Anybody good at chemistry?

[Move The Chains]

lol. I'm trying to figure out if a molecule or ion has a "dipole moment."


:thinking:

[lulu]

Not me:(

[Move The Chains]

(Actual question)


Which one of the following molecules or ions has a dipole moment?

a. IO4-
b. ICl2-
c. SF4
d. XeF4
e. CO2

[Sweetwater Red]

Ask a Brownwood poster. Given the nightly news surely one of them knows
something about chemistry.:thinking:

[BLACK&GOLD4LIFE]

not 100% sure but i think xef4 has 4 dipoles, but they all cancel each other out, but don't just listen to me.

[Move The Chains]

Originally posted by BLACK&GOLD4LIFE
not 100% sure but i think xef4 has 4 dipoles, but they all cancel each other out, but don't just listen to me. I chose that as my answer earlier and got it wrong. I get 3 attempts, so I've got 2 more. I thought for sure that was the answer.:confused:

[BLACK&GOLD4LIFE]

ok i believe it is sf4, check out this http://www.brainmass.com/homework-help/chemistry/general-chemistry/90376

[Move The Chains]

Originally posted by BLACK&GOLD4LIFE
ok i believe it is sf4, check out this http://www.brainmass.com/homework-help/chemistry/general-chemistry/90376 Ok, so that tells me that the answer is SF4 but I need to know HOW to get that. Know what I'm saying? I got a test over this junk Thursday.

[BLACK&GOLD4LIFE]

sry man not much help i can give u but try some forums like this one they seem like they have some good answers http://www.physicsforums.com/showthread.php?t=14245

[ziggy29]

Originally posted by Move The Chains
Ok, so that tells me that the answer is SF4 but I need to know HOW to get that. Know what I'm saying? I got a test over this junk Thursday.
I think it has something to do with the shape of the molecule. If a molecule like this one were in the plane and fully symmetrical, there would be no dipole moment. But I think SF4 is shaped more like a tetrahedron where all the fluorine atoms all bend out of the plane in the third dimension. So if all the F atoms are on the same side of the Z axis, they would be producing a somewhat negative charge on the side of the Z axis where the F atoms are oriented.

The other side of the Z axis (the S atom) would have a counterbalancing positive ionic charge.

I think that's about it but it's been *mumbledy mumble* years since I've had chemistry in college, so I'm very, very rusty.

[Move The Chains]

Originally posted by BLACK&GOLD4LIFE
sry man not much help i can give u but try some forums like this one they seem like they have some good answers http://www.physicsforums.com/showthread.php?t=14245 Nevermind man. I figured it out. You gotta find out which molecule or ion has an unstable form, or an excess of electrons. All of those work out really nice except SF4. It has 2 excess electrons. :clap:

[Move The Chains]

Originally posted by ziggy29
I think it has something to do with the shape of the molecule. If a molecule like this one were in the plane and fully symmetrical, there would be no dipole moment. But I think SF4 is shaped more like a tetrahedron where all the fluorine atoms all bend out of the plane in the third dimension. So if all the F atoms are on the same side of the Z axis, they would be producing a somewhat negative charge on the side of the Z axis where the F atoms are oriented.

The other side of the Z axis (the S atom) would have a counterbalancing positive ionic charge.

I think that's about it but it's been *mumbledy mumble* years since I've had chemistry in college, so I'm very, very rusty. I see what you're saying, those Flourine atoms bend because there are 2 extra electrons that form a lone electron pair, pushing the flourine atoms away and causing the atom to be polar.

[RedWhiteBlue]

I do believe BEAST and Rocket have some very good 'chemistry' going on..... just sayin':rolleyes: :thinking:

[Sweetwater Red]

Originally posted by RedWhiteBlue
I do believe BEAST and Rocket have some very good 'chemistry' going on..... just sayin':rolleyes: :thinking:

Wow! That would give a whole new meaning to "Brotherly love" now wouldn't it? Then again it is Brownwood(banjos playing) so I guess it is to be expected. Lol

[ASUFrisbeeStud]

Chemistry was my minor, but now I sell insurance so I'm a little fuzzy but if you have anymore questions just PM me and I'll see if I can help

[sahen]

sorry i didnt see this last night...i have a BS in Biochemistry and work w/ chemistry every day...but yall are correct SF4 is the answer, if i could draw it i would show you, kinda hard to just explain....best way to find dipole moments is to look for free electrons though....

