Spiral galaxies have three main components: a bulge, disk, and halo (see right). The bulge is a spherical structure found in the center of the galaxy. This feature mostly contains older stars. The disk is made up of dust, gas, and younger stars. The disk forms arm structures. Our Sun is located in an arm of our galaxy, the Milky Way. The halo of a galaxy is a loose, spherical structure located around the bulge and some of the disk. The halo contains old clusters of stars, known as globular clusters<span>.
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Elliptical galaxies are shaped like a spheriod, or elongated sphere. In the sky, where we can only see two of their three dimensions, these galaxies look like elliptical, or oval, shaped disks. The light is smooth, with the surface brightness decreasing as you go farther out from the center. Elliptical galaxies are given a classification that corresponds to their elongation from a perfect circle, otherwise known as their ellipticity. The larger the number, the more elliptical the galaxy is. So, for example a galaxy of classification of E0 appears to be perfectly circular, while a classification of E7 is very flattened. The elliptical scale varies from E0 to E7. Elliptical galaxies have no particular axis of rotation.
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Answer:
Hope it helped
Explanation:
For neutral atoms, the number of valence electrons is equal to the atom's main group number. The main group number for an element can be found from its column on the periodic table. For example, carbon is in group 4 and has 4 valence electrons. Oxygen is in group 6 and has 6 valence electrons.
It was due to the metal foil in which the alpha particles can't even pass through. This experiment conducted by Rutherford led to the discovery of protons.
Answer is: a beaker contains <span>
heterogeneous mixture.
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A heterogeneous mixture<span> have compounds that remain separate in the sample.</span>
Heterogeneous
mixture is not uniform in composition (in this mixture different sand and small pebbles), but proportions of its components (in this
mixture particles of different colors and size) vary throughout
the sample.
Answer:
solubility of X in water at 17.0 
is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0 can be calculated using the information given.
Let's assume solubility of X in water at 17.0 is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y = 
= 0.11
Hence solubility of X in water at 17.0 is 0.11 g/mL.
