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3 years ago

would you expect potassium to have a high electronegativity or a low electronegativity? explain your answer

Chemistry

1 answer:

GalinKa [24]3 years ago

4 0

A low electronegativity

Explanation:

Potassium is a metal that is expected to have a very low electronegativity value.

Electronegativity is the relative tendency by which an atom attracts valence electrons in a chemical bond.

Potassium is an element in the first group on the periodic table.

The common trend is that electronegativity increases from left to right and decreases down a group.

Potassium as metal will prefer to lose electrons rather than attracting because that will make it achieve the octet configuration that will ensure its stability.
This is why it will have low electronegativity.

Learn more:

Electronegativity brainly.com/question/11932624

#learnwithBrainly

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3 years ago

What is the bohr model for pottasium

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3 years ago

a sample of ammonia contains 9g hydrogen and 42g nitrogen. another sample contains 5g hydrogen .calculate the amount of nitrogen

balandron [24]

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5 g of hydrogen - x g of nitrogen

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3 years ago

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

8090 [49]

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is $P_A$ and vapour pressure of component B is $P_B$ .

$P^0_A$ = Vapour pressure of component A in pure form

$P^0_B$ = Vapour pressure of component B in pure form

$x_A$ =Mole fraction of component A

$x_B=$ =Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

$P_A >P^0_A\cdot x_A$ , $P_B>P^0_B\cdot x_B$

Therefore, $P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B$

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

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3 years ago

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vazorg [7]

I don’t know This one is there a picture ?

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2 years ago