chemistry tutor? | INFJ Forum

chemistry tutor?

subwayrider

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Sep 26, 2011
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Does anyone want to e-tutor me?
I'm taking a general chemistry class. I'm getting by all right, but i do have questions every now and again. Anyone?
 
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What's the question? i took gen chem so I could pry help you with some problems.
 
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PM.
 
Depends. I'm willing to work contract for $100/hr. That may be more expensive than you are willing to go.
 
I was always better at physics than chemistry, less exceptions for every rule when it came to physics! :) But, I did do university chemistry for materials engineering, including physical chemistry which we all decided was the most **** off hardest subject of all time. It's got to be buried in there somewhere... happy to try and help when Google fails you! :)
 
Thanks, guys. I'll post on this thread or PM if I have ever have a problem. Didn't have anything in mind for the moment, but I always have questions that spring into my mind. I'll remember to write them down because now I know I have somewhere they'll be answered.
 
Depends. I'm willing to work contract for $100/hr. That may be more expensive than you are willing to go.

[MENTION=3473]InvisibleJim[/MENTION]

You'd better be a damn good tutor.
 
Don't hesitate to ask for help on here... we have quite a few people on here who are adept, or at least decent at general chemistry.

I can also be of assistance should you need it. : )
 
Thanks. Well, it just so happens that I have a few questions tonight.
[MENTION=2873]Serenity[/MENTION] [MENTION=4718]Sebastian[/MENTION] [MENTION=2926]Bird[/MENTION]


1) Combustion Reactions and CO

What kinds of combustion reactions release CO? I understand whenever you smoke a cigarette, you are getting a good amount of CO in your lungs? Does this hold true for marijuana smoke? What extent of exposure to CO would it take for it to start killing brain cells? In most situations, smoking a cigarette in a well-ventilated room, or outside, would there be any harmful effects from the CO?

2) Why does Force have units of kg x m/s^2 ?

3) Why does Work have units of kg x m^2/s^2?

4) Why are ⌂T, ⌂V, ⌂P, etc. state functions, when q(heat) and w(work) are not?

5) Elaborate on enthalpy further than "a thermodynamic variable" or "H= E + PV"
 
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1. Combustion Reactions and CO

In real world situations there is not usually enough oxygen available when combustion occurs for complete combustion... so you end up with incomplete combustion... which ends up producing CO instead of CO2.

So, when you learn about the combustion equation for Propane (for example)... the real world situation is that it's not just the perfect combustion that occurs, it's the incomplete combustion that happens at the same time.

From Wiki:

Perfect combustion of Propane:
C3H8 + 5 O2 → 3 CO2 + 4 H2O + heat

Incomplete combustion of Propane:
2 C3H8 + 7 O2 → 2 CO2 + 2 CO + 2 C + 8 H2O + heat


I don't know much about the medical side re: how much damage is being done with CO in the blood stream from cigarettes / joints though... I'm glad I gave up smoking though :)



2. The standard measurement of Force in SI is a Newton... which is the amount of Force required to accelerate one kilogram of mass at one metre per second per second. It's a standard of measurement but it's not the only one... so really it's not true to say Force has units of kg x m/s^2... it's Newtons that are kg x m/s^2... which happens to be one of any possible standard ways of measuring Force :)



3. Similar answer to 2.... except Work is Force x Distance so you've got an extra "m" in there, so it's kg x m^2/s^2



*runs away from all the hard questions* :)
 
LOL. Thanks. Ok, so...

2. The standard measurement of Force in SI is a Newton... which is the amount of Force required to accelerate one kilogram of mass at one metre per second per second. It's a standard of measurement but it's not the only one... so really it's not true to say Force has units of kg x m/s^2... it's Newtons that are kg x m/s^2... which happens to be one of any possible standard ways of measuring Force :)

the s^2 (what on earth is a second squared?) comes into play with acceleration because you're increasing the velocity of 1 kg by 1 meter per second every second that it is in motion. Do I have this right? My book can be vague
 
LOL. Thanks. Ok, so...

the s^2 (what on earth is a second squared?) comes into play with acceleration because you're increasing the velocity of 1 kg by 1 meter per second every second that it is in motion. Do I have this right? My book can be vague

Yeah it's a bit weird that "second squared" business... I think of it in just the way you have written it though... every time another second passes you've increased the acceleration by another metre / second..... SO, if it was going 5 metres every second, then one second later it would be going 6 metres every second.


The Wiki page is a bit confusing... but, says the same thing really:

As acceleration, the unit is interpreted physically as change in velocity or speed per time interval, i.e. metre per second per second.

http://en.wikipedia.org/wiki/Metre_per_second_squared
 
[MENTION=4717]subwayrider[/MENTION]

1) Combustion Reactions and CO


Carbon monoxide is released when carbon material is reacted, or burnt with sub-stoichiometric levels of Oxygen. This is most typically seen in car engines which can produce a very high ppmv of CO in their exhaust gas.

Here is the reaction scheme for carbon and oxygen:

(1) Cn + nO2 -> nCO2
(2) Cn + (n/2)O2 -> nCO

Therefore reaction (2) is preferred when less Oxygen is supplied.

As an interesting aside, the Claus process is used industrially to burn H2S (a highly toxic and unwanted chemical present in gas production) to create benign pure sulphur and water.

2) Why does Force have units of kg x m/s^2 ?

F = m x a

where, Force = F, mass = kg, acceleration = m/s2

Acceleration is the derivative of speed, the rate at which speed in m/s changes per second. i.e. m/s/s = m/s2

Therefore multiplying the units , F in kg.m/s2

3) Why does Work have units of kg x m^2/s^2?

