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arlik [135]
3 years ago
9

irregular Galaxy: appear as _________________ , _________________ shaped “blobs”; very active areas of star formation

Chemistry
1 answer:
marusya05 [52]3 years ago
7 0

Answer: They are typically smaller than spiral galaxies.

Some show evidence for prior collision or close encounter with another galaxy.

They exhibit vigorous star-forming activity.

Irregular galaxies are typically smaller than spiral or normal ellipticals, of irregular form, and without spiral arms. Irr I galaxies are often characterized by lots of gas, dust, and associated star formation. Irr II galaxies have forms that indicate prior collisions or close encounters with other galaxies. Irregular galaxies are one of the most common types of galaxies.

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Balancing Chemical Equations & Identifying Chemical Reactions
Stells [14]
The answer is 17 and a half
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3 years ago
Draw the products of the complete hydrolysis of an acetal. Draw all products of the reaction.
jolli1 [7]

Answer: the product is ketone or aldehyde

Explanation:

The first step is the conversion of acetal to hemiacetal in the presence of H3O+/ ROH, and then the final conversion of hemiacetal to ketone/aldehyde using

H3O+/ ROH...

Attached is the structural conversion

8 0
3 years ago
The mass of a solid substance is 21.112 g. If the volume of the solid substance is 19.5 cm3, calculate the density of the substa
Ostrovityanka [42]

Answer:

ρ = 1.08 g/cm³

Explanation:

Step 1: Given data

Mass of the substance (m): 21.112 g

Volume of the substance (V): 19.5 cm³

Step 2: Calculate the density of the substance

The density (ρ) of a substance is equal to its mass divided by its volume.

ρ = m / V

ρ = 21.112 g / 19.5 cm³

ρ = 1.08 g/cm³

The density of the substance is 1.08 g/cm³.

6 0
3 years ago
16.
mixas84 [53]

Answer:

Explanation:

When going from top to bottom in any group of the periodic table, the atomic radius always tends to increase

8 0
3 years ago
In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quan
lidiya [134]

Answer:

Second element(Titanium); [Ar] 3d2 4s2

Third element(Vanadium):Ar 3d3 4s2

Explanation:

Given that there are only three d orbitals in universe L instead of five, the electronic configuration of the second and third elements in the first transition series will now look thus;

Second element(Titanium); [Ar] 3d2 4s2

Third transition element(Vanadium):Ar 3d3 4s2

Hence, the electronic configuration of Titanium and Vanadium in universe L is just the same as what it is on earth.

6 0
3 years ago
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