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Bingel [31]
3 years ago
9

Which of these is an ionic compound? NO2 LiCl SO3 CO

Chemistry
2 answers:
kompoz [17]3 years ago
5 0
LiCl because it contains one metal and a nonmetal element
Pavlova-9 [17]3 years ago
4 0
It is "LICI" I believe. I could be wrong please let me know if I am wrong.
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3 years ago
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What is the definition of relief in science terms
SOVA2 [1]
The apparent topography exhibited by minerals in thin section as a consequence of refractive index.
6 0
3 years ago
The radius of a single atom of a generic element X is 169 picometers (pm) and a crystal of X has a unit cell that is body-center
Virty [35]

The volume of the unit cell is 2.67 x 10⁻²⁸ m³.

<h3>What is the volume of a unit cell of a body-centered cubic crystal?</h3>

In a body-centered cubic unit cell, the volume occupied by the particles of the substance is about 68% of the total unit cell.

Assuming that a single atomic a sphere, the volume is:

Volume(atom) = 4/3 x π x r³

Volume(atom) = 4/3 x π x (169 x 10⁻¹²)³

Volume(atom) = 2.02 x 10⁻²⁹ m³

There are a total of 9 atoms in a body-centered unit cell, so the total volume occupied by atoms is:

2.02 x 10⁻²⁹ x 9

= 1.82 x 10⁻²⁸ m³

Volume of cell = (1.15 x 10⁻²⁸ ) / 0.68

Volume of cell = 2.67 x 10⁻²⁸ m³

Therefore, the volume of the unit cell is 2.67 x 10⁻²⁸ m³.

Learn more volume of unit cells at: brainly.com/question/1594030

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8 0
2 years ago
4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o
Vilka [71]

<u>Answer:</u> The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = \frac{1}{1}\times 0.0711=0.0711mol of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g

To calculate the percentage yield of 3-hexanol, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

4 0
3 years ago
A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
sammy [17]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

5 0
3 years ago
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