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vaieri [72.5K]
3 years ago
10

A large amount of dust and ash from a recent volcanic eruption settles in the region, which is most likely to survive, deer, fis

h in a pond, grass or trees
Chemistry
1 answer:
Ugo [173]3 years ago
7 0

Answer:

A deer

Explanation:

Volcanic Ash and dust would cover the surface of the pond. This would prevent the growth of planktons which serves as food for them. Sunlight would not be able to penetrate the pond to furnish it with a very important ingredient for photosynthesis. As we know that most fishes are heterotrophs.

For grasses and trees, the dust and Ash particles would cover the leaves which is the site of photosynthesis. Plants are autotrophs that manufactures their own food through sunlight and carbondioxie. Most of the plants would die off since they can't remove the Ash and dust covering on them.

A deer is an heterotroph and a motile animal. It can easily leave that region and source for food elsewhere in the ecosystem.

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When the atmospheric pressure is equal to the vapor pressure of the liquid, boiling will begin. The bubbles formation started to the liquid molecules which have gained enough energy to change to the gaseous phase.

The vapor pressure of ethanol is 5.95 kPa at 20.0 °C, and its vapor pressure is 53.3 kPa at 63.5 °C.

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7 0
2 years ago
How many moles of water are in 1/4 cup of water
il63 [147K]

Answer:

3.4752 moles of water

Explanation:

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5 0
3 years ago
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1. Complete the following
Marat540 [252]

Answer:

\large \boxed{\text{0.603 mol}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of O₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

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m/g:        18.1

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\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{18.1 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1005 mol C$_{6}$H$_{12}$O}_{6}

b) Moles of O₂

\text{Moles of O}_{2} =\text{0.1005 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol O}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.603 mol O}_{2}\\\\\text{The reaction requires $\large \boxed{\textbf{0.603 mol}}$ of oxygen}  

3 0
3 years ago
22.4 L is the volume of any gas regardless of atmospheric conditions.<br><br> O True<br><br> O False
Contact [7]
This is false. One mole of a gas occupies 22.4 L at STP, which is taken to be 0°C (273 K) and 1 atm. If atmospheric conditions depart from these values, this assumption cannot be used.
8 0
3 years ago
Can someone convert 10,000,000 sec to weeks for me? Thank you so much!
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Hi,

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Hope this helps.
r3t40
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3 years ago
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