What is a resistor that can stop most

Discuss how to separate a mixture of iron fillings, salt, sand, and wood chips.

dexar [7]

Wrap a magnet in plastic lunch wrap and move it through the mixture of the three solids. The iron filings will stick to the magnet. The filings can be removed by unwrapping the plastic from the magnet carefully! Mix the remaining salt and sand in water and stir.

3 0

4 years ago

In molecules, energy is stored in? Atoms Electrons The Nucleus Bonds

Nikolay [14]

Answer:

Atoms

Explanation:

Energy, potential energy, is stored in the covalent bonds holding atoms together in the form of molecules. This is often called chemical energy.

4 0

3 years ago

When copper crystallizes, it forms face-centered cubic cells. The unit cell edge length is 361.5 pm. Calculate the density of co

Olegator [25]

Density of copper can be calculated using the following formula:

$d=\frac{m}{V}$

Here, m is mass and V is volume, thus, to calculate density first calculate mass and volume of copper.

In FCC, number of atoms in a unit cell are 4, atomic mass of Cu is 63.546 g/mol and number of atoms in 1 mole is $6.023\times 10^{23}$ .

Mass of Cu will be:

$m=4 atoms\times \frac{1 mol}{6.023\times 10^{23}}\times \frac{63.546 g}{mol}=4.22\times 10^{-22}g$

Now, volume can be calculate as:

$V=a^{3}$

Here, a is edge length.

First convert edge length from pm to cm

$1pm=10^{-10}cm$

Thus,

$361.5 pm=3.615\times 10^{-8}cm$

Putting the value,

$V=(3.615\times 10^{-8} cm)^{3}=4.72\times 10^{-23} cm^{3}$

Now, from mass and volume density can be calculated as follows:

$d=\frac{m}{V}=\frac{4.22\times 10^{-22}g}{4.72\times 10^{-23} cm^{3}}=8.932 g/cm^{3}$

Therefore, density of copper will be $8.932 g/cm^{3}$ .

8 0

4 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 $Al_2O_3$