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3 years ago

How might emission spectra be used in studying stars?

Chemistry

2 answers:

nordsb [41]3 years ago

5 0

Answer:

Emission spectra may be used in studying stars in order to determine what atoms makeup the individual star produces, due to the fact that each atom’s emission spectra biunique, one can observe the spectra emitted by the star and identify the atoms that are released by the light the star produces.

malfutka [58]3 years ago

4 0

Answer:

^yes

Explanation:

what they said was right

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What element has 47 protons in its nucleus?

sesenic [268]

Answer:

b). silver (Ag)

Explanation:

If you look at the periodic table, you just need to look at the atomic number of the element, because the atomic number tells you how many protons there are in the nucleus of the element.

But do be careful because some periodic tables have the molar mass at the top left corner, but the one I use has the atomic number at the top left corner, so make sure you look for the atomic number and not the molar mass.

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3 years ago

Is c10H8 a conductor

Lunna [17]

Answer:

Explanation:

Napthalene cannot conduct electricity

5 0

3 years ago

HELP WITH CHEMISTRY PLEASE!

maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

In our problem:

P1 = ??? (is needed to be calculated) and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.

2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

In our problem:

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? (is needed to be calculated)

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.

3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

In our problem:

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? (is needed to be calculated) and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.

4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; 101.325 kPa = 1.0 atm, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.

5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

In our problem:

P1 = 2200.0 mmHg and T1 = ??? (is needed to be calculated) .

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

6 0

3 years ago

Read 2 more answers

The caloric content of 0.50 cup of cottage cheese is 1.10 x 102 kcal. This is the amount of energy released when 0.50 cup of cot

V125BC [204]

Answer:

A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.

What is the specific heat of the substance?

If it is one of the substances found in Table 8.1.1, what is its likely identity?

3 0

3 years ago

Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 6.9 g of hexane is mi

MrRa [10]

Answer:

There, there are no leftover for C6H14 instead, an additional mass of 4g of C6H14 is needed to completely react with 38.4g of O2.

Explanation:

Step 1:

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

2C6H14 + 19O2 —> 12CO2 + 14H2O

Step 2:

Let us calculate the masses of C6H14 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Molar Mass of C6H14 = (12x6) + (14x1) = 72 + 14 = 86g/mol

Mass of C6H14 from the balanced equation = 2 x 86 = 172g

Molar Mass of O2 = 16x2 =32g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608g

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Step 3.

Now, let us determine the mass of C6H14 that will react with 38.4 g of oxygen. This is illustrated below:

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Therefore, Xg of C6H14 will react with 38.4g of O2 i.e

Xg of C6H14 = (172 x 38.4) /608

Xg of C6H14 = 10.9g

From the calculations made above, we can see clearly that the mass of C6H14 is limited as the reaction requires 10.9g of C6H14 and only 6.9g was given. There, there are no leftover for C6H14 instead, an additional mass ( 10.9 - 6.9 = 4g) of 4g of C6H14 is needed to completely react with 38.4g of O2.

5 0

3 years ago