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3 years ago

The molecular mass of sodium oxide (Na2O) is

1 answer:

6 0

<span>The molecular mass of sodium oxide (Na2O) is A. 61.97894. The molecular mass of a molecule (Mr) is the sum of atomic masses of its atoms (Ar). The molecular mass of sodium oxide is: Mr(Na2O) = 2 * Ar (Na) + Ar(O). From the periodic table, Ar(Na) = 22.989769 and Ar(O) = 15.9994. The molecular mass of sodium oxide is: Mr(Na2O) = 2 * 22.989769 + 15.9994 = 45.979538 + 15.9994 = 61.97894.</span>

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Sodium hydroxide dissolves in water ....

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n the reaction will be Exothermic, where heat will be released. .

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3 years ago

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.

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repartExiplanation:cionde atomos

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Polar liquids have both negative and positive ends.

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3 years ago

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When pcl5 solidifies it forms pcl4 cations and pcl6–anions. according to valence bond theory, what hybrid orbitals are used by p

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When pcl5 solidifies it forms pcl4 cations and pcl6–anions. according to valence bond theory, hybrid orbitals that are used by phosphorus in the pcl4 cations are one orbital of s and three orbital of p as it is sp³hydridised.

<h3>What is sp³ hybridization?</h3>

Hybridization is a process or system which specifies the shape and geometry of any element or molecule with bond angles too.

The pcl4 cation is sp³ hybridized because of the phosphorus behave as a central atom here and the 4 chloride molecules are attached with the p- orbitals to the phosphorus molecule.

Therefore, pcl4 cation is sp³ hybridized.

Learn more about sp³ hybridization, here :

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