All categories

3 years ago

What is the electron configuration for potassium (K)?

Chemistry

1 answer:

Bond [772]3 years ago

3 0

Answer:

1s2 2s2 2p6 3s2 3p6 4s1

Explanation:

Aufbau's principal of the arrangement of electronic configuration states that, electrons fill orbitals of lower energy first before the higher energy. 1s2 2s2 2p6 3s2 3p6 4s2 3d10...

This arrangement explains the Aufbau's principal, hence, or therefore, the 1s2 2s2 2p6 3s2 3p6 4s1 is the right answer for the representation of the electron configuration for potassium (K).

hope this was helpful!

You might be interested in

Do you expect potassium iodide to be soluble in cyclohexane

Allushta [10]

Answer is: no, potassium iodide has low solubility in cyclohexane.
Potassium iodide (KI) is ionic compound, salt, that dissolve good in polar solvents (for example water), but it is very low soluble in non-polar solvents (in this example cyclohexane C₆H₁₂).
Cyclohexane has zero net polarity, because it is a symmetric molecule, with sp2 hybridization of carbon.

8 0

4 years ago

Read 2 more answers

What kind of air do winds coming from land masses bring?

Ludmilka [50]

Low air pressure cause it was on my test last year

8 0

3 years ago

Read 2 more answers

Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl

Archy [21]

Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

Mass of helium (He) gas = 6.24 g

Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

$\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles$

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

$V\propto n$

or,

$\frac{V_1}{V_2}=\frac{n_1}{n_2}$

where,

$V_1$ = initial volume of gas = V

$V_2$ = final volume of gas = 2V

$n_1$ = initial moles of gas = 1.56 mole

$n_2$ = final moles of gas = ?

Now we put all the given values in this formula, we get

$\frac{V}{2V}=\frac{1.56mole}{n_2}$

$n_2=3.12mole$

Now we have to calculate the mass of helium gas at doubled volume.

$\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He$

$\text{Mass of }He=3.12mole\times 4g/mole=12.48g$

Therefore, the mass of helium gas added must be 12.48 grams.

4 0

3 years ago

The elements of a delta Hf of ___ kj\mole at standard temp and pressure

EastWind [94]

Answer:

One mole of a substance forms Instead of standard atmospheric pressure and temperature from its pure elements{ conditions being the same} then this change and enthalpy is known as enthalpy of formation.

Explanation:

The equation of this formation can let us know what actually happens

ΔH = ΔHf(products) - ΔHf (reactants)

Hence we can say this change in enthalpy formation is the difference between the sum of product and sum of reactants.

6 0

3 years ago

Natural gas (CH4) has a molar mass of 16.0 g/mole. You started out the day with a tank containing 200.0 g of natural gas. At the

hodyreva [135]

Considering the definition of molar mass, the moles of gas used are 10.625 moles.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Amount of moles used</h3>

Natural gas has a molar mass of 16.0 g/mole.

You started out the day with a tank containing 200.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 200 grams are contained in how many moles?

$amount of moles at the beginning=\frac{200 gramsx1 mole}{16 grams}$

amount of moles at the beginning= 12.5 moles

At the end of the day, your tank contains 30.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 30 grams are contained in how many moles?

$amount of moles at the end=\frac{30 gramsx1 mole}{16 grams}$