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lapo4ka [179]
3 years ago
15

The figure above shows the Earth at two different positions in its orbit around the Sun. Which position corresponds to summer in

the Northern Hemisphere?
A.

position B

B.

position A

C.

both position A and position B

D.

neither position A nor position B

Chemistry
2 answers:
iren [92.7K]3 years ago
7 0

Answer:

A

Explanation:

In position A, the Southern Hemisphere is exposed to more direct rays from the Sun than the Northern Hemisphere. This corresponds to summer in the Southern Hemisphere.

In position B, the Northern Hemisphere is exposed to more direct rays from the Sun than the Southern Hemisphere. This corresponds to summer in the Northern Hemisphere.

Since Texas is in the Northern Hemisphere, the correct answer is position B.

hope I helped :)

Lelu [443]3 years ago
3 0

Answer:

A. Position B.

Explanation:

In the summer the northern hemisphere is closer to the Sun so it's Position B.

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The question requires us to explain the differences in radii of neutral atoms, cations and anions.

To answer this question, we need to keep in mind that a neutral atom presents the same number of protons (positive particles) and electrons (negative particles). Another important information is that the protons are located in the nucleus of the atom, while the electrons are around the nucleus. Also, there is an electrostatic force between protons and electrons, which means that they the protons tend to attract the electrons to the nucleus.

While a neutral atom presents the same number of protons and electrons, a cation is an ion with positive charge, which means it has lost one or more electrons. In a cation, the balance between protons and electrons doesn't exist anymore: now, there is more positive than negative charge (more protons than electrons), and the overall attractive force that the protons have for the electrons is increased. As a result, the electrons stay closer to the nucleus and the radius of a cation is smaller than the neutral atom from which it was derived.

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We can use the following equation to calculate the volume of more concentrated solution required:

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where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

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<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

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