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hichkok12 [17]
3 years ago
9

Which of the following types of reproduction is most likely to result in a rapid growth in population?

Chemistry
2 answers:
salantis [7]3 years ago
8 0
Asexual reproduction<span>
</span>
Iteru [2.4K]3 years ago
5 0
If there's any answer that mentions asexual reproduction that's probably it but what are the choices

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Consider a possible triatomic molecule containing a nitrogen, oxygen, and fluorine atoms for the following questions.
STALIN [3.7K]

Answer:

a) nitrogen

b) nitrogen =5

Oxygen = 6

Fluorine =7

Explanation:

Usually, if we have two or more elements in a compound, the central atom in the compound is the atom having the least value of electro negativity.

If we consider fluorine, oxygen and nitrogen; nitrogen is the least electronegative of the trio hence it should be the central atom of the triatomic molecule.

The number of valence electrons on the valence shell of each atom is shown below;

nitrogen =5

Oxygen = 6

Fluorine =7

5 0
3 years ago
If you stacked Avogadro's number of pennies one on top of the other on Earth's surface, how far would the stack extend (in kilom
attashe74 [19]

Answer:

Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km

<em>Note: The question is missing some parts. The complete question is as follows;</em>

<em>A penny has a thickness of approximately 1.0 mm . If you stack ed Avogadro's number of pennies one on top of the other on Earth 's surface, how far would the stack extend (in km)? [For comparison, the sun is about 150 million km from Earth and the nearest star (Proxim a Centauri) is about 40 trillion km from Earth.]</em>

Explanation:

Avogadro number = 6.02 * 10²³

thickness of a penny = 1.0 mm

I mm = 0.001 m

Thickness of Avogadro number of pennies stacked one upon another will be:

6.02 * 10²³ * 0.001 m = 6.02 * 10²⁰ m

Distance in km;

1 m = 0.001 km

therefore, 6.02 * 10²⁰ m = 6.02 * 10²⁰ * 0.001 km = 6.02 * 10¹⁷ km

Avogadro number of pennies will extend to a distance of 6.02 * 10¹⁷ km

5 0
3 years ago
77 grams of an unknown metal at 99ᵒC is placed in 225 grams of water which is initially at 22ᵒC. The water is inside a 44 gram a
BigorU [14]

Answer:

The specific heat capacity of the unknown metal is C = 0.6991 J/g°C = 0.1671 cal/g°C

Explanation:

Heat lost by the unknown metal is equal to the heat gained by the water and aluminium cup.

Given,

Mass of unknown metal = 77 g

Initial Temperature of unknown metal = 99°C

Mass of water = 225 g

Initial Temperature of water = 22°C

Mass of Aluminium cup = 44 g

Specific heat capacity of Aluminium cup = 0.22 cal/gᵒC = 0.92048 J/g°C

Final temperature of the setup = 26°C

Note that, specific heat capacity of water = 4.186 J/g°C

Let the specific heat of the unknown metal be C

Heat lost from the unknown metal

= (77)(C)(99 - 26) = (5,621C) J

Heat gained by water

= (225)(4.186)(26 - 22) = 3,767.4 J

Heat gained by Aluminium cup

= (44)(0.92048)(26 - 22) = 162.00448 J

Heat lost by unknown metal = (Heat gained by water) + (Heat gained by Aluminium cup)

5621C = 3,767.4 + 162.00448 = 3,929.40448

5621C = 3,929.40448

C = (3,929.40448 ÷ 5621) = 0.6991 J/g°C

C = 0.6991 J/g°C = 0.1671 cal/g°C

Hope this Helps!!!

6 0
3 years ago
Read 2 more answers
How many moles are in 6 x 10^23 molecules of H2O
goldfiish [28.3K]
One mole represents 6.022∙1023 separate entities, just like one dozen represents 12 objects. So, if there are 6.022∙1023 H2O molecules, that is the same as one mole of water.
7 0
3 years ago
Five the IUPAC names for the following
White raven [17]

Answer:

the answer is b ch2 ch ch3 cl ch2 ch2 ch3

4 0
3 years ago
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