vera_ewing
  • vera_ewing
Which of the following illustrates an everyday-life example of how London forces are induced? A. The attraction of a magnet with a metal object B. The attraction of your hair to a balloon after the balloon is rubbed against it C. The attraction between sodium and chloride in table salt D. The attraction of an object to Earth through gravity
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
which do u think it is @vera_ewing
anonymous
  • anonymous
I think its D
vera_ewing
  • vera_ewing
@sweetburger

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

sweetburger
  • sweetburger
I feel like you asked this question before or someone did. I also remember that no one could figure out an answer for it as none of them make any firm sense.
sweetburger
  • sweetburger
sweetburger
  • sweetburger
London Dispersion forces from what I remember is when electrons become to close to each other and end up repelling. This in turn induces an attraction between molecules.
vera_ewing
  • vera_ewing
Correct answer is B. The attraction of your hair to a balloon after the balloon is rubbed against it
sweetburger
  • sweetburger
Alright I feel like you would have to have some underlying knowledge of some areas of physics to answer that...
vera_ewing
  • vera_ewing
How does entropy change when an ionic solid dissolves in water? A. The entropy can only decrease since the system becomes more disordered. B. The entropy can only increase since the system becomes more disordered. C. The entropy typically increases since the crystal lattice and the ordered water become less ordered, but it can decrease if the hydrated ions become more ordered. D. The entropy typically decreases since the crystal lattice and the ordered water become less ordered, but it can increase if the hydrated ions become more ordered.
sweetburger
  • sweetburger
C. makes sense as the first statement is true and the second statement is true if a non polar compound is dissolved the entropy will actually decrease as the hydrated ions become more ordered. Then again the question is asking specifically about ionic solids dissolving. That leaves B a possibility.
vera_ewing
  • vera_ewing
sweetburger
  • sweetburger
Things that will affect boiling point are the mass of the substance and usually what kind of forces are acting on the substance.
sweetburger
  • sweetburger
The higher the mass the higher the boiling point and the stronger the forces acting on the substance the higher the boiling point.
vera_ewing
  • vera_ewing
So which one do you think it is?
sweetburger
  • sweetburger
I think it might be D,B,C,A
sweetburger
  • sweetburger
D has the highest molar mass and experiences hydrogen bonding, B has hydrogen bonding, c has a higer molar mass than A, and A is nonpolar experiencing only weak intermolecular forces with low mass
vera_ewing
  • vera_ewing
If a balloon is filled with helium and has a volume of 1.56 L at 25°C, what will happen to the balloon if it is placed in a refrigerator at 4°C? (Assume the pressure remains constant.) A. The volume of the balloon will decrease by 1.45 L. B. The volume of the balloon will decrease by 1.31 L. C. The volume of the balloon will decrease by 0.110 L. D. The volume of the balloon will increase by 0.118 L.
sweetburger
  • sweetburger
convert to kelvins first
sweetburger
  • sweetburger
so 1.56L/298K= XL/274K then you can solve for X which will give you a value in Liters
sweetburger
  • sweetburger
this is using the equation of V1/T1=V2/T2
sweetburger
  • sweetburger
anyways because the temperature decreased we know that the volume of the gas will decrease as well as the root mean square speed of the gas is smaller our answer becomes 1.45L
vera_ewing
  • vera_ewing
Oh it was C.
sweetburger
  • sweetburger
I should have done 1.56-1.45 = .110
sweetburger
  • sweetburger
it asked how much did it decrease by. My bad...
vera_ewing
  • vera_ewing
:) It's okay. If a gas tends to exhibit ideal gas behavior under certain laboratory conditions, which statement best describes the gas at these conditions? A. It contains few covalent bonds. B. It experiences a large number of intermolecular interactions. C. It contains many covalent bonds. D. It experiences few intermolecular interactions.
sweetburger
  • sweetburger
ideal gas behaviour will usually have gases experiencing few intermolecular interactions
anonymous
  • anonymous
Ideal gas is the one that has few attractive forces in the molecules, has high and constant kinetic energy and they behave as small spheres. Therefore, I would go for D.
sweetburger
  • sweetburger
Well said @Hoslos
vera_ewing
  • vera_ewing
Nice, thanks @Hoslos
vera_ewing
  • vera_ewing
anonymous
  • anonymous
The trick was that, it cannot be A, which would be your answer @sweetburger , because if for instance you have nitrogen molecules in a container, they are full of covalent bonds between every 2 N atoms. The interaction between these is what is scarce.
vera_ewing
  • vera_ewing
Ohh. Okay thank you @Hoslos ! How about this one?
anonymous
  • anonymous
An organic compound with a large chain, rather than a branched one has a higher boilinh point. Thus, C.
vera_ewing
  • vera_ewing
C. A is the answer or D. C ?
sweetburger
  • sweetburger
