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ELEN [110]
2 years ago
12

Sodium chloride (table salt,) whose formula is NaCl: Ionic Covalent

Chemistry
2 answers:
Vesna [10]2 years ago
7 0
Since sodium chloride contains both a metal AND a nonmetal, the combination of those would result in an ionic bond.
lys-0071 [83]2 years ago
6 0

Answer: Ionic

Explanation:

Ionic bonds are formed by transfer of electrons between metal and non metals.  Covalent bonds are formed by sharing of electrons between non metals

In Sodium chloride

Electronic configuration of sodium:

[Na]=1s^22s^22p^63s^1

Sodium atom will loose one electron to gain noble gas configuration and form sodium cation with +1 charge.

[Na^+]=1s^22s^22p^63s^0

Electronic configuration of chlorine:

[Cl]=1s^22s^22p^63s^23p^5

Chlorine atom will gain one electron to gain noble gas configuration and form chloride ion with -1 charge.

[Cl^-]=1s^22s^22p^63s^23p^6

In sodium chloride the one electron from sodium metal gets transferred to chlorine atom and thus is ionic.

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890 has 2 significant figures

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A balloon has a volume of 6.2 liters at 23.2 C. The balloon is then heated to a temperature of 144.0 C. What is the volume of th
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8.7 L

Explanation:

T2(V1/T1) = V2

417.15 K(6.2 L/296.45 K) = 8.7 L

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3 years ago
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Read 2 more answers
Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
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