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2 years ago

discuss how variations in electronegativity result in the unequal sharing of electrons in polar molecules.

1 answer:

6 0

Variations in electronegativity prompt in the unequal halves of electrons in polar molecules because when one atom is more electronegative than the other, it becomes more polar than the other.

It results in the more electronegative atom to have a slightly negative (-ve) charges, and the other atom to have partial or slightly positive(+ve) charges.

Polar molecules have unequal sharing of electrons because the atoms have unequal attraction for electrons so the sharing is unequal.

The larger the difference in electronegativity between the two atoms, the more the polar the bond.

Hydrogen bonds are involved in unequal sharing of electrons between two atoms.

To know more about variations in electronegativity in polar molecules here :

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What is hybridization in chemistry? Give some example

dmitriy555 [2]

Answer:

Co2,so2

Explanation:

Hybridization is the mixing of atomic orbitals to form new hybrid orbitals.

Molecule shapes arise from how many hybrid orbitals the central atom uses and how they are arranged.

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3 years ago

Which blocks will move under these conditions and explain

Nonamiya [84]

Answer:

Correct: B

Explanation:

The first block wont move because:

Σf=3-3=0N.

The second block will move because:

Σf:5-3=2N (direction to the right).

The third block will move because:

Σf=3N (direction to the right).

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3 years ago

Give an example of an oxidisation reaction!

konstantin123 [22]

<span>When an atom or compound is oxidized, its properties change. For example, when an iron object undergoes oxidation, it is transformed because it has lost electrons. Unoxidized iron is a strong, structurally sound metal, while oxidized iron is a brittle, reddish powder. The diagram below illustrates what happens to an atom of iron as it is oxidized

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4 years ago

A sample consisting of n mol of an ideal gas undergoes a reversible isobaric expansion from volume Vi to volume 3Vi. Find the ch

kolezko [41]

Answer:

The change in entropy of gas is $\Delta S= nC_{P}ln3$

Explanation:

n= Number of moles of gas

Change in entropy of gas = $ds= \int \frac{dQ}{T}$

$dQ= nC_{p}dT$

From the given,

$V_{i}=V$

$V_{f}=3V$

Let "T" be the initial temperature.

$\frac {V_{i}}{T_{i}}=\frac {V_{f}}{T_{f}}$

$\frac {V}{T}=\frac {3V}{T_{f}}$

${T_{f}} = 3T$

$\int ds = \int ^{T_{f}}_{T_{i}} \frac{nC_{P}dT}{T}$

$\Delta S = nC_{p}ln(\frac{T_{f}}{T_{i}})$

$\Delta S = nC_{p}ln3$

Therefore, The change in entropy of gas is $\Delta S= nC_{P}ln3$

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3 years ago

How much heat energy is needed to raise the temperature of 59.7g of cadmium from 25°C to 100°C? The specific heat of cadmium is

labwork [276]

Using the Fundamental Equation of Calorimetry, we have:

$Q=mc\Delta T \\ Q=59.7.0.231.(100-25) \\ \boxed {Q=1034.3025J}$

If you notice any mistake with my english, please know me, because I am not native.

7 0

4 years ago