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pochemuha
3 years ago
5

How is modeling useful to scientists?

Chemistry
1 answer:
baherus [9]3 years ago
6 0
C it illustrates possible outcomes
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Which best explains the relationship between evaporation and temperature?
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A liquid evaporates faster at higher temperatures because more particles have a higher speed and can overcome attractions in the liquid.
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The half of the moon facing the sun is always lit, but the different phases happen because:
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Answer:

we only see parts of the lit side as the moon goes around the earth

Explanation:

Unlike the sun, the moon orbits the Earth. This is the reason why we see the <em>different phases of the moon.</em> The reflection of the moon is being illuminated back to us with the help of the sun. So, as the moon circles the Earth, we only see parts of the lit side. Such changes helps us see the moon in different phases such as<em> </em>the <em>Third Quarter, Crescent, New Moon, Full Moon, etc.</em>

For example, during "Full Moon," <em>the moon's entire face is lit up by the sun</em>. Thus, we see the entire moon's lit portion.

Thus, this explains the answer.

7 0
3 years ago
One or more substances changing into one or more substances is an example of a
lana [24]

Hello There!

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3 years ago
Which statement describes covalent bases?
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Gaseous ethane (CH,CH,) will react with gaseous oxygen (0,) to produce gaseous carbon dioxide (CO) and gaseous water (H2O). Supp
notka56 [123]

Answer:

0.00 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.  

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ            30.07      32.00  

              2CH₃CH₃ + 7O₂ ⟶ 4CO₂ + 6H₂O

Mass/g:      1.50          11.

2. Calculate the moles of each reactant  

\text{moles of C$_{2}$H}_{6} = \text{1.50 g C$_{2}$H}_{6} \times \dfrac{\text{1 mol C$_{2}$H}_{6}}{\text{30.07 g C$_{2}$H}_{6}} = \text{0.04988 mol C$_{2}$H}_{6}\\\\\text{moles of O}_{2} = \text{11. g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.34 mol O}_{2}

3. Calculate the moles of CO₂ we can obtain from each reactant

From ethane:

The molar ratio is 4 mol CO₂:2 mol C₂H₆

\text{Moles of CO}_{2} = \text{0.04988 mol C$_{2}$H}_{6} \times \dfrac{\text{4 mol CO}_{2}}{\text{2 mol C$_{2}$H}_{6}} = \text{0.09976 mol CO}_{2}

From oxygen:

The molar ratio is 4 mol CO₂:7 mol O₂

\text{Moles of CO}_{2} =  \text{0.34 mol O}_{2}\times \dfrac{\text{4 mol CO}_{2}}{\text{7 mol O}_{2}} = \text{0.20 mol CO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is ethane, because it gives the smaller amount of CO₂.

The excess reactant is oxygen.

5. Mass of ethane left over.

Ethane is the limiting reactant. It will be completely used up.

The mass of ethane left over will be 0.00 g.

8 0
3 years ago
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