[Move The Chains]

Originally posted by sahen
best way to find dipole moments is to look for free electrons though....


Originally posted by Move The Chains
Nevermind man. I figured it out. You gotta find out which molecule or ion has an unstable form, or an excess of electrons. All of those work out really nice except SF4. It has 2 excess electrons. :clap: I got it man. Thanks. :)

[SweenyBulldog]

Originally posted by Move The Chains
(Actual question)


Which one of the following molecules or ions has a dipole moment?

a. IO4-
b. ICl2-
c. SF4
d. XeF4
e. CO2 The answer is B always is in tough questions......naw I'm joking I have not one clue :D

[carter08]

so, if it has a free electron it has a dipole moment?

my community college chem class didn't teach me that. it did teach me how to find free electrons though.

[Move The Chains]

Originally posted by carter08
so, if it has a free electron it has a dipole moment?

my community college chem class didn't teach me that. it did teach me how to find free electrons though. Basically if there are free electrons or electrons that create a lone pair, it makes the ion or molecule have a dipole moment.

[BILLYFRED0000]

Gee and I got here to late to contribute. I am rusty too but at one time had a clue. Although I was more into plate techtonics and Astrophysics as a rule. Free electrons are the starting place.
yee haw. I never pay for mine.

[ctown]

Sorry, I thought this thread was about Brownwood posters having BIPOLAR moments. If you're trying to balance an equation with a Brownwood element, all you have to add is 5:1 amounts of logic and reason. Be careful cause the mixture can become quite volatile on its way to getting settled down. Makes good ROCKET fuel...

[Move The Chains]

more chemistry:


What volume of O2(g), measured at 17.7 °C and 0.978 atm reacts with 15.1 g C4H10 to produce CO2(g) and H2O(l)? (R = 0.08206 L·atm/mol·K)

a. 2.51 L
b. 6.34 L
c. 23.5 L
d. 41.2 L
e. 239 L



*** I think you have to use the pv=nrt formula.... but not sure how that works when were combining different things. I've only used that formula when working with single elements.

[Move The Chains]

Originally posted by Move The Chains
more chemistry:


What volume of O2(g), measured at 17.7 °C and 0.978 atm reacts with 15.1 g C4H10 to produce CO2(g) and H2O(l)? (R = 0.08206 L·atm/mol·K)

a. 2.51 L
b. 6.34 L
c. 23.5 L
d. 41.2 L
e. 239 L



*** I think you have to use the pv=nrt formula.... but not sure how that works when were combining different things. I've only used that formula when working with single elements. ^^^^ C'mon all you people who claim to be good at this mess.

[WOS87]

Well, it's a two part question...

First you have to find the right formula for the reaction... which I believe should look like this


13 O2 + 2 C4H10 ---> 8 CO2 + 10 H2O

Then calculate how many moles 15.1g of C4H10 is (which is the n in the pV=nRT equation)

You will need 13/2 times as many moles of O2 for the reaction to work

Once you have that, solve for V (V=nRT/p)

You will need to convert the Celsius to Kelvin if you are going to use their given value for R

[DU_stud04]

why, yes i am! thanks for asking!

[Move The Chains]

Originally posted by WOS87
Well, it's a two part question...

First you have to find the right formula for the reaction... which I believe should look like this


13 O2 + 2 C4H10 ---> 8 CO2 + 10 H2O

Then calculate how many moles 15.1g of C4H10 is (which is the n in the pV=nRT equation)

You will need 13/2 times as many moles of O2 for the reaction to work

Once you have that, solve for V (V=nRT/p)

You will need to convert the Celsius to Kelvin if you are going to use their given value for R dude. thanks so much.

[Move The Chains]

Originally posted by WOS87
Well, it's a two part question...