Work is the instantaneous energy being applied to a system, in Watts,

W = F x d

Where F is the force applied in kg.m/s2 and distance is in m, thus kg.m2/s2

4) Why are ⌂T, ⌂V, ⌂P, etc. state functions, when q(heat) and w(work) are not?

I object of the use of the term q as heat when it is duty, heat is more enthalpy versus pressure because I'm an engineer, scientists disagree because they are morons, but this is a different problem which causes fist fights between engineers and scientists so lets stick to the question.

A state function is a property of the system which is only dependant upon the state of the system at an instantaneous point in time, they can be gauged at any point in time as a consequence and are at that instant in equilibrium. By contrast mechanical work and heating duty are process quantities that are dependant upon the transition between two equilibrium states.

5) Elaborate on enthalpy further than "a thermodynamic variable" or "H= E + PV"

Enthalpy allows you to track the impact of adding heat to a system whilst taking into account pressure and volume changes.

Note, E = internal energy
P = Pressure and
V = Volume

In effect a kW of heat added to the system increases the enthalpy of the system by said kW.

However, this kW will only directly translate into a thermal effect by increasing the internal energy of the system if it is held at constant pressure and volume. (isobaric and fixed volume). Therefore the equation is a method of explaining why adding a kW of energy to a system does not increase the heat by a kW.

Conversely - and perhaps more usefully - it allows one to gauge how manipulating the volume or pressure of a system will change it's internal energy without applying additional heating duty to the system.

It should be noted that internal energy is made up of a number of factors besides sensible heat such as entropy.

Enthalpy is a state function of the system even though it cannot be directly measured with an 'enthalpy gauge' therefore it must be calculated by experiment with a reference point of absolute zero temperature/pressure.
 
IJ's got it on lock.
 
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@subwayrider

1) Combustion Reactions and CO


Carbon monoxide is released when carbon material is reacted, or burnt with sub-stoichiometric levels of Oxygen. This is most typically seen in car engines which can produce a very high ppmv of CO in their exhaust gas.

Here is the reaction scheme for carbon and oxygen:

(1) Cn + nO2 -> nCO2
(2) Cn + (n/2)O2 -> nCO

Therefore reaction (2) is preferred when less Oxygen is supplied.

As an interesting aside, the Claus process is used industrially to burn H2S (a highly toxic and unwanted chemical present in gas production) to create benign pure sulphur and water.

2) Why does Force have units of kg x m/s^2 ?

F = m x a

where, Force = F, mass = kg, acceleration = m/s2

Acceleration is the derivative of speed, the rate at which speed in m/s changes per second. i.e. m/s/s = m/s2

Therefore multiplying the units , F in kg.m/s2

3) Why does Work have units of kg x m^2/s^2?

Work is the instantaneous energy being applied to a system, in Watts,

W = F x d

Where F is the force applied in kg.m/s2 and distance is in m, thus kg.m2/s2

4) Why are ⌂T, ⌂V, ⌂P, etc. state functions, when q(heat) and w(work) are not?

I object of the use of the term q as heat when it is duty, heat is more enthalpy versus pressure because I'm an engineer, scientists disagree because they are morons, but this is a different problem which causes fist fights between engineers and scientists so lets stick to the question.

A state function is a property of the system which is only dependant upon the state of the system at an instantaneous point in time, they can be gauged at any point in time as a consequence and are at that instant in equilibrium. By contrast mechanical work and heating duty are process quantities that are dependant upon the transition between two equilibrium states.

5) Elaborate on enthalpy further than "a thermodynamic variable" or "H= E + PV"

Enthalpy allows you to track the impact of adding heat to a system whilst taking into account pressure and volume changes.

Note, E = internal energy
P = Pressure and
V = Volume

In effect a kW of heat added to the system increases the enthalpy of the system by said kW.

However, this kW will only directly translate into a thermal effect by increasing the internal energy of the system if it is held at constant pressure and volume. (isobaric and fixed volume). Therefore the equation is a method of explaining why adding a kW of energy to a system does not increase the heat by a kW.

Conversely - and perhaps more usefully - it allows one to gauge how manipulating the volume or pressure of a system will change it's internal energy without applying additional heating duty to the system.

It should be noted that internal energy is made up of a number of factors besides sensible heat such as entropy.

Enthalpy is a state function of the system even though it cannot be directly measured with an 'enthalpy gauge' therefore it must be calculated by experiment with a reference point of absolute zero temperature/pressure.

@_@ You are smart.
 
Big thanks to everyone. It's making more sense now. More to come, if you're up to it ;)
 
Total ionic equations: I noticed there are certain compounds that are expressed in their molecular format in ionic equations. What do these compounds have in common with each other? Why are they expressed this way?
 
Total ionic equations: I noticed there are certain compounds that are expressed in their molecular format in ionic equations. What do these compounds have in common with each other? Why are they expressed this way?

Could you be more specific?

As a hint, generally the idea is to match the number of + anions and - cations as well as the number of atoms of each type on the left and the right hand side of the equation. This lets you work out various states of electrochemical potential e.g. pH.
 
Total ionic equations: I noticed there are certain compounds that are expressed in their molecular format in ionic equations. What do these compounds have in common with each other? Why are they expressed this way?

I assume that you are talking about gases/halogens?

Br2, F2, O2, etc...The reason they are expressed this way is because they are in a neutral state as a molecule. A neutral molecule/atom differs from a charged atom (anion/cation). When you are crossing out the spectator ions, if the molecule becomes ionized in the product, then it remain in the net ionic equation as one of the reactants.