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tekilochka [14]
3 years ago
6

Describe the Sun’s interior. Include references to the main physical processes that occur at various depths within the sun.

Chemistry
1 answer:
ASHA 777 [7]3 years ago
6 0

Answer:

The sun is the large astronomical body that is located at the center of the solar system. The interior structure of the sun are as follows-

  • Core-  It represents the extreme interior portion of the sun. This is the hottest region of the sun and there occurs the process of constant nuclear fusion reactions that fuel the sun and helps in burning. Here the nucleus of two hydrogen atoms is fused together to form a heavy nucleus of a helium atom. It has a thickness of about 500 km.
  • Radiative zone- This is the layer that surrounds the core of the sun. From this layer, the energy is radiated outward where the protons carry these energies in the form of thermal radiation. Here, the process of radiation takes place.
  • Convective zone- This is the outer layer of the sun's interior. It is about 200,000 km thick, and in this layer, the energy constantly flows, allowing the heat to move upward by undergoing the process of convection.
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You've probably heard that some types of bacteria can cause infections and make you sick.
Novosadov [1.4K]

Answer:

<u>(A) Plants would probably not have enough nitrogen.</u>

Explanation:

<em>According to the passage, bacteria help us digest our food and make yogurt. But it is the bacteria in the soil specifically that "Cycle nitrogen through the ecosystem, which plants rely on"</em>

6 0
2 years ago
10. When a 13.6 g sample of a compound containing only magnesium and oxygen is decomposed, 5.4 g of oxygen is obtained. What is
irinina [24]

Answer:5.4 g / 13.6 g *100

Explanation:Its is the correct answer

7 0
3 years ago
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Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution
Zigmanuir [339]

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

Explanation:

Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.

The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.

7 0
3 years ago
The structure of a water molecule (H20) can be written as H-O-H, with each dash representing a chemical bond.
svet-max [94.6K]

Answer:

oxygen shows 2 bonds with hydrogen

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3 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
3 years ago
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