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Kruka [31]
1 year ago
14

Repulsion of electrons within two interacting molecules produces changes in electron distribution. This change in electron distr

ibution creates temporary dipole moments.Which of the following does this explain? Choose one or more: A. This explains how two noble gases' molecules can have an attractive force between them. B. This explains why ammonia and nitrogen gas exhibit an attractive force between them.C. This explains why long hydrocarbon chains have relatively high boiling points. D. This explains how the molecules hydrogen fluoride and methanol can exhibit uncharacteristically strong intermolecular forces. E. This explains why the dipole-dipole attractive force between dimethyl ether and acetone does not entirely account for the attractive force between these molecules.
Chemistry
1 answer:
ycow [4]1 year ago
6 0

Answer:

D

Explanation:

This explains how two noble gases molecules can have an attractive force between them.

This force is called as van dar Waals forces.

It plays a fundamental role in fields in as diverse as supramolecular chemistry structural biology .

If no other forces are present, the point at which the force becomes repulsive rather than attractive as two atoms near one another is called the van der Waals contact distance. This results from the electron clouds of two atoms unfavorably coming into contact.[1] It can be shown that van der Waals forces are of the same origin as the Casimir effect, arising from quantum interactions with the zero-point field.[2] The resulting van der Waals forces can be attractive or repulsive.[3] It is also sometimes used loosely as a synonym for the totality of intermolecular forces.[4] The term includes the force between permanent dipoles (Keesom force), the force between a permanent dipole and a corresponding induced dipole (Debye force), and the force between instantaneously induced dipoles

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A piece of lead loses 78.0 J ofheat and experiences a decrease in temperature of 9.0C the specific heat of lead is .130J/gC what
Ratling [72]
The  mass of  the piece  of  lead   is    calculated  using the  below  formula
Q(heat)= mC delta T

Q = 78.0  j
M=mass =?
C=specific heat capacity (   0.130 j/g/c
delat T=change in temperature =  9.0 c

by  making  M  the  subject  of  the formula
M = Q/ c delta T

M=  78.0 j/ 0.130 j/g/c  x  9.0 c =  66.7 g  of  lead
3 0
2 years ago
B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 reacts
n200080 [17]

Answer:

Heat realesed is - 2192.7 kJ

Explanation:

First lets have the chemical equation balanced, and then solve the question based on the fact that the enthalpy change for a reaction is the sum of the enthalpies of formation of products minus reactants.

    B₅H₉ (l) +     O₂ (g) ⇒       B₂O₃ (s) +      H₂O(l)

B atoms we have 5 reactants and 2 products, so the common mltiple is 10 and we have

  2  B₅H₉ (l) +     O₂ (g) ⇒      5 B₂O₃ (s) +      H₂O(l)

Now balance H  by multiplying by 9 the H₂O

  2  B₅H₉ (l) +     O₂ (g) ⇒      5 B₂O₃ (s) +    9 H₂O(l)

Finally, balance the O since we have 24 in products by multiplying by 12 the

O₂ ,

        2  B₅H₉ (l) +  12 O₂ (g) ⇒      5 B₂O₃ (s) +    9 H₂O

ΔHrxn = 5 x ΔHºf B₂O₃  + 9 x   ΔHºf H₂O  -  ( 2 x ΔHºf  B₅H₉ + 12 x  ΔHºf O₂  )

We have all the  ΔHºf s except oxygen but remember the enthalpy of formation of a pure element in its standard estate is cero.

ΔHrxn =  5 mol x ( -1272 kJ/mol )+ 9 mol  x ( -285.4 kJ/mol )   - (  2mol x 732 kJ/mol )

= -8928.6 kJ - 1464  kJ = -10,392 kJ

Now this enthalpy change was based in 2 mol  reacted according to the balanced equation, so for 0.211 mol of  B₅H₉  we will have:

-10,392 kJ/ 2 mol B₅H₉   x 0.211 mol B₅H₉ = - 2192.7 kJ

8 0
1 year ago
Read 2 more answers
Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of
gtnhenbr [62]

Answer:

The concentration of Li (in wt%) is 3,47g/mol

Explanation:

To obtain the 2,42g/cm³ of density:

2,42g/cm³ = 2,71g/cm³X + 0,534g/cm³Y <em>(1)</em>

<em>Where X is molar fraction of Al and Y is molar fraction of Li.</em>

X + Y = 1 <em>(2)</em>

Replacing (2) in (1):

Y = 0,13

Thus, X = 0,87

The weight of Al and Li is:

0,87*26,98g/mol = 23,4726 g of aluminium

0,13*6,941g/mol = 0,84383 g of lithium

The concentration of Li (in wt%) is:

0,84383g/(0,84383g+23,4726g) ×100= <em>3,47%</em>

6 0
1 year ago
How much 2 M HBr is needed to neutralize 380 mL of 0.1 M NH4OH?
Nata [24]

Answer:

19ml

Explanation:

trust me

8 0
1 year ago
The K w for water at 0 ∘ C is 0.12 × 10 − 14 M 2 . Calculate the pH of a neutral aqueous solution at 0 ∘ C . p H = Is a pH = 7.2
labwork [276]

Answer:

pH → 7.46

Explanation:

We begin with the autoionization of water. This equilibrium reaction is:

2H₂O  ⇄   H₃O⁺  +  OH⁻            Kw = 1×10⁻¹⁴         at 25°C

Kw = [H₃O⁺] . [OH⁻]

We do not consider [H₂O] in the expression for the constant.

[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴   →  1×10⁻⁷ M

Kw depends on the temperature

0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻]  → [H₃O⁺] = [OH⁻]         at 0°C

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M

- log [H₃O⁺] = pH

pH = - log 3.46×10⁻⁸ → 7.46

5 0
2 years ago
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