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2 years ago

What is the mass of 0.5 moles of carbon tetrafluoride, CF4?

Chemistry

2 answers:

klasskru [66]2 years ago

7 0

Answer:

44.0015

Explanation:

we go and look at the mass for carbon which is 12.011 then at the mass of fluorine which is 18.998. Since there are 4 Fluorines the we multiply 4 by 18.998 which we get 75.992. Then we add 75.992 + 12.011 which will give us 88.003 then multiply 88.003 by 0.5 which gives us 44.0015

VashaNatasha [74]2 years ago

4 0

Answer:

44 g

Explanation:

The formula for the number of moles (n) is equal to $n=\frac{mass}{molecular weight}$ .

Since we need to find the mass, we derive it from the formula of the number of moles and we get that mass = n x molecular weight .

The molecular weight of $CF_{4}$ = 12 g/mol (from the carbon) + 19x4 g/mol (from the 4 fluorine atoms)= 88 g/mol

We plug in the numbers in the derived formula for the mass and we get :

mass = n x molecular weight = 0.5 mol x 88 g/mol = 44 g

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In two or more complete sentences describe all of the van der Waals forces that exist between molecules of sulfur

zavuch27 [327]

Answer:

Dipole-Dipole attraction

Explanation:

Dipole-dipole attraction is a type of vander waals forces found in the molecules of sulfur dioxide.

Vander waals forces are weak attractions joining non-polar and polar molecules together. They are of two types:

London dispersion forces which are weak attractions found between non-polar molecules.
Dipole-Dipole attraction are the forces of attraction which exists between polar molecules. Such molecules have permanent dipoles. This implies that the positive pole of one molecule attracts the negative pole of another. This is what happens between the oxygen and sulfur molecules.

3 0

2 years ago

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

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

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

2 years ago

Read 2 more answers

A sample of an unknown compound was decomposed and found to be composed of 1.36 mol oxygen, 4.10 mol hydrogen, and 2.05 mol carb

Flura [38]

Answer:

C3H6O2

Explanation:

To find the empirical formula of the compound, we divide the amount in moles of each of the elements by the amount in mole of the element with the smallest number of mole. In this question, the element with the smallest number of moles is oxygen with 1.36 mole. Hence, we divide the number of moles of each element by this.

H = 4.10/1.36 = 3

O = 1.36/1.36 = 1

C = 2.05/1.36 = 1.5

We then multiply through by 2 to yield the compound with the empirical formula C3H6O2

7 0

2 years ago

Rahul and Manav each were given a mixture of iron filings and sulphur powder. Rahul heated the mixture strongly and a new substa

AlexFokin [52]

Answer:

sure

Explanation:

The substance formed after heating the mixture of that of Rahul is caleed a compound. Whereas, Manav's mixture still remains in its current stae that is a heterogeneous mixture.

The compound formed is in black in color whereas the mixture is a mix of brownish-red and yellow.

The compound is a homogeneous mixture whereas the mixture is a heterogenous mixture because of its uneven distribution.

3 0

2 years ago

How many chloride ions are in 0.486 moles of chloride ions?

svetlana [45]

Answer:

Since in a chloride ion, we have an additional electron

you might think that it will affect the mass but the mass of an electron is almost negligible so we will ignore that

Amount of ions in 1 mol = 6.022 * 10^23

Amount of ions in 0.486 moles = 0.486 * (6.022*10^23)

Amunt of ions in 0.486 moles = 2.9 * 10^23 ions

Hence, option 1 is correct

6 0

2 years ago