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mrs_skeptik [129]
3 years ago
13

Question 13: Consider the strength of the Hβ absorption line in the spectra of stars of various surface temperatures. This is th

e amount of light that is missing from the spectra because Hydrogen electrons have absorbed the photons and jumped from level 2 to level 4. How do you think the strength of Hβ absorption varies with stellar surface temperature?
Chemistry
1 answer:
finlep [7]3 years ago
5 0

Answer:

The absorption and strength of the H-beta lines change with the temperature of the stellar surface, and because of this, one can find the temperature of the star from their absorption lines and strength. To better comprehend, let us look into the concept of the atom's atomic structure.  

Atoms possess distinct energy levels and these levels of energy are constant, that is, the temperature has no influence on it. However, temperature possesses an influence on the electron numbers found within these levels of energy. Therefore, to generate an absorption line of hydrogen in the electromagnetic spectrum's visible band, the electrons are required to be present in the second energy level, that is when it captivates a photon.  

Therefore, after captivating the photons the electrons jump from level 2 to level 4, which shows that there is an increase in the stellar surface temperature and at the same time one can witness a decline in the strength of the H-beta lines. In case, if the temperature of the surface increases too much, then one will witness no attachment of electron with the hydrogen atom and thus no H lines, and if the temperature of the surface becomes too low, then the electrons will stay in the ground state and no formation of H lines will take place in that condition too.  

Hence, to generate a very robust H line, after captivating photons the majority of the electrons are required to stay in the second energy level.  

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Un tecnico di laboratorio deve preparare una soluzione di carbonato di sodio decaidrato, Na2CO3⋅ 10 H2O per eseguire alcune anal
fiasKO [112]

Answer:

1.  3.70 g Na₂CO₃·10H₂O

2. 50.0 mL of the first solution

Explanation:

1. Prepare the solution

(a) Calculate the molar mass of Na₂CO₃·10H₂O

\begin{array}{rrr}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 2\times22.99 & 45.98\\\text{1C} & 1\times 12.01 & 12.01\\\text{13O}&13 \times16.00 & 208.00\\\text{20H}&20 \times 1.008 & 20.16\\&\text{TOTAL =} & \mathbf{286.15}\\\end{array}

The molar mass of Na₂CO₃·10H₂O is 286.15 g/mol.

(b) Calculate the moles of Na₂CO₃·10H₂O

\text{Moles of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\= \text{0.250 L solution} \times \dfrac{\text{0.0500 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 L solution}}\\\\= \text{0.0125 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}

(c) Calculate the mass of Na₂CO₃·10H₂O

\text{Mass of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O }\\= \text{0.012 50 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O } \times \dfrac{\text{296.15 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}\\\\= \text{3.70 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\\text{You need $\large \boxed{\textbf{3.70 g}}$ of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}

2. Dilute the solution

We can use the dilution formula to calculate the volume needed.

V₁c₁ = V₂c₂

Data:

V₁ = ?;            c₁ = 0.0500 mol·L⁻¹

V₂ = 100 mL; c₂ = 0.0250 mol·L⁻¹

Calculation:

\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\V_{1}\times \text{0.0500 mol/L} & = & \text{100 mL} \times\text{0.0250 mol/L}\\0.0500V_{1}& = & \text{2.500 mL}\\V_{1}&=& \dfrac{\text{2.500 mL}}{0.0500}\\\\& = &  \text{50.0 mL}\\\end{array}\\\text{You need $\large \boxed{\textbf{50.0 mL}}$ of the concentrated solution.}

7 0
3 years ago
a sample of sucrose was burned in a calorimeter, and 41,600 joules oh heat energy were released. which statement is true about t
Gala2k [10]

It absorbed 41,600 joules.

Explanation:

If the sample of sucrose burnt in the calorimeter releases 41600Joules of heat, then the water in the calorimeter will absorb 41600J.

 The heat lost by burning the sucrose in the calorimeter is also the heat gained by the water therein.

A calorimeter is a device by prevents heat loss to the environment in any form.

It is specially designed to minimize heat loss by conduction, convection radiation.

Therefore, since no heat is lost, the heat released by burning the sucrose is used in heating the water there.

learn more:

Specific heat brainly.com/question/7210400

#learnwithBrainly

5 0
3 years ago
Which law is based on the graph that is shown below? A graph is shown with pressure on the horizontal axis and volume on the ver
DIA [1.3K]

<u>Answer:</u>

"Boyle's Law" is based on the graph that is shown below.

<u>Explanation:</u>

Boyle's law or Boyle – Mariotte law or Mariotte's law, is an experimental gas law that discusses how a gas's pressure tends to rise as the container volume start declining. This shows the relationship between pressure and volume for a fixed mass at a constant temperature, i.e., number of a gas molecules.This rule visualizes the actions of gas molecules in a confined space. This law can be understood from following equation:

p₁V₁ = p₂V₂

Above the product of the initial volume and pressure is equal to the product of the volume and pressure after a change.

5 0
3 years ago
Read 2 more answers
In the lab, ammonia was mixed with water to form ammonium hydroxide. What is/are the reactant(s)?
Nataly_w [17]

The reactants are what is put into the reaction equation to get the product. In this case water and ammonia are put into the equation to get the product ammonium hydroxide.

The first choice is the correct answer:

Water and ammonia

Hope this helped!

~Just a girl in love with Shawn Mendes

7 0
3 years ago
Read 2 more answers
Enter a balanced equation for the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II
Dominik [7]

Answer:

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)

Explanation:

Let's consider the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II) chloride and aqueous lithium nitrite.

Pb(NO₂)₂(aq) + LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)

This is a double displacement reaction. We will start balancing Cl by multiplying LiCl by 2.

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)

Now, we have to balance Li by multiplying LiNO₂ by 2.

Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)

The equation is now balanced.

4 0
3 years ago
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