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gogolik [260]
2 years ago
6

What type of elements make up a metallic bond

Chemistry
1 answer:
zhenek [66]2 years ago
7 0

Answer:

Explanation:

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51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
What is the empirical formula for a compound if a sample contains 3.72 g of P and 21.28 g of Cl
fenix001 [56]
M(P)=3.72 g
M(P)=31 g/mol

m(Cl)=21.28 g
M(Cl)=35.5 g/mol

n(P)=m(P)/M(P)
n(P)=3.72/31=0.12 mol

n(Cl)=m(Cl)/M(Cl)
n(Cl)=21.28/35.5=0.60 mol

P : Cl = 0.12 : 0.60 = 1 : 5

PCl₅ - is the empirical formula
6 0
3 years ago
Which of the following is the correct sequence for the discovery of subatomic particles? a Electrons> Protons > Neutrons b
laila [671]

Answer:

C

Explanation;

Maybe is C

4 0
3 years ago
Read 2 more answers
Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical
melisa1 [442]

Answer:

-125.4

Explanation:

Target equation is 4C(s) + 5H2(g) = C4H10

These are the data equations for enthalpy of combustion

  1. C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
  2. H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
  3. 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse

To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...

Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.

7 0
3 years ago
Which of the following describes the formation of an ionic bond?
Marysya12 [62]
Redox Reaction is an ionic bond
4 0
3 years ago
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