Match the following.

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

finlep [7]

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.

5 0

4 years ago

s e vapor pressure of the waterretnylene glycol solution at 32 "C? 5. What is the Boiling Point of the solution resulted from th

lyudmila [28]

Answer:

What is the Boiling Point of the solution resulted from the dissolving of 32.5 g of NaCl in 250.0 g of water? 102,31°C

Explanation:

This question involves the Elevation of boiling point

ΔT = Kb . m . i

(T°solution - T°solvent pure) = Ebulloscopic constant . molality . Van 't Hoff factor

Van 't Hoff factor is 2 for NaCl because you have two ions on disociation.

Kb for water is, 0,52 °C . kg/mol - It is a known value.

You know that water pure boils at 100°C so let's build the formula

(T°solution - 100°C) = 0,52 °C.kg/mol . molality . 2

We have to find out the molality (moles of solute/1kg solvent)

In 250 g H2O we have 32,5 g NaCl so the rule of three will be

250 g H2O ______32,5 g NaCl

1000g __________ (1000g . 32,5g) / 250g = 130 g NaCl

Molar mass NaCl : 58,45 g/m

Moles NaCl : mass/ molar mass --> 130g /58,45 g/m = 2,22 m

(T°solution - 100°C) = 0,52 °C.kg/mol . 2,22 mol/gk . 2

T°solution - 100°C = 2,31 °C

T° at boiling point in the solution = 2,31°C + 100°C = 102,31°C

3 0

3 years ago

How much energy is needed to raise the temperature of a 200 g piece of lead by 250 o C?

devlian [24]

Answer: The amount of energy required to raise the given amount of temperature is 6400 J

Explanation:

To calculate the heat absorbed by the lead, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of lead = 200 g

c = heat capacity of lead = 0.128 J/g°C

$\Delta T$ = change in temperature = 250°C

Putting values in above equation, we get:

$q=200g\times 0.128J/g^oC\times 250^oC=6400J$

Hence, the amount of energy required to raise the given amount of temperature is 6400 J

4 0

3 years ago

Nitric acid is usually purchased in a concentrated form that is 70.3% HNO3 by mass and has a density of 1.41 g/mL. How much conc

Marina CMI [18]

Answer:

The answer to your question is: 9 .43 ml of HNO3 70.3%

Explanation:

Data

Purity: 70.3%

density = 1.41g/ml

Volume = 1 l

Concentration = 0.12 M

Molecular weight = 1 + 14 + (16 x 3) = 63 gr

63 gr of HNO3 ------------------1 mol of HNO3 or IM

x ------------------- 0.12 M

x = 0.12M x 63 /1 = 7.65 g of HNO3

Now, calculate volume

density = mass / volume and volume = mass/density

volume = 7.65 / 1.41 = 6.63 ml of HNO3

Now, consider purity (it's and inver rule of three)

6.63 ml ----------------- 100%

x ----------------- 70.3 %

x = (6,63 x 100)/70.3 = 9.43 ml of HNO3

8 0

4 years ago

A sample of gas occupies a volume of 450.0 ML at 740 mm Hg and 16c determine the volume of this sample at 760mm Hg and 37C show

algol13

Answer:

474.92 mL.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n is constant, and have two different values of (P, V and T):

P₁V₁T₂ = P₂V₂T₁

P₁ = 740 mmHg, V₁ = 450 mL, T₁ = 16°C + 273 = 286 K.

P₂ = 760 mmHg, V₂ = ??? mL, T₂ = 37°C + 273 K = 310 K.

∴ V₂ = P₁V₁T₂/P₂T₁ = (740 mmHg)(450 mL)(310 K)/(760 mmHg)(286 K) = 474.92 mL.

5 0

3 years ago

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