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ycow [4]
3 years ago
14

Iron(II) sulfide has a primitive cubic unit cell with sulfide ions at the lattice points.

Chemistry
1 answer:
masya89 [10]3 years ago
4 0

We have that the  the density of FeS  is mathematically given as

  •  \phi=2.56h/cm^3

From the question we are told

Iron(II) sulfide has a primitive <em>cubic</em> unit cell with <em>sulfide</em> ions at the <em>lattice points.</em>

The ionic radii of iron(II) ions and sulfide ions are 88 pm and 184 pm, respectively.

What is the density of FeS (in g/cm3)?

<h3>Density</h3>

Generally the equation for the Velocity  is mathematically given as

V_c=a^3=(2\pi Fe^{2+}+2\pi S^{2-})^3\\\\Therefore\\\\V=a^3(2\pi*0.088+2\pi 0.184)^3\\\\V=16.98*10^{-23}\\\\Therefore\\\\\phi=n\frac{PFeion+PSion}{VNa}\\\\\phi=3*\frac{55.85+32}{16.9*10^{-23}*6.023*10^{23}}

  • \phi=2.56h/cm^3
  • \phi=2.56h/cm^3

For more information on ionic radii visit

brainly.com/question/13981855

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The heat of vaporization of water at the normal boiling point, 373.2 K, is 40.66 kJ/mol. The molar heat capacity of liquid water
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Answer:

\Delta _{vap}H(300.2K)=43,658\frac{J}{mol}=43.66\frac{kJ}{mol}

Explanation:

Hello!

In this case, according to the Kirchhoff's law for the enthalpy change, it is possible to compute the heat of vaporization at 300.2 K by considering the following thermodynamic route:

\Delta _{vap}H(300.2K)=Cp_{liq}(T_b-T\°)+\Delta _{vap}H\°+Cp_{vap}(T-T_B)

Whereas the first term stands for the effect of taking the liquid from 298.15 K to 373.15 K, the second term stands for the standard enthalpy of vaporization and the last term that of the vapor from the boiling point to 300.2 K; thus we plug in to obtain:

\Delta _{vap}H(300.2K)=75.37\frac{J}{mol*K} (373.2K-298.15K)+40,660\frac{J}{mol} +36.4\frac{J}{mol*K}(300.2K-373.2K)\\\\\Delta _{vap}H(300.2K)=43,658\frac{J}{mol}=43.66\frac{kJ}{mol}

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When silver crystallizes, it forms face-centered cubic cells. The unit cell edge length is 408.7 pm. Calculate the density of si
tester [92]

The density of silver can be calculated by using the following formula:

d=\frac{m}{V}

Here, m is mass and V is volume.

To calculate density, mass of silver needs to be calculated first.

In FCC, the number of Ag atoms in a unit cell will be 4, atomic mass of Ag is 107.87 g/mol and number of atoms in 1 mol are 6.023\times 10^{23} atoms, thus, mass can be calculated as follows:

m=4 atoms\times \frac{1 mol}{6.023\times 10^{23}atoms}\times \frac{107.87 g}{mol}=7.163\times 10^{-22} g.

Now, volume can be calculated as follows:

V=a^{3}

First convert pm to cm:

1 pm=10^{-10} cm

Thus,

408.7 pm=4.087\times 10^{-8} cm

Putting the value to calculate volume,

V=(4.087\times 10^{-8} cm)^{3}=6.83\times 10^{-23} cm^{3}

Now, calculate density as follows:

d=\frac{7.163\times 10^{-22}g}{6.627\times 10^{-23}cm^{3}}=10.5 g/cm^{3}

Therefore, density of silver is 10.5 g/cm^{3}.

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4 years ago
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