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almond37 [142]
3 years ago
10

Stellar fusion produces atoms up to and including ___________ Question 51 options: Helium (atomic number 2) Carbon (atomic numbe

r 6) Iron (atomic number 26) Lead (atomic number 82) Question 52 (1 point)
Chemistry
1 answer:
Ganezh [65]3 years ago
3 0
<h2>Iron (Z = 26)</h2>

Explanation:

  • Stellar fusion is the nuclear reaction that occurs in stars.
  • In this reaction, the nuclei build to form heavier elements.
  • Stars fuse elements of lighter ones to heavier ones
  • This fusion reaction occurs in their cores.
  • During this process, a lot of energy is released.
  • This process is also known as stellar nucleosynthesis as there is a synthesis of the different nucleus.
  • Stellar fusion produces lighter elements of many types including iron(Fe) and nickel(Ni) in many of the giant stars.
  • These elements remain trapped in the cores of stars.
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Answer:

No.

Explanation:

The reason comes the <em>Law of Conservation of Mass</em>.

In an ordinary chemical reaction, <em>you cannot create or destroy atoms</em>.

So, you must have as many atoms at the beginning of a reaction (in the reactants) as at the end (in the products)

We use this principle to balance chemical equations.

For example, the equation for the formation of water from hydrogen and oxygen is

2H₂ + O₂ ⟶ 2H₂O

There are four atoms of H and two of O both before and after the reaction.

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Describe what happens in a condensation reactions
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8 0
1 year ago
C-12, c-13, and c-14 have the same number of protons but different numbers of neutrons so they are
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3 years ago
Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

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}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
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