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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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3.00 cm

Explanation:

The absorbance can be expressed using <em>Beer-Lambert's law</em>:

A = ε*b*c

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Now we <u>match the absorbance values for both solutions</u>, because we want the absorbance value to be the same for both solutions:

A = ε * 1.00 cm * 7.68x10⁻⁶M = ε * b * 2.56x10⁻⁶ M

And <u>solve for b:</u>

 ε * 1.00 cm * 7.68x10⁻⁶M = ε * b * 2.56x10⁻⁶ M

1.00 cm * 7.68x10⁻⁶M = b * 2.56x10⁻⁶ M

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5 0
1 year ago
Consider the electrolysis of aqueous agno3. Refer to the table of standard reduction potentials as needed. What should form at t
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The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
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<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

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

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
1 year ago
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