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3 years ago

What are the two factors that affect that density of matter ?

Chemistry

1 answer:

Morgarella [4.7K]3 years ago

6 0

Mass and volume are two factors that affect that density of matter!

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What was Newtons first law of motion??

Ksivusya [100]

Answer:

Basically an object in motion or at rest will remain that way unless an outside force acts against it.

Explanation:

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia.

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3 years ago

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What properties do molecules have that allow them to cross the bilayer without any help?

IceJOKER [234]

Hydrophobic. Since bilayer has hydrophobic tail as their core region, molecules can pass through the hydrophobic tail if they are also hydrophobic.

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3 years ago

4.<br> Think and discuss: Why is there an attraction between the two ions in this chemical bond?

Iteru [2.4K]

Answer:

One atom will give electrons to another atom to fill its shell, this is an ionic bond. The atom giving away electrons becomes positive and the one recieving electrons becomes negative.

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Explanation:

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3 years ago

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Calculate the number of formula units in 2.50 mol NaNO

marta [7]

The number of formula units in 2.50 mol of the compound is 15.1 * 10^23.

The question is unclear whether NaNO2 or NaNO3 is implied. However,in either case, the solution applies equally.

6.02 * 10^23 formula units of the compound are contained in 1 mole

x formula units are contained in 2.5 moles of the compound

x = 6.02 * 10^23 formula units * 2.5 moles/ 1 mole

x = 15.1 * 10^23 formula units of the compound.

Learn more; brainly.com/question/9743981

3 0

3 years ago

Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at

ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

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

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $O_2$ 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, 3 mole of $O_2$ react to give 2 mole of $Al_2O_3$

So, 0.188 moles of $O_2$ react to give $\frac{2}{3}\times 0.188=0.125$ moles of $Al_2O_3$

Now we have to calculate the amount of heat released in the combustion.

As, 1 mole of $Al_2O_3$ releases amount of heat = 1676 kJ

So, 0.125 mole of $Al_2O_3$ releases amount of heat = $0.125\times 1676kJ=209.5kJ$

Thus, the amount of heat released in the combustion is, 209.5 kJ

4 0

3 years ago