Using mathematical expressions drive the compression factor

A lead atom has a mass of 3.14 x 10 to the negative 22nd g.How

frozen [14]

Explanation:

The given data is as follows.

Mass of a lead atom = $3.14 \times 10^{-22}$

Volume = 2.00 $cm^{3}$

Density = 11.3 $g/cm^{3}$

As it is mentioned that 1 cubic centimeter contains 11.3 grams of lead.

So, in 2 cubic centimeter there will be $2 \times 11.3 g = 22.6 g$ of lead atoms.

One lead atom has a mass of $3.14 \times 10^{-22}$ . Therefore, number of atoms present in 22.6 g of lead will be as follows.

$\frac{22.6 g}{3.14 \times 10^{-22}}$

= $7.197 \times 10^{22} atoms$

Thus, we can conclude that there are $7.197 \times 10^{22} atoms$ of lead are present.

4 0

3 years ago

Which energy change occurs during boiling?

dybincka [34]

Answer:

heat energy because u are boiling something

5 0

3 years ago

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An astronaut is free-floating in space near a space station and it is not tethered to anything. He has some tools and a fire ext

worty [1.4K]

Hey there

the answers are

A.

and

D. He could spray the fire extinguisher in the opposite direction of the space station

hope this helps

7 0

3 years ago

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Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03%

nlexa [21]

Answer: The empirical and molecular formula of chrysotile is $Mg_3Si_2H_3O_4$ and $Mg_6Si_4H_6O_{16}$

Explanation:

We are given:

Percentage of Mg = 28.03 %

Percentage of Si = 21.60 %

Percentage of H = 1.16 %

Percentage of O = 49.21 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of Mg = 28.03 g

Mass of Si = 21.60 g

Mass of H = 1.16 g

Mass of O = 49.21 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Magnesium = $\frac{\text{Given mass of Magnesium}}{\text{Molar mass of Magnesium}}=\frac{28.03g}{24g/mole}=1.17moles$

Moles of Silicon = $\frac{\text{Given mass of Silicon}}{\text{Molar mass of Silicon}}=\frac{21.06g}{28g/mole}=0.752moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.16g}{1g/mole}=1.16moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{49.21g}{16g/mole}=3.07moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.752 moles.

For Magnesium = $\frac{1.17}{0.752}=1.5$

For Silicon = $\frac{0.752}{0.752}=1$

For Hydrogen = $\frac{1.16}{0.752}=1.5$

For Oxygen = $\frac{3.07}{0.485}=4.08\approx 4$

To convert the mole ratios into whole numbers, we multiply individual mole ratios by 2

Mole ratio of Magnesium = (2 × 1.5) = 3

Mole ratio of Silicon = (2 × 1) = 2

Mole ratio of Hydrogen = (2 × 1.5) = 3

Mole ratio of Oxygen = (2 × 4) = 8

Step 3: Taking the mole ratio as their subscripts.

The ratio of Mg : Si : H : O = 3 : 2 : 3 : 8

The empirical formula for the given compound is $Mg_3Si_2H_3O_8$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 520.8 g/mol

Mass of empirical formula = [(24 × 3) + (28 × 2) + (1 × 3) + (16 × 8)] = 259 g/mol

Putting values in above equation, we get:

$n=\frac{520.8g/mol}{259g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$Mg_{(3\times 2)}Si_{(2\times 2)}H_{(3\times 2)}O_{(8\times 2)}=Mg_6Si_4H_6O_{16}$

Hence, the empirical and molecular formula of chrysotile is $Mg_3Si_2H_3O_4$ and $Mg_6Si_4H_6O_{16}$

5 0

3 years ago

How many moles of sodium ions are present in 0.25 kg of a 2.0 m solution of sodium chloride in water

LenKa [72]

0.50 is the answer on edge

8 0

3 years ago

Using mathematical expressions drive the compression factor​

Using mathematical expressions drive the compression factor