Increasing the airplane's speed or wing size does which of the following

Explain why the ionization energy to remove a second electron from potassium is higher than the ionization energy to remove four

never [62]

Explanation:

It is more difficult to remove electrons from the second shell or energy level because of the imbalance between the positive nuclear charge and the remaining electrons.

The amount of energy required to remove electrons in ground state of an atom is the ionization energy.
The first ionization energy is the energy needed to remove the most loosely bound electron of an atom in the gas phase in ground state.
The second energy has a greater nuclear pull as it is closer to the nucleus.
Both potassium and silicon have the same number of energy levels.

6 0

3 years ago

How can the actions of people affect runoff and absorption in a positive way? How do they affect them in a negative way?

lora16 [44]

Answer:

pollution and waste every where like in are ouceans

Explanation:

4 0

3 years ago

The chemical formula for diiodine nonaflouride

stira [4]

Answer:

The chemical formula is I2F9

Explanation:

3 0

4 years ago

How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH4NO3 in order to prepare a 0.452 m solution?

Vanyuwa [196]

Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

$Molality=\frac{n\times 1000}{W_s}$

where,

n= moles of solute

Moles of $NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles$

$W_s$ = weight of the solvent in g = ?

$0.452=\frac{0.348\times 1000}{W_s}$

$W_s=770g$

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate $NH_4NO_3$ in order to prepare a 0.452 m solution

5 0

3 years ago

Select correct examples of linear molecules for five electron groups. Select correct examples of linear molecules for five elect

Shalnov [3]

Explanation:

Formula to calculate hybridization is as follows.

Hybridization = $\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of $BeCl_{2}$ is as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[2 + 2]$

= 2

Hybridization of $BeCl_{2}$ is sp. Therefore, $BeCl_{2}$ is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of $XeF_{2}$ is calculated as follows.

Hybridization = $\frac{1}{2}[V + N - C + A]$

= $\frac{1}{2}[8 + 2]$

= 5

Therefore, hybridization of $XeF_{2}$ is $sp^{3}d. Therefore, [tex]XeF_{2}$ is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

Thus, we can conclude that out of the given options $XeF_{2}$ is the correct examples of linear molecules for five electron groups.

7 0

3 years ago