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2 years ago

6

Binary Covalent (molecular) Compounds

1 answer:

prisoha [69]2 years ago

5 0

Answer:

why is ur name different sexxxy............................................................

Explanation:

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Which of the following statements incorrectly describes the assumptions of the kinetic molecular theory? Group of answer choices

Flauer [41]

Sorry
J vre fdtnyscdvf. H Fr,I

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3 years ago

What is the best reason for the difference in properties of LiCL and C6H14O

statuscvo [17]

The substances have different atomic masses, one is organic and one is inorganic, and Cl and Oxygen belong to different groups. There are more differences but these are the most basic ones.

7 0

3 years ago

. A container with a volume of 0.50 L contains air at a pressure of 1.0 atm. If the volume is reduced to 0.10 L at constant temp

marin [14]

Answer:

The answer to your question is Pressure = 5 atm

Explanation:

Data

Volume 1 = V1 = 0.5 l

Pressure 1 = P1 = 1 atm

Volume 2 = V2 = 0.1 l

Pressure 2 = P2 = x

Formula

To solve this problem use the Boyle's equation

V1P1 = V2P2

Solve for P2

P2 = V1P1/V2

Substitution

P2 = (0.5 x 1) / 0.1

Simplification

P2 = 0.5/0.1

Result

P2 = 5 atm

3 0

3 years ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Delicious77 [7]

<u>Answer:</u>

<u>For a:</u> The edge length of the crystal is 533.5 pm

<u>For b:</u> The atomic radius of potassium is 231.01 pm

<u>Explanation:</u>

<u>For a:</u>

To calculate the lattice parameter or edge length of the crystal, we use the equation:

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

where,

$\rho$ = density = $0.855g/cm^3$

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

M = atomic mass of metal = 39.09 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:

$0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm$

<u>Conversion factor:</u> $1cm=10^{10}pm$

Hence, the edge length of the crystal is 533.5 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 = 533.5 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm$

Hence, the atomic radius of potassium is 231.01 pm

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3 years ago

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sveta [45]

Answer:

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3 years ago