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3 years ago

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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B. Radiation

Vika [28.1K]

First one, occurs fast !

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3 years ago

Read 2 more answers

I'm not sure on the answer for this question

Natalka [10]

If it has a metal and a nonmetal bonded together it is an ionic bond. you are looking for non-metal non-metal bond. those are all covalent bonds. answer is B

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3 years ago

The number placed in front of a compound to balance a chemical reaction is called

masha68 [24]

Answer: The number placed in front of a compound to balance a chemical reaction is called coefficient.

Explanation:

Assume this general form for a chemical equation:

aA + bB → cC + dD

The letters a, b, c, and d, in front of each compound A, B, C, and D, are called coefficients and indicate the number of formula units (molecules or ions) that take part in the equation.

Those coefficients are needed to balance the equation and ensure compliance with the law of mass conservation.

This example shows it:

Word equation: hydrogen + oxygen yields water

Chemical equation: H₂ (g) + O₂(g) → H₂O(g)

Balance, adding the coefficients so that the number of each kind of atoms is the same on the left and the right of the chemical equation:

H₂ (g) + 2O₂(g) → 2H₂O(g)

In that equation:

The coefficient of H₂ (g) on the left is 1 (it is not written)

The coefficient of O₂(g) on the left is 2

The coefficient of H₂O(g) on the right is 2

You read it as: 1 mole of gaseous hydrogen and 2 moles of gaseous oxygen yield 2 moles of water vapor.

4 0

3 years ago

Consider the following unbalanced redox reactions. In each case, separate the whole reactions into half-reactions, balance the h

Marina86 [1]

Answer & Explanation:

(a)

$Fe^{2+} +NO_{3}^{-} => Fe{(OH)}_3 + N_2$

reducing agent = Fe²⁺

Oxidizing agent = NO₃⁻

oxidation

Fe²⁺ ⇒ Fe(OH)₃

reduction

NO₃⁻ ⇒ N₂

Oxidation Half Reaction

(redox reactions are balanced by adding appropriate H⁺ and H₂O atoms)

Fe²⁺ ⇒ Fe(OH)₃

Balance O atoms

Fe²⁺ + 3H₂O ⇒ Fe(OH)₃

Balance H atoms

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺

balance Charge

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺ + e⁻..............(1)

reduction Half Reaction

NO₃⁻ ⇒ N₂

Balance N atoms

2NO₃⁻ ⇒N₂

Balance O atoms by adding appropriate H₂O

2NO₃⁻ ⇒ N₂ + 6H₂O

Balance H atoms

2NO₃⁻ + 12H⁺ ⇒ N₂ + 6H₂O

Balance Charge

2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O.................(2)

Combine Equation (1) and (2)

(1) × 10: 10Fe² + 30H₂0 ⇒ 10Fe(OH)₃ + 30H⁺ + 10e⁻

(2) × 1: 2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O

(1) + (2): 10Fe² + 30H₂0 + 2NO₃⁻ + 12H⁺ + 10e⁻ ⇒10Fe(OH)₃ + 30H⁺ + 10e⁻ +

N₂ + 6H₂O

10Fe² + 24H₂0 + 2NO₃⁻ ⇒ 10Fe(OH)₃ + 18H⁺ + N₂

this is the balanced reaction

REDUCTION POTENTIAL

10Fe²⁺(aq) + 10e⁻ ⇒ 10Fe(OH)₃(aq) E°ox = 10(-0.44) = -4.4V

2NO₃⁻(aq) - 2e⁻ ⇌ N₂(g) + 18H⁺ E°red = 2(+0.80) = +1.6

10Fe² + 24H₂0 + 2NO₃⁻ ⇒ 10Fe(OH)₃ + 18H⁺ + N₂ E°cell = -2.8V

E°cell = E°red + E°ox

4 0

4 years ago

What are the basis of making symbols????

telo118 [61]

Answer:

1/ Using the first letter of the name of the elements.

2/ If the name of two or more elements begins with the same letter,the second letter of their name is also taken with first letter.

3/ Symbols are made by using the letters of the name of elements in other languages also.

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4 years ago

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

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