@Hoslos My answer to the previous question was also D. as I said "ideal gas behaviour will usually have gases experiencing few intermolecular interactions" which is the exact wording of D. I was just complementing on how well you explained it :)
vera_ewing
  • vera_ewing
@Hoslos Is it C. A or D. C ?
anonymous
  • anonymous
C.A
vera_ewing
  • vera_ewing
Which of the following would cause a gas to deviate most significantly from ideal behavior? A. Metallic bonds B. Non polar bonds C. Hydrogen bonds D. London forces
sweetburger
  • sweetburger
C.
anonymous
  • anonymous
Oh, no problem @sweetburger !
sweetburger
  • sweetburger
if you ever see a metal in a gaseous state its probably pretty hot lol
vera_ewing
  • vera_ewing
Which molecule has the strongest intermolecular forces? A. HOCH2CHOHCHOHCH2OH B. CH3CH2CH2OH C. CH3CH2CH2CH3 D. HOCH2CH2CH2OH
sweetburger
  • sweetburger
I think A. looks like it has the most O-H bonds which would correspond with the most hydrogen bonding sites on the molecule
anonymous
  • anonymous
Correct @sweetburger .
vera_ewing
  • vera_ewing
Last one! Two atoms with charges of +1 and -1 are separated by a distance of 0.5 angstroms. What would happen to the lattice energy between ions if the distance increases to 2 angstroms? A. The lattice energy decreases by a factor of 16. B. The lattice energy decreases by a factor of 4. C. The lattice energy increases by a factor of 4. D. The lattice energy increases by a factor of 16.
sweetburger
  • sweetburger
Im not sure if this a application of Coloumbs Law, but I am thinking it is.
vera_ewing
  • vera_ewing
@Hoslos What do you think the answer is?
sweetburger
  • sweetburger
F=k(q1)(q2)/r so if The distance is increasing between the molecules then the radius would become larger which would in turn decrease the lattice energy by what I am assuming is a factor of 4
sweetburger
  • sweetburger
I am not completely sure somewhat might want to confirm or have another idea.
sweetburger
  • sweetburger
I do know from this application though that the lattice energy/ energy of attraction would decrease with a larger radius so it probably will not be C. or D.
vera_ewing
  • vera_ewing
@chmvijay What do you think?
sweetburger
  • sweetburger
You could look at this from another stand point Lets say this isnt 2 different atoms and say its the earth and the sun The earth and the sun have set force of attraction between themselves. If you distance between each other became increased the force of attraction between the earth and the sun would become smaller. I remember learning this in Physics as a source.
sweetburger
  • sweetburger
If you increase the distance between the earth and the sun the force of attraction between the earth and the sun would become smaller.
vera_ewing
  • vera_ewing
So @sweetburger the answer is B?
chmvijay
  • chmvijay
The lattice energy decreases by a factor of 4.
chmvijay
  • chmvijay
apply by born lande equation
vera_ewing
  • vera_ewing
The answer was A.
chmvijay
  • chmvijay
ok let me check out again
sweetburger
  • sweetburger
4/.25=16 the coloumbs equation is r^2 in the bottom which of course i forgot to add
chmvijay
  • chmvijay
coulombs equation wont apply for lattice energy
sweetburger
  • sweetburger
Oh yea this should be the Born-Lande equation, but I am not sure how this works out.
chmvijay
  • chmvijay
its inversely proportion to Distance and not the square of it that is so its 4 and not 16
anonymous
  • anonymous
The formula of attraction of ions based from their relative distance is given by\[F=\frac{ q _{1}*q _{2} }{ r ^{2} }\] Given the charges and distances of them, we find the first answer, which is \[F=\frac{ +1*-1 }{ (0.5*10^{-10})^{2} }=-4*10^{20}\] Let us assume that energy can be like that of kinetic, Ke=Fv and that v or velocity is constant. The other case is \[F=\frac{ -1*1 }{ (2*10^{-10})^{2} }=-2.5*10^{19}\] Finding the ratio between them , will be \[(\frac{ -2.5*10^{19} }{ -4*10^{20} })^{-1}=0.062^{-1}=16\] This is how I take it. Therefore, the answer shall be A.
vera_ewing
  • vera_ewing
Yep, that's correct @Hoslos :) Thank you!
anonymous
  • anonymous
My pleasure. Nice question though.
sweetburger
  • sweetburger
Nice job @Hoslos :)
Photon336
  • Photon336
For columbs law we see that F = kQ1Q2/r^2 where r is the distance between the two charges. If we look at the formula we can see that the force is inversely proportional 1/r^2. If you increase the distance r between two charges the force should go down by a factor of 1/r^2
anonymous
  • anonymous
Thanks @sweetburger !
rvc
  • rvc
good job @sweetburger @Hoslos :)
sweetburger
  • sweetburger
These are all based off the equation that F=(k)(m1)(m2)/(r)^2 which appears to have many applications in which k is a specific constant and m1 and m2 are 2 objects/charges and r will always be the distance between them. This allows for a force of attraction to be found between objects/charges.
anonymous
  • anonymous
Yes.
Photon336
  • Photon336
If you look at columbs law you see that F is inversely proportional to 1/r^2 The distance goes up by a factor of 4 So 1/(4)^2 = 16 the force decreases by a factor of 16 when the distance goes up by a facto of 4.

Looking for something else?

Not the answer you are looking for? Search for more explanations.