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2 years ago

Among the intermolecular forces,which forces are typically the weakest

Chemistry

1 answer:

nekit [7.7K]2 years ago

8 0

Answer:

The London dispersion force is the weakest intermolecular force. The London dispersion force is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles.

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Use the periodic table to identify the element with the electron configuration 1s²2s²2p⁴. Write its orbital diagram, and give th

natta225 [31]

Answer:

1. Orbital diagram

2p⁴ ║ ↑↓ ║ "↑" ║ ↑

2s² ║ ↑↓ ║

1s² ║ ↑↓ ║

2. Quantum numbers

n = 2,
l = 1,

$m_{l}$ = 0,

$m_{s}$ = +1/2

Explanation:

The fill in rule is:

Follow shell number: from the inner most shell to the outer most shell, our case from shell 1 to 2

Follow the The Aufbau principle, 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
Hunds' rule: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).

So, the orbital diagram of given element is as below and the sixth electron is marked between " "

2p⁴ ║ ↑↓ ║ "↑" ║ ↑

2s² ║ ↑↓ ║

1s² ║ ↑↓ ║

The quantum number of an electron consists of four number:

n (shell number, - 1, 2, 3...)
l (subshell number or orbital number, 0 - orbital s, 1 - orbital p, 2 - orbital d...)
$m_{l}$ (orbital energy, or "which box the electron is in"). For example, orbital p (l = 1) has 3 "boxes", it was number from -1, 0, 1. Orbital d (l = 2) has 5 "boxes", numbered -2, -1, 0, 1, 2
$m_{s}$ (spin of electron), either -1/2 or +1/2

In our case, the electron marked with " " has quantum number

n = 2, shell number 2,
l = 1, subshell or orbital p,

$m_{l}$ = 0, 2nd "box" in the range -1, 0, 1

$m_{s}$ = +1/2, single electron always has +1/2

3 0

3 years ago

A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and

lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

$F_{solution}$ = mass of solvent + mass of oil (i.e A+C)

Feed Phase:

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

= 25kg

Mass ratio of oil to solution $Y_{F}$ = $\frac{Mass C}{Mass (A+C)}$

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg

therefore $Y_{F}$ = $\frac{2.5}{25}$

= 0.1

mass ratio of solid to solution $Y_{A}$ = $\frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}$

Solvent Phase:

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

Underflow:

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = $\frac{mass B}{mass(A+C)}$

solution in underflow E₁ = Mass (A+C)

Overflow:

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75= E₁ × N₁

75 = E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

5 0

3 years ago

Upper n subscript 2 (g) plus 3 upper H subscript 2 (g) double-headed arrow 2 upper N upper H subscript 3 (g). At equilibrium, th

artcher [175]

Answer:

The equilibrium constant is:

$k_c=0.0030M^{-2}$

Explanation:

The correct equation is:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.

The equation for the equilibrium constant is:

$k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}$

Substituting:

$k_c=\dfrac{(0.105M)^2}{(1.1M)\cdot (1.50M)^3}$

$k_c=0.0030M^{-2}$

6 0

3 years ago

2. What is the current running through a toaster if the resistance is 15

pogonyaev

I’ll help you 8n a bit

8 0

2 years ago

What can the student predict if she finds the mass of the whole candy bar, including the wrapper?

Vesna [10]

Answer:

A student can predict Amount of calories present if she finds the mass of whole candy bar,including the wrapper.

Explanation:

Based on the mass of a whole candy bar a student can predict the calories contained in the candy bar. If the mass of candy bar is more than it means it has more calories and if the candy bar has less mass than it shows the presence of fewer calories.

More calories mean more sugar content in the candy bar. With sugar, the candy bar has carbohydrates also. The sugar present in the bar gives us energy.

6 0

3 years ago