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3 years ago

The molecules in a solid do not move. True False

Chemistry

2 answers:

weqwewe [10]3 years ago

7 0

Answer:

True

Explanation:

Solid molecules don't move.

Dennis_Churaev [7]3 years ago

5 0

Molecule of a solid are tightly packed there is no space between them so there is no movement of the molecule.

The particles (atoms or molecules) are attracted to each other. The particles (atoms or molecules) vibrate but do not move past one another. The solid retains its shape.

true

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3 years ago

Thermal Energy and Kinetic Molecular Theory Quick Check

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1 year ago

Which of the following steps will slow down the reaction between coal and oxygen?

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3 years ago

Read 2 more answers

How many square centimeters are in an area of 6.80 inch^2

Vedmedyk [2.9K]

43.87cm²

Explanation:

Given parameter;

Area = 6.80in²

Problem:

conversion of in² to cm²

1in = 2.54cm

Given 6.8in²:

Converting:

6.8 x in x in x $\frac{2.54cm}{1in} x \frac{2.54cm}{1in}$ ;

The inches cancels out to give:

= 43.87cm²

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3 years ago

The reaction betwen aluminum and iron (III) oxide can

Alona [7]

Answer : The mass of $Al_2O_3$ formed will be, 468.18 grams.

Solution : Given,

Mass of Al = 124 g

Mass of $Fe_2O_3$ = 601 g

Molar mass of Al = 27 g/mole

Molar mass of $Fe_2O_3$ = 160 g/mole

Molar mass of $Al_2O_3$ = 102 g/mole

First we have to calculate the moles of Al and $O_2$ .

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{124g}{27g/mole}=4.59moles$

$\text{ Moles of }Fe_2O_3=\frac{\text{ Mass of }Fe_2O_3}{\text{ Molar mass of }Fe_2O_3}=\frac{601g}{160g/mole}=3.76moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe$

From the balanced reaction we conclude that

As, 2 mole of $Al$ react with 1 mole of $Fe_2O_3$

So, 4.59 moles of $O_2$ react with $\frac{4.59}{2}=2.295$ moles of $Fe_2O_3$

From this we conclude that, $Fe_2O_3$ is an excess reagent because the given moles are greater than the required moles and $Al$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$

From the reaction, we conclude that

As, 2 mole of $Al$ react to give 2 mole of $Al_2O_3$

So, 4.59 moles of $O_2$ react to give $\frac{2}{2}\times 4.59=4.59$ moles of $Al_2O_3$

Now we have to calculate the mass of $Al_2O_3$

$\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3$

$\text{ Mass of }Al_2O_3=(4.59moles)\times (102g/mole)=468.18g$

Therefore, the mass of $Al_2O_3$ formed will be, 468.18 grams.

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3 years ago