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Vadim26 [7]
2 years ago
10

Sara has made this list to summarize what she learned in science class today: 1. Core 2. Radiation zone 3. Convection zone What

is the best title for her list?
Chemistry
2 answers:
gizmo_the_mogwai [7]2 years ago
3 0
Answer: Layer's of the Sun.

Explanation:

The three words of the list correspond to the layers of the interior and atmosphere of the Sun.

This briefly tells you what every layer is:

Core: it is the central zone of the zone.

There is where the nuclear reactions (fusion) that transform hydrogen into helium and produce vast amounts of energy occur.

Radiation zone:

Is a very hot region in the interior of the sum where electromagnetic radiation is the mechanism of heat energy transference.

Convection zone

<span>It is the outermost layer of the interior of the Sun, which is less dense than the radiation zone and the main heat transference mechanism is convection.
</span>


n200080 [17]2 years ago
3 0

Answer:

Answer: Layer's of the Sun.

Explanation:

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Which of these was not proposed by john dalton in 1805?
wel

Answer:

Atoms are made of protons, neutrons and electrons.

Explanation:

The Dalton's atomic theory was an early attempt at describing the properties of atoms. It stipulated that atoms were the smallest indivisible particle of a substance. Chemical reactions occur as a result of a combination or separation of atoms. Atoms of the same element are exactly alike and differ from atoms of other elements. Atoms can neither be created nor destroyed.

As time went on, modern scientific evidence began to modify the original postulates of the Dalton's atomic theory. It was not postulated in 1805 that atoms were composed of subatomic particles; electrons, neutrons and protons. Dalton's theory held the atom to be 'indivisible'. However in 1897, JJ Thompson discovered the electron. Subsequently, the proton and neutrons were discovered. This shows that the atom in itself consisted of even smaller particles.

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1 year ago
If a cell is isotonic with a 0.88% nacl solution, how would an extracellular fluid with 1% nacl affect the cell?
EleoNora [17]
<span>The extracellular fluid is high in NaCl so the cell would be dehydrated further and the two solutions would equilibrate. Ultimately water would leave the cell and passes to </span>extracellular fluid and equilibrium is reached.
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How could installing new technology, such as scrubber machines, affect the factories required to install them? Name a positive a
Sindrei [870]

Answer:

Installing new technology, such as scrubbers, in factories will decrease their harmful emissions. This helps improve the safety of the surrounding community and the workers. But this technology is expensive and requires time and effort to install.

Explanation:

Hope this helped :)

5 0
1 year ago
How many sigma and pi bonds in propionic bond
Ierofanga [76]
Propanoic acid formula is ch ch 2 so it has 8 bonds
7 0
2 years ago
A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
Colt1911 [192]

Answer:

M_{Na^+}=1.36M

M_{Br^-}=1.58M

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

Once we've got the moles we compute the final volume via:

V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L

Thus, the molarity of the sodium atoms turn out into:

M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

Best regards.

7 0
1 year ago
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