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2 years ago

1. What would happen if you aimed light from the magenta part of the Horsehead Nebula through a spectrograph?

1 answer:

4 0

Given what we know, we can confirm that if you aimed light from the magenta part of the Horsehead Nebula through a spectrograph we would be able to determine more precisely the structure and details of the cloud.

<h3>How do we use Spectrums in order to understand stars?</h3>

The spectrums recorded by scientists, such as those of stars or nebulas like the horsehead nebula can tell us a great deal about the composition of said entities. Studying the spectrum can tell scientists about the chemical composition of stars or nebulas, such as information about the elements that form them, like their temperatures and densities.

<h3 /><h3>How would a discontinuous emission of hydrogen gas look in the spectrum?</h3>

This would appear as pauses in the lines of the spectrum. If the emission of the hydrogen gas were constant, there would be a continuous line on the spectrum graph to indicate the illuminated hydrogen, though if this line were discontinuous, we would be able to assume that its source is emission from another gas instead.

Therefore, we can confirm that spectrography is an essential part of scientific discovery pertaining to our universe. It allows us to study the chemical composition of stars and nebulas, and determine the sources of certain emissions like that of hydrogen gases.

To learn more about spectrographs visit:

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How did Bohr refine the model of the atom? How did Bohr refine the model of the atom? He developed electrochemistry. He discover

BartSMP [9]

Answer:

He developed the concept of concentric electron energy levels

Explanation:

Before Bohr's model, Rutherford's model was proposed. This model explains most of the properties of the atom but failed to explain the stability of the atom.

As per Rutherford's model, electrons revolve around the nucleus in the orbit.

But revolving electron in their orbit around nucleus would give up energy and so gradually move towards the nucleus and therefore, eventually collapse.

Bohr's proposed that the electrons around the nucleus move orbit of fixed energy called "stationary states". Electrons in these stationary states do not radiate energy.

Therefore, proposal of concentric electron energy levels refine the atomic models.

5 0

3 years ago

How many electrons are there in carbonate ion, CO3^-2?

NISA [10]

It should be 24 electrons

3 0

3 years ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

<u>Moles of potassium iodide = 0.48 moles </u>

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

<u>Moles of lead(II) nitrate = 0.342 moles </u>

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

<u>Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.</u>

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

<u>So, Concentration = 0.102/0.585 M = 1.174 M</u>

<u>Statement A is correct.</u>

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

<u>Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g </u>

<u>Statement B is correct.</u>

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

<u>So, Concentration = 0.48/0.585 M = 0.821 M</u>

<u>Statement C is correct.</u>

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

<u>So, Concentration = 0.684/0.585 M = 1.169 M</u>

<u>Statement D is incorrect.</u>

4 0

3 years ago

1. Which of the following is an SI base unit for time? (2 points)

oee [108]

Number one would be Decades

6 0

3 years ago

Water will move from a ____________ salt solution to a ____________ salt solution when they are across a differentially permeabl

const2013 [10]

The answers are low concentrated (dilute) and high concentrated respectively.

As the low concentrated salt solution has a higher water potential than that of the high concentrated salt solution, water molecules will flow from the region of higher water potential to the region of lower water potential, thus from the dilute salt solution to the high concentrated salt solution. This is due to the movement called osmosis. Note that osmosis also requires water to flow through a differentially permeable membrane, which means the membrane can allow certain substances (not all) to go in or out. If the differentially permeable membrane is not present, the movement of water molecules may be regarded as diffusion.

Therefore, the answers for the blanks are low concentrated and high concentrated.

6 0

3 years ago