Write the half-reaction for ribose conversion to CO2. Is it an oxidation- or reduction- half reaction

Answers

Answer 1
Answer:

Answer:

5H_2O+C_5H_(10)O_5\rightarrow 5CO_2+20H^++20e^-

Explanation:

Hello.

In this case, when ribose (C₅H₁₀O₅) yields carbon dioxide (CO₂) we write:

C_5H_(10)O_5\rightarrow CO_2

Which needs to be balanced by adding water and hydrogen ions:

5H_2O+C_5H_(10)O_5\rightarrow 5CO_2+20H^++20e^-

You can also see that there are 20 transferred electrons, since the carbon atoms in the ribose have 0 as their oxidation state and the carbon atoms in the carbon dioxide have +4 as the oxidation state, thus, each carbon transfers  4 electrons, a five carbon atoms transfer 20 electrons overall.

In such a way, since the carbon is increasing its oxidation state, such half reaction is an oxidation half reaction.

Best regards.


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How many moles of h2 are need to produce 11.21 g of nh3
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Based on your knowledge of waves and fider optic cables, which type of wave would be found in a fiber optic cable? Explain how you got the answer and i will mark brainliest!!!! Thank you sooooo much!

Answers

The answer is; Transverse waves

Fiber optic cables transmit information using monochromatic light pulses. All electromagnetic waves (light included) are transverse waves.  This means that the particles move perpendicular to the direction of the wave. This is unlike sound waves that are longitudinal waves (particles move parallel to the direction of the wave).


Final answer:

Fiber optic cables transmit data using light waves, which are essentially electromagnetic waves. The wave characteristics of light, especially total internal reflection, interference, and diffraction, facilitate effective data transmissions through these fibers. Factors like high bandwidth, low signal loss, and reduced crosstalk further contribute to their advantage over traditional cables.

Explanation:

Based on the principles of optics, electromagnetic waves, particularly light waves, are what you could find in a fiber optic cable. Fiber optic cables work by transmitting data as pulses of light through strands of fiber made from glass or plastic. This process utilizes the characteristic phenomenon of total internal reflection. When light rays enter the fiber, they bounce off the walls of the fiber cable, undergoing multiple total internal reflections, which ensures that no light escapes the fiber and all signals are conveyed effectively.

Light's wave characteristics are crucial in enabling this functionality. The wave nature of light helps explain properties such as interference and diffraction, essential for the transmission of data in fiber optic networks. These principles are especially relevant when light interacts with small objects such as the core/cladding of the fiber, a subject area often referred to as wave or physical optics.

Another advantage is the high bandwidth of fiber optics, made possible because lasers can emit light with characteristics that allow far more data transmission than electric signals on a single conductor. Meanwhile, properties like low loss and reduced crosstalk enhance the functional superiority of fiber optic cables over traditional copper cable systems.

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For each of the following redox reactions in the, identify the oxidizing agent and the reducing agent.2Na(aq)+2H2O(l)→2NaOH(aq)+H2(g)
C(s)+O2(g)→CO2(g)
2MnO−4(aq)+5SO2(g)+2H2O(l)→2Mn2+(aq)+5SO2−4(aq)+4H+(aq)

Answers

anywhere between 5sogt2 and 776sogt2

What is the symbol of the element that is classified as an alkali metal and is in period 4?

Answers

Answer:

Potassium (K) [First element in period 4]

Write the balanced equation for the equilibrium reaction for the dissociation ofsilver chloride in water, and write the K expression for this reaction. Then create an ICE chart. Since we know the equilibrium concentration of the silver ion, we can solve for Ksp.Does it agree with the literature value

Answers

Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:

Ksp=[Ag^+][Cl^-]

And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

       \ \ \ \ \ \ \ \ \ \ \ \ \ \ AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)

I          -                   0             0

C        -                   +x           +x

E        -                    x             x

Which leads to the following modified equilibrium expression:

Ksp=x^2

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.

Regards!

A gas mixture with a total pressure of 770 mmHg contains each of the following gases at the indicated partial pressures: 128 mmHg CO2, 232 mmHg Ar, and 183 mmHg O2. The mixture also contains helium gas. What is the partial pressure of the helium gas?

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The partial pressure of  Helium gas can be calculated by applying Daltons law of partial pressure and that comes out to be 227 mmHg.

What is  Dalton's law of partial pressure?

Dalton's law of partial pressure helps in calculating the total pressure exerted by all the gases that are present in a mixture that is equal to the sum of partial pressure of all individual gases.

Partial pressure is the pressure that is exerted by one gas in a mixture of gases on the walls of the container. This law is applicable to only mixture of gases that are ideal gases. Ideal gases are the one which do not have any forces of attraction or repulsion between the particles of thee gas.

In our question

p_(total) =p_{CO_(2) }+p_(Ar)+p_{O_(2) } +p_{{He} }

p_{{He} } =p_(total)-[p_{CO_(2) }+p_(Ar)+p_{O_(2) } ]

substituting all given values

p_{{He} } =770-(128+232+183)

      =227mmHg

Thus the partial pressure of Helium gas is 227mmHg

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Answer:

227 mmHg

Explanation:

Ptotal= PCO2 +PAr+PO2+PHe

PHe= Ptotal-PCO2 -PAr-PO2

PHe= (770 mmHg)-(128 mmHg)-(232 mmHg)-(183 mmHg)

PHe=227 mmHg

Among these element groups, which is least likely to form ions?: 7A 3A 1A 4A

Answers

The  element  group  which  is least  likely  to form ions  is 4A  group


 explanation

Element   of   group  4A   has   4  valence  electrons.  for example Carbon which has   [He] 2S2 2P2  electron configuration has 4 valence electrons.

Since 4A elements has 4 valence electrons ,  it means  that  their valence shell  is half  filled which is  relatively stable. For  this reason 4A element readily  form  ions.