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Soloha48 [4]
2 years ago
8

A hypothetical AX type of ceramic material is known to have a density of 3.55 g/cm3 and a unit cell of cubic symmetry with a cel

l edge length of 0.39 nm. The atomic weights of the A and X elements are 90.5 and 37.3 g/mol, respectively. On the basis of this information, which one of the following crystal structures is possible for this material?a) fluoriteb) zinc blendc) cesium chloride
Chemistry
1 answer:
postnew [5]2 years ago
3 0

Explanation:

For AX type ceramic material, the number of formula per unit cells is as follows.

         \rho = \frac{n'(A_{c} + A_{A})}{V_{C}N_{A}}

or,     n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}

where,   n' = no. of formula units per cell

           A_{c} = molecular weight of cation = 90.5 g/mol

           A_{a} = molecular weight of anion = 37.3 g/mol

          V_{c} = volume of cubic cell = 3.55 g/cm^{3}

           a = edge length of unit cell = 3.9 \times 10^{-8} cm

        N_{A} = Avogadro's number = 6.023 \times 10^{23}

          \rho = density = 3.55 g/cm^{3}

Now, putting the given values into the above formula as follows.

           n' = \frac{\rho a^{3}N_{A}}{(A_{c} + A_{A})}

                      = \frac{3.55 g/cm^{3} \times (3.9 \times 10^{-8})^{3} \times 6.023 \times 10^{23}}{(90.5 + 37.3)}

                     = 0.9

                    = 1 (approx)

Therefore, we can conclude that out of the given options crystal structure of cesium chloride is possible for this material.

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arsen [322]

Answer:

92.65256 cm^3

Explanation:

To find this, we can simply multiply all three dimensions to get the answer in cubic centimeters, and we get the answer above. If you want to be more specific, we can go by the sigfig rule and the answer would be rounded to 93 cm^3.

5 0
2 years ago
Metals are considered this if they can be made into sheets
Korvikt [17]

They are considered malleable. They can be made into sheets

Happy to help! Please mark me as the brainliest!

6 0
2 years ago
Olympic cyclist fill their tires with helium to make them lighter. Calculate the mass of air in an air filled tire and the mass
inn [45]

<u>Answer:</u> The mass difference between the two is 7.38 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

PV=nRT

where,

P = pressure = 125 psi = 8.50 atm    (Conversion factor:  1 atm = 14.7 psi)

V = Volume = 855 mL = 0.855 L    (Conversion factor:  1 L = 1000 mL)

T = Temperature = 25^oC=[25+273]K=298K

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

8.50atm\times 0.855L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{8.50\times 0.855}{0.0821\times 298}=0.297mol

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

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

  • <u>For air:</u>

Moles of air = 0.297 moles

Average molar mass of air = 28.8 g/mol

Putting values in equation 1, we get:

0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol\times 28.8g/mol)=8.56g

Mass of air, m_1 = 8.56 g

  • <u>For helium gas:</u>

Moles of helium = 0.297 moles

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol\times 4g/mol)=1.18g

Mass of helium, m_2 = 1.18 g

Calculating the mass difference between the two:

\Delta m=m_1-m_2

\Delta m=(8.56-1.18)g=7.38g

Hence, the mass difference between the two is 7.38 grams.

5 0
2 years ago
If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct
MA_775_DIABLO [31]

Answer:

1. sp  = XY₂

2. sp²  = XY₂Z, XY₃

3. sp3³ = XY₄, XY₂Z₂, XY₃Z

4. sp³d  = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

5. sp³d² = XY₆, XY₄Z₂, XY₅Z

Explanation:

this is quite dicey, so it should be looked into carefully.

we would classify each of the abbreviation according to their  hybridization and it electron domain.

⇒ sp hybridization = XY₂

in this, we can see that the central atom X is bonded to two outer atoms Y.

this makes the no of hybrid orbitals and the no of sigma bonds both 2.

electron domain = 2.

⇒ sp² hybridization = XY₂Z, XY₃

Here we can see the central atom X bonded with three outer atoms Y in XY₂Z and in XY₃. For XY₂Z molecule, the no of sigma bonds is 2 and the no of hybrid orbitals is 3. While for XY₃ molecule, the no of sigma bonds is 3 while the no of hybrid orbital is 3.

electron domain = 3.

⇒ sp³ hybridization = XY₄, XY₂Z₂, XY₃Z

for XY₄ molecule, the central atom X is bonded with four outer atoms Y. It has 4 numbers of both the sigma and orbital atoms.

In XY₂Z₂, the central atom X is bonded to 2 outer atoms Y, and has 2 lone pairs Z. From this, the no of hybrid orbitals is 4 and the no of sigma bonds is 2, with 2 lone pairs causing the sp³ hybridization.

⇒ sp³d hybridization = XY₅, XY₂Z₃, XY₃Z₂, XY₄Z

for all the molecules listed above, the sum of both the lone pairs and the outer atoms both give a total of 5, hence have the sp³d structure, viz;

XY₂Z₃:

total electron domain = 2+3 = 5  

XY₃Z₂:

total electron domain = 2+3 = 5

XY₄Z:

total electron domain = 1+4 = 5

⇒ sp³d² hybridization = XY₆, XY₄Z₂, XY₅Z

Same thing goes for the above molecule, where the sum of both the outer atoms and the lone pairs gives a total of 6 as can be seen in the example below.

XY4Z2:

total electron domain = 2+4 = 6

cheers, i hope this helps.

5 0
2 years ago
Hydrogen was collected over water using the approach in the manual. The water temperature was 220C and the measured pressure ins
olasank [31]

Answer:

Pressure of hydrogen gas = 695.2 mmHg

Explanation:

Given:

Water temperature = 22°C

Pressure inside the tube = 715 mmHg

Find:

Pressure of hydrogen gas

Computation:

Using vapor pressure of water table

Water pressure at 22°C = 19.8 mmHg

Pressure inside the tube = Pressure of hydrogen gas + Water pressure at 22°C

715 = Pressure of hydrogen gas + 19.8

Pressure of hydrogen gas = 715 - 19.8

Pressure of hydrogen gas = 695.2 mmHg

3 0
2 years ago
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