Answer:
6.75 × 10⁻⁸is the value of the equilibrium constant at this temperature.
Explanation:
2H₂O(g) ⇄ 2H₂(g) + O₂(g)
Partial pressure of H₂O = 0.0500 atm
Partial pressure of H₂ = 0.00150 atm
Partial pressure of O₂ = 0.00150 atm
The expression of Kp for the given chemical equation is:
6.75 × 10⁻⁸is the value of the equilibrium constant at this temperature
The equation for carbon-14 emission by Radium-223 nuclei is given below:
Radioactivity is the spontaneous decay of a substance with emission of radiation.
The equation for carbon-14 emission by Radium-223 nuclei is given below:
In conclusion, the emission of carbon-14 by Radium-223 nuclei produces Lead-209 nuclei.
Learn more about radioactivity at: brainly.com/question/3603596
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Answer: Reducing agent in the given reaction is .
Explanation:
A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.
In the given reaction, oxidation state of sulfur in is +2 and
has 0 oxidation state.
In oxidation state of S is 2.5 and in
oxidation state of I is -1.
Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.
Thus, we can conclude that reducing agent in the given reaction is .
<u>Answer:</u>
<u>For a:</u> The edge length of the unit cell is 314 pm
<u>For b:</u> The radius of the molybdenum atom is 135.9 pm
<u>Explanation:</u>
To calculate the edge length for given density of metal, we use the equation:
where,
= density =
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal (molybdenum) = 95.94 g/mol
= Avogadro's number =
a = edge length of unit cell =?
Putting values in above equation, we get:
Conversion factor used:
Hence, the edge length of the unit cell is 314 pm
To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:
where,
R = radius of the lattice = ?
a = edge length = 314 pm
Putting values in above equation, we get:
Hence, the radius of the molybdenum atom is 135.9 pm
Answer:
4) Each cytochrome has an iron‑containing heme group that accepts electrons and then donates the electrons to a more electronegative substance.
Explanation:
The cytochromes are <u>proteins that contain heme prosthetic groups</u>. Cytochromes <u>undergo oxidation and reduction through loss or gain of a single electron by the iron atom in the heme of the cytochrome</u>:
The reduced form of ubiquinone (QH₂), an extraordinarily mobile transporter, transfers electrons to cytochrome reductase, a complex that contains cytochromes <em>b</em> and <em>c₁</em>, and a Fe-S center. This second complex reduces cytochrome <em>c</em>, a water-soluble membrane peripheral protein. Cytochrome <em>c</em>, like ubiquinone (Q), is a mobile electron transporter, which is transferred to cytochrome oxidase. This third complex contains the cytochromes <em>a</em>, <em>a₃</em> and two copper ions. Heme iron and a copper ion of this oxidase transfer electrons to O₂, as the last acceptor, to form water.
Each transporter "downstream" is <u>more electronegative</u><u> than its neighbor </u>"upstream"; oxygen is located in the inferior part of the chain. Thus, the <u>electrons fall in an energetic gradient</u> in the electron chain transport to a more stable localization in the <u>electronegative oxygen atom</u>.
