The weather behind a cold front is dominated by which of the following?

Suppose a new group 5A element is discovered. Photoelectron spectroscopy (PES) reveals that its last two peaks occur at 1370 kJ/

iren2701 [21]

Answer: The average valence electron energy (AVEE) of this element =

1014.2 KJ/ mol or 1.0142mJ/mol.

Explanation:

The average valence electron energy = (number of electrons in s subshell x Ionization energy of that subshell) + (number of electrons in p subshell x Ionization energy of that subshell) / total number of electrons in both subshells of the valence shells.

The 5A elements are non-metals like Nitrogen and Phosphorus with the metallic character increasing as you go down the group, So a new 5A element will have characteristics of its group with 5 valence electron in its outermost shell represented as ns2 np3

Therefore the average valence electron energy (AVEE) of this element will be calculated as

The average valence electron energy = (2 x 1370 kJ/mol + 3 x 777 kJ/mol.) / 5

2740+2331/ 5 =5071/5

=1014.2 KJ/ mol or 1.0142mJ/mol.

5 0

3 years ago

The state of matter in which molecules are attracted to each other but can change position is a

slavikrds [6]

i believe the answer is a liquid

4 0

3 years ago

Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0

2 years ago

At a cost of $1600/oz, how much would you have to pay for a solid cubic foot of gold?

Alex777 [14]

We'll begin by calculating the mass in ounce (oz) of a cube foot (ft³) of gold. This can be obtained as follow:

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Density of gold = 19298 oz/ft³

Volume of gold = 1 ft³

Density = mass /volume

19298 = mass / 1

Finally, we shall determine the cost of 19298 oz of gold. This can be obtained as follow:

1 oz = $ 1600

Therefore,

19298 oz = 19298 × 1600

19298 oz = $ 30876800

Therefore, a solid cube foot of gold (i.e 19298 oz) will cost $ 30876800

Learn more: brainly.com/question/15407624

4 0

2 years ago

1. When you first opened the bottle of coke the pressure of gas in the coke (increased/decreased) and the dissolved gas( stays i

Serga [27]

Answer:

1. When you first opened the bottle of coke the pressure of gas in the coke (increased) and the dissolved gas(leaves) the coke. 2. When you placed the coke in hot water , the pressure of gas in the coke (increased) and the dissolved gas(leaves) the coke 3. Therefore to increase the solubility of a gas in a liquid ( that is to make a gas more soluble in a liquid.

Explanation:

Hope it helps.

3 0

2 years ago