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2 years ago

Can a molecule with nonpolar bonds ever be polar? Why why not?

Chemistry

1 answer:

Tanya [424]2 years ago

5 0

Answer:

A molecule that has only nonpolar bonds and no polar bonds cannot be polar.

Explanation:

However, a molecule that CONTAINS nonpolar bonds is different, because it can contain polar bonds. A molecule that contains nonpolar bonds can be polar as long as it also contains polar bonds.

Pls, choose me as brainliest!

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Salt is an ionic compound that is produced when a cation and an anion from an acid-base reaction combine, and solutions of salts

Andrej [43]

Answer:

a. True

Explanation:

The statement given regarding salt is correct,

Salt is ionic in nature, as it is composed of a positive ion, i.e. , cation and a negative ion, i.e. , anion.

Salt can be made by the combination of respective acid and base.

There are three nature of salt, acid, base and neutral.

Hence, from the given statement of the question,

The correct option is a. True.

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3 years ago

Describe the difference beetween an extensive and intesive property

olasank [31]

Extensive properties, such as mass and volume, depend on the amount of matter being measured. Intensive properties, such as density and color, do not depend on the amount of the substance present. Physical properties can be measured without changing a substance's chemical identity.

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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

Nimfa-mama [501]

Answer:

lattice parameter = 5.3355x10^-8 cm

atomic radius = 2.3103x10^-8 cm

Explanation:

known data:

p=0.855 g/cm^3

atomic mass = 39.09 g/mol

atoms/cell = 2 atoms

Avogadro number = 6.02x10^23 atom/mol

a) the lattice parameter:

Since potassium has a cubic structure, its volume is equal to:

v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]

substituting values:

v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3

but as the cell volume is

a^3 =v

$a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8$ cm

for a BCC structure, the atomic radius is equal to

$r=\frac{ax\sqrt{3} }{4}=\frac{5.3355x10^{-8}x\sqrt{3} }{4}=2.3103x10^{-8}cm$

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3 years ago

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k0ka [10]

Answer:

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Explanation:

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3 years ago

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