First you have to find the right formula for the reaction... which I believe should look like this


13 O2 + 2 C4H10 ---> 8 CO2 + 10 H2O

Then calculate how many moles 15.1g of C4H10 is (which is the n in the pV=nRT equation)

You will need 13/2 times as many moles of O2 for the reaction to work

Once you have that, solve for V (V=nRT/p)

You will need to convert the Celsius to Kelvin if you are going to use their given value for R Where does the 13/2 come in?

[Move The Chains]

If a gas effuses 1.618 times faster than Kr, what is its molar mass?

a. 18.02 g/mol
b. 28.01 g/mol
c. 32.01 g/mol
d. 51.79 g/mol
e. 65.88 g/mol

[WOS87]

Graham's Law of Effusion

Did you get c. 32.01?

[Move The Chains]

Originally posted by WOS87
Did you get c. 32.01? yes.

[WOS87]

Rate1/Rate2 = Square root (Molar Mass 2/Molar Mass 1)

Rate1/Rate(Kr) = 1.618 which is the ratio of unknown gas/Kr

Molar Mass (Kr) = 83.798

So...

1.618 = Square root (83.798/Unknown Molar Mass)

Square both sides...

1.618x1.618 = 2.618
2.618 = 83.798/ Unknown

Solve for unknown...

Unknown = 83.798/2.618 = 32.01

There's your answer

Goodnight!

[Move The Chains]

Originally posted by WOS87
Rate1/Rate2 = Square root (Molar Mass 2/Molar Mass 1)

Rate1/Rate(Kr) = 1.618 which is the ratio of unknown gas/Kr

Molar Mass (Kr) = 83.798...

So...

1.618 = Square root (83.798/Unknown Molar Mass)

Square both sides...

1.618x1.618 = 2.618
2.618 = 83.798/ Unknown

Solve for unknown...

Unknown = 83.798/2.618 = 32.01 yea.... lol. if you got some time b4 tomorrow at noon... can you just give me the steps here? I'll do all the work.



How long will it take 10.0 mL of Cl2 gas to effuse through a porous barrier if it has been observed that 76.3 minutes are required for 10.0 mL of HCl gas to effuse through the same barrier?

a. 54.7 min.
b. 106 min.
c. 148 min.
d. 202 min.
e. 289 min.

[WOS87]

Originally posted by Move The Chains
yea.... lol. if you got some time b4 tomorrow at noon... can you just give me the steps here? I'll do all the work.



How long will it take 10.0 mL of Cl2 gas to effuse through a porous barrier if it has been observed that 76.3 minutes are required for 10.0 mL of HCl gas to effuse through the same barrier?

a. 54.7 min.
b. 106 min.
c. 148 min.
d. 202 min.
e. 289 min.

What's the ratio of the Molar Mass of Cl2 over Molar Mass of HCl?

Use the same equation..

What is rate?

Rate is amount/time... since the amount is the same for both gasses in the equation you can use the equation:

10ml/time1 divided by 10ml/time2 = square root (Molar mass 2/Molar mass 1)

This works out to Time2/Time1 = Square root(Molar Mass2/Molar Mass1)

[Move The Chains]

Originally posted by WOS87
What's the ratio of the Molar Mass of Cl2 over Molar Mass of HCl? 1.939

[Move The Chains]

Originally posted by WOS87
What's the ratio of the Molar Mass of Cl2 over Molar Mass of HCl?

Use the same equation..

What is rate?

Rate is amount/time... since the amount is the same for both gasses in the equation you can use the equation:

10ml/time1 divided by 10ml/time2 = square root (Molar mass 2/Molar mass 1)

This works out to Time2/Time1 = Square root(Molar Mass2/Molar Mass1) Thanks. Get some sleep. :)

[WOS87]

Square root 1.939 = 1.39

1.39 x 76.3 minutes = 106 minutes

(Answer = b)

[Move The Chains]

Originally posted by WOS87
Square root 1.939 = 1.39

1.39 x 76.3 minutes = 106 minutes

(Answer = b) Thanks. I had my 1 and 2 different.... but I got the same answer.