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Nikitich [7]
3 years ago
9

Which statement most accurately explains why group 8 elements do not react with other elements​

Chemistry
1 answer:
Hunter-Best [27]3 years ago
5 0

Explanation:

group 8 elements don't react with other elements because they are noble element which means that they can't lose or gain electron.

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Describe how the molecular models you assembled are similar to real molecules.
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Answer:

Atoms models are bigger but they don not match the true or positive way of atoms. to make them visible scientists make their models that resembled with the real ones.

Explanation:

Molecules are microscopic entities and only can be seen under a  microscope.

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The invention of the ____ allows scientists to learn about cells.
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The Microscope allowed scientists to learn about cells.
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Which of the following choices identifies the forms of mechanical energy?
WITCHER [35]
Kinetic and potential energy would be the correct answer to your question
3 0
4 years ago
A pendulum has 721 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
Sloan [31]

Answer:

721 J  

Step-by-step explanation:

At the bottom of its swing, all the potential energy of the pendulum has been converted to kinetic energy.

KE = 721 J

8 0
3 years ago
Read 2 more answers
Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2
Flauer [41]

Answer:

\boxed{28.81}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      58.12                   44.01

           2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g:     9.511

1. Moles of C₄H₁₀

\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}

3. Mass of CO₂

\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}

8 0
3 years ago
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