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3 years ago

How does solar radiation affect the thermosphere? God bless all! :)

Chemistry

2 answers:

DENIUS [597]3 years ago

5 0

Answer:

Thermospheric temperatures increase with altitude due to absorption of highly energetic solar radiation. ... Radiation causes the atmosphere particles in this layer to become electrically charged particles, enabling radio waves to be refracted and thus be received beyond the horizon.

Explanation:

AURORKA [14]3 years ago

3 0

Answer: How does solar radiation affect the thermosphere?

Explanation: During periods of high solar activity, the X-ray and ultraviolet radiation from the Sun increase, and the thermosphere swells as it sops up this increase in energy from the Sun. As the Sun approaches solar minimum, the thermosphere cools and shrinks as the intensity of the X-ray and ultraviolet radiation decreases.

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According to the following reaction, how many grams of calcium sulfate will be formed upon the complete reaction of 30.1 grams o

gtnhenbr [62]

Answer:

55.32 g of CaSO₄.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₂SO₄ + Ca(OH)₂ —> CaSO₄ + 2H₂O

Next, we shall determine the mass of Ca(OH)₂ that reacted and the mass of CaSO₄ produced from the balanced equation. This can be obtained as follow:

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

Molar mass of CaSO₄ = 40 + 32 + (16×4)

= 40 + 32 + 64

= 136 g/mol

Mass of CaSO₄ from the balanced equation = 1 × 136 = 136 g

SUMMARY:

From the balanced equation above,

74 g of Ca(OH)₂ reacted to produce 136 g of CaSO₄.

Finally, we shall determine the mass of CaSO₄ produced by the reaction of 30.1 g of Ca(OH)₂. This can be obtained as follow:

From the balanced equation above,

74 g of Ca(OH)₂ reacted to produce 136 g of CaSO₄.

Therefore, 30.1 g of Ca(OH)₂ will react to produce = (30.1 × 136)/74 = 55.32 g of CaSO₄.

Thus, 55.32 g of CaSO₄ were obtained from the reaction.

7 0

3 years ago

What product(s) forms at the cathode in the electrolysis of an aqueous solution of k2so4?

fenix001 [56]

Catod(-)
K⁺
2H20 +2e⁻ ---> H2 +2OH⁻

We can also say, that
K⁺ +OH⁻ +H2 = KOH +H2
At the cathode KOH and H2 are formed.

8 0

3 years ago

Calculate the concentration 3.8g of copper sulfate, CuSO4 dissolved in 250cm4 of water

11Alexandr11 [23.1K]

Answer:

Molarity = 0.08 M

Explanation:

Given data:

Mass of copper sulfate = 3.8 g

Volume of water = 250 cm³ (250/1000 = 0.25 L)

Concentration of solution = ?

Solution:

Number of moles of copper sulfate:

Number of moles = mass/molar mass

Number of moles = 3.8 g/ 159.6 g/mol

Number of moles = 0.02 mol

Concentration:

Molarity = Number of moles / volume in L

By putting values,

Molarity = 0.02 mol / 0.25 L

Molarity = 0.08 mol/L

Molarity = 0.08 M

6 0

3 years ago

A 5.00g piece of metal is heated to 100.0°C, then placed in a beaker containing 20.0 of water at 10.0°C. The temperature of the

Ahat [919]

Answer:

This metal has a specific heat of 0.9845J/ g °C

Explanation:
Step 1: Given data
q = m*ΔT *Cp
⇒with m = mass of the substance
⇒with ΔT = change in temp = final temperature T2 - initial temperature T1
⇒with Cp = specific heat (Cpwater = 4.184J/g °C) (Cpmetam = TO BE DETERMINED)

Step 2: Calculate specific heat
For this situation : we get for q = m*ΔT *Cp
q(lost, metal) = q(gained, water)

- mass of metal(ΔT)(Cpmetal) = mass of water (ΔT) (Cpwater)
-5 * (15-100)(Cpmetal) = 20* (15-10) * (4.184J/g °C =
-5 * (-85)(Cpmetal) = 418.4

Cpmetal = 418.4 / (-5*-85) = 0.9845 J/g °C

This metal has a specific heat of 0.9845J/ g °C

7 0

3 years ago

Refer to the periodic table tool and write the electron configurations of the following elements in both long and short terms

ahrayia [7]

Answer:-

Carbon

[He] 2s2 2p2

1s2 2s2 2p2.

potassium

[Ar] 4s1.

1s2 2s2 2p6 3s2 3p6 4s1

Explanation:-

For writing the short form of the electronic configuration we look for the nearest noble gas with atomic number less than the element in question. We subtract the atomic number of that noble gas from the atomic number of the element in question.

The extra electrons we then assign normally starting with using the row after the noble gas ends. We write the name of that noble gas in [brackets] and then write the electronic configuration.

For carbon with Z = 6 the nearest noble gas is Helium. It has the atomic number 2. Subtracting 6 – 2 we get 4 electrons. Helium lies in 1st row. Starting with 2, we get 2s2 2p2.

So the short term electronic configuration is [He] 2s2 2p2

Similarly, for potassium with Z = 19 the nearest noble gas is Argon. It has the atomic number 18. Subtracting 19-18 we get 1 electron. Argon lies in 3rd row. Starting with 4, we get 4s1.

So the short electronic configuration is [Ar] 4s1.

For long term electronic configuration we must write the electronic configuration of the noble gas as well.

So for Carbon it is 1s2 2s2 2p2.

For potassium it is 1s2 2s2 2p6 3s2 3p6 4s1

5 0

3 years ago