the following is the balanced chemical equation for photosynthesis?

Chemistry
2 answers:
LuckyWell [14K]3 years ago
8 0
The first one, <span>6H2O + 6CO2 → 6O2 + C6H12O6 is.</span>
Sauron [17]3 years ago
4 0

Aerobic cellular respiration is the process by which all organisms use C6H12O6 (glucose) and 6O2 to make 6CO2, 6H2O and 36 ATP. It is called aerobic because it requires oxygen.

soo that means that the anwser is

a.

6H2O + 6CO2   ® 6O2 + C6H12O6

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Calculate the oxidation numbers of the elements in this equation<br> 2 Al2O3 → 4 Al + 3 O2
riadik2000 [5.3K]

Explanation:

Answer

Open in answr app

The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.

Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O )  = 0

2× ( Oxidation number of Al) +3(−2)=0

2× ( oxidation number of Al) +6

∴ Oxidation number of Al =+3

5 0
3 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
How can i toast a marshmallow using convection ?
Elina [12.6K]
Convection is where the heat source heats the air, and the air transfers the heat to object. Some people prefer to heat their marshmallow through convection, this is achieved by holding the marshmallow high above the flames where the heated air is rising, transforing the heat into the marshmallow.
7 0
2 years ago
How many atoms are in the 3.5 moles of arsenic atoms?
Dmitry_Shevchenko [17]
Number of atom in one mole = 6.022 × 10²³
So, in 3.5 moles, it would be: = 6.022 × 10²³ * 3.5 = 2.1 × 10²⁴

In short, Your Answer would be 2.1 × 10²⁴ atoms

Hope this helps!
6 0
3 years ago
Read 2 more answers
Germanium is classified as a<br> (1) metal (3) nonmetal<br> (2) metalloid (4) noble gas
hodyreva [135]
Germanium is classified as a metalloid. 
6 0
3 years ago
Read 2 more answers
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