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4 years ago

What is used to classify galaxies?star typesestimated agecolorshape

1 answer:

4 0

The correct answer among the choices given is the last option. Galaxies are classified according to their shapes or visual morphology. There three main types of galaxies currently. They are the elliptical, spiral and irregular. Elliptical galaxies are like a spheroid or an elongated sphere. Spiral galaxies are characterized and distinguished by having a bulge, a disk and a halo. Lastly, irregular galaxies do not have regular or symmetrical shape.

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Join my Kahoo t Pin-04454287 join if u like anime

aniked [119]

Answer:

okayy

Explanation:

5 0

3 years ago

Help!!!!!!!!!!!!!!!!!!

skad [1K]

The density of the sample : 0.827 g/L

<h3>Further explanation</h3>

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm , N/m²

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)

T = temperature, Kelvin

n= 1 mol

MW Neon = 20,1797 g/mol

mass of Neon :

$\tt mass=mol\times MW\\\\mass=1\times 20,1797 =20.1797~g$

The density of the sample :

$\tt \rho=\dfrac{m}{V}\\\\\rho=\dfrac{20,1797}{24.4}=0.827~g/L$

or We can use the ideal gas formula ta find density :

$\tt \rho=\dfrac{P\times MW}{RT}\\\\\rho=\dfrac{1\times 20.1797}{0.082\times 298}\\\\\rho=0.826~g/L$

8 0

3 years ago

Calculate the ratio of the velocity of helium atoms to the velocity of neon atoms at the same temperature.

o-na [289]

Answer:

vHe / vNe = 2.24

Explanation:

To obtain the velocity of an ideal gas you must use the formula:

v = √3RT / √M

Where R is gas constant (8.314 kgm²/s²molK); T is temperature and M is molar mass of the gas (4x10⁻³kg/mol for helium and 20,18x10⁻³ kg/mol for neon). Thus:

vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol

vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

The ratio is:

vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol

vHe / vNe = 2.24

I hope it helps!

8 0

3 years ago

How many grams of O2 are present in 44.1 L of O2 at STP?

ycow [4]

Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

Then, in this case:

P= 1 atm
V= 44.1 L
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 $\frac{atmL}{molK}$ × 273 K

Solving:

$n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}$

n=1.97 moles

Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:

$1.97 molesx\frac{32 g}{1 mole}$ = 63.04 g ≈ 63 g

Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.

Learn more about the ideal gas law:

brainly.com/question/4147359?referrer=searchResults

7 0

3 years ago

What is an ideal gas? how many ideal gases are there in the universe?

tiny-mole [99]

Ideal gases are hypothetical gases whose molecules occupy negligible space and have no interactions, and that consequently obeys the gas laws exactly.
Not exactly sure about the amount...
I hope this helps! :)

6 0

4 years ago