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satela [25.4K]
3 years ago
13

The main sugar that is used as a reactant in cellular respiration is called

Chemistry
2 answers:
vekshin13 years ago
6 0

Answer:

I believe that the answer is glucose?

Explanation:

if not, tell me the answer choices and i will help u from there

Llana [10]3 years ago
3 0
Answer:

Glucose


Explanation:

The two cellular processes illustrated by the test tubes are cellular respiration and photosynthesis. During cellular respiration, the reactants—glucose (sugar) and oxygen—combine together to form new products: carbon dioxide molecules and water molecules.
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Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.1
inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V

Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

This is less than zero[0]. Therefore,  it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.  

7 0
3 years ago
Commercial agriculture can often lead to water-quality problems. In one to two sentences, explain how two of those problems occu
anygoal [31]

Commercial agriculture can often lead to water-quality problems in the following ways:

  • The washing of fertilizers and pesticides into water bodies from farms.

<h3>What is water quality?</h3>

Water quality refers to the state of a water body that encompasses it's physical, chemical and biological characteristics.

The water quality of a water body is crucial to its suitability for domestic or drinking purpose.

Commercial agriculture greatly affects water quality in the following ways:

  • The washing of fertilizers and pesticides into water bodies from farms.

Learn more about water quality at: brainly.com/question/20848502

#SPJ1

4 0
2 years ago
In the following reaction, how many liters of oxygen will react with 270 liters of ethene (C2H4) at STP?
lawyer [7]
According to the equation you have 1 mole of C2H4 and 3 moles of O2.

1 • (22.4L / 270L) = 3 • (22.4L / x)

1/270L = 3/x

x = 3(270) / 1

x = 810 L

810 Liters of oxygen will react with 270 liters of ethene (C2H4) at STP
4 0
3 years ago
Prelab Questions:
liberstina [14]

Answer:

I know only first answer.....

Explanation:

when we say a substance is magnetic we meant that the substance can attract the metals or it has property like a magnet....

hope it helps...

4 0
3 years ago
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for
olga_2 [115]

<u>Answer:</u> The E^o_{cell}\text{ and }K_{eq} of the reaction is 0.78 V and 2.44\times 10^{26} respectively.

<u>Explanation:</u>

For the given half reactions:

Oxidation half reaction: Fe(s)\rightarrow Fe^{2+}+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V

Reduction half reaction: Cu^{2+}+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.34V

Net reaction: Fe(s)+Cu^{2+}\rightarrow Fe^{2+}+Cu(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.34-(-0.44)=0.78V

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

\Delta G^o=-nfE^o_{cell}

and,

\Delta G^o=-RT\ln K_{eq}

Equating these two equations, we get:

nfE^o_{cell}=RT\ln K_{eq}

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell = 0.78 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = 25^oC=[273+25]=298K

K_{eq} = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

2\times 96500\times 0.78=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=2.44\times 10^{26}

Hence, the E^o_{cell}\text{ and }K_{eq} of the reaction is 0.78 V and 2.44\times 10^{26} respectively.

3 0
4 years ago
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