Homework Help Thread

Discuss literature (e.g. books, newspapers), educational studies (getting help or opinions on homework or an essay), and philosophy.

Homework Help Thread

Unread postby Rhiannon » Mon Jun 02, 2003 12:24 pm

Okay, not too helpful at the moment, since it's almost summer...

This is a thread where you can ask for help with any homework you have, or help understanding a concept in something you're learning. This is also a good place to ask for references/links if you're researching something. If you have a deadline, please let us know in your post, so that we don't put off answering you until it's too late. :wink:
"For us to have self-esteem is truly an act of revolution and our revolution is long overdue."
— Margaret Cho
User avatar
Rhiannon
Joy & Oblivion
Joy & Oblivion
 
Posts: 5268
Joined: Sat Nov 23, 2002 10:10 pm

Unread postby Shi Jing Xu » Mon Jun 16, 2003 7:17 pm

Of course...I understand that there is no school right now, but that didn't mean my teachers couldn't give us homework over the summer-- right?

Of course not ^_^
Ehh...my Chemistry teacher (yes, I will be in another year of Chemistry-- not because I failed mind you ^_~) gave us work to do out of the Steven S. Zumdahl Chemistry 3rd Edition book. Any help would be appreciated on this problem. I'm stuck. This actually gives the answer in the back but there is no use copying it if I don't understand how to do it. So, I will give the question and the answer and please someone help me figure out how to get the supposed answer.

Question: (pg 74)
17.) Early tables of atomic weights (masses) were generated by measuring the mass of a substance that reacts with 1g of oxygen. Given the following data and taking the atomic mass of hydrogen to be 1.00, generate a table of relative atomic masses for oxygen, sodium, and magnesium. [This table below is a replica of that created by John Dalton]

Element------Mass that Combines with 1.00g Oxygen----Assumed Formula
Hydrogen-------------0.1260g----------------------------------HO
Sodium--------------2.8750g-----------------------------------NaO
Magnesium-----------1.5000g----------------------------------MgO
- How do your values compare with those in the periodic table?
- How do you account for any differences?

Answer (from the back):
O, 7.94; Na, 22.8; Mg, 11.9; O and Mg are incorrect by a factor of ≈2; correct formulas are H20 (water), Na2O, and MgO.

Any help is appreciated!! Damn this AP crap! :cry:
"As if you could kill time without injuring eternity." ~ Henry David Thoreau
User avatar
Shi Jing Xu
Hermione G
 
Posts: 1659
Joined: Sat Jun 15, 2002 6:05 pm
Location: The Middle Way

Unread postby Seven at One Stroke » Mon Jun 16, 2003 7:30 pm

Shi Jing Xu wrote:Element------Mass that Combines with 1.00g Oxygen----Assumed Formula
Hydrogen-------------0.1260g----------------------------------HO
Sodium--------------2.8750g-----------------------------------NaO
Magnesium-----------1.5000g----------------------------------MgO
- How do your values compare with those in the periodic table?
- How do you account for any differences?

Answer (from the back):
O, 7.94; Na, 22.8; Mg, 11.9; O and Mg are incorrect by a factor of ≈2; correct formulas are H20 (water), Na2O, and MgO.

Ok, I'm terrible at explaining things but I'll try. First of all, let's assume that hydrogen has atomic mass of 1.00, then the unit mass of hydrogen is 1g/unit. If 0.1260g of Hydrogen reacts with 1.0g of oxygen to form HO, then there is 0.1260g H × (1 unit H /1g H) =0.1260 unit of hydrogen in that much HO. Since HO has hydrogen:oxygen ratio of 1:1, there is just as much, meaning 0.1260 units of oxygen in that 1.0g of oxygen. So 1.0g O/ 0.1260 unit=7.94 g/unit. The rest follows. The reason why the atomic weight is off by a factor of two is because the molecular formula is wrong.

If you've taken chemistry before, then consider the unit to be moles, g/mol is the molar mass. If not, just remember atomic weight to be mass/unit quantity.

I hope this helped.
Moderation in pursuit of actual work is no vice.
User avatar
Seven at One Stroke
Sei's Slave
 
Posts: 1852
Joined: Wed Oct 02, 2002 4:16 am

Unread postby GuanYu » Mon Jun 16, 2003 7:58 pm

Can anybody help with my geography essay?
User avatar
GuanYu
Langzhong
 
Posts: 488
Joined: Mon May 05, 2003 9:45 pm

Unread postby Shi Jing Xu » Mon Jun 16, 2003 8:08 pm

Hisui wrote:Ok, I'm terrible at explaining things but I'll try. First of all, let's assume that hydrogen has atomic mass of 1.00, then the unit mass of hydrogen is 1g/unit. If 0.1260g of Hydrogen reacts with 1.0g of oxygen to form HO, then there is 0.1260g H × (1 unit H /1g H) =0.1260 unit of hydrogen in that much HO. Since HO has hydrogen:oxygen ratio of 1:1, there is just as much, meaning 0.1260 units of oxygen in that 1.0g of oxygen. So 1.0g O/ 0.1260 unit=7.94 g/unit. The rest follows. The reason why the atomic weight is off by a factor of two is because the molecular formula is wrong.

If you've taken chemistry before, then consider the unit to be moles, g/mol is the molar mass. If not, just remember atomic weight to be mass/unit quantity.

I hope this helped.


This part I understood, about Oxygen and Hydrogen. But when I tried to do the sodium and magnesium problems, I dont get the answers they give, following the same mathematical procedures.
"As if you could kill time without injuring eternity." ~ Henry David Thoreau
User avatar
Shi Jing Xu
Hermione G
 
Posts: 1659
Joined: Sat Jun 15, 2002 6:05 pm
Location: The Middle Way

Unread postby Seven at One Stroke » Mon Jun 16, 2003 8:20 pm

So far we have Oxygen: 7.94g/unit, hydrogen: 1.00g/unit.

Sodium--------------2.8750g-----------------------------------NaO
Magnesium-----------1.5000g----------------------------------MgO

1g O × (1 unit O/ 7.94g O) × (1 unit Na/ 1 unit O)=0.1259 unit Na
2.8750g Na/0.1259 unit Na=22.83g/unit Na

1g O × (1 unit O/7.94g O) × (1 unit Mg/1 unit O)=0.1259 unit Mg
1.500g Mg/ 0.1259 unit Mg=11.91 g/unit Mg

You should realize by now, all that matters here is the units. The units should cancel out in each operation, and in the end you'll get the unit of g/unit, and that's the atomic weight.
Moderation in pursuit of actual work is no vice.
User avatar
Seven at One Stroke
Sei's Slave
 
Posts: 1852
Joined: Wed Oct 02, 2002 4:16 am

Unread postby Shi Jing Xu » Mon Jun 16, 2003 8:24 pm

OH! I get it now! I have been out of school for too long and I seem to forget everything (summer will do that to you ^_~). Thank you!
"As if you could kill time without injuring eternity." ~ Henry David Thoreau
User avatar
Shi Jing Xu
Hermione G
 
Posts: 1659
Joined: Sat Jun 15, 2002 6:05 pm
Location: The Middle Way

Unread postby Seven at One Stroke » Mon Jun 16, 2003 8:35 pm

I know how it is. The problem stumped me for a second too.
Moderation in pursuit of actual work is no vice.
User avatar
Seven at One Stroke
Sei's Slave
 
Posts: 1852
Joined: Wed Oct 02, 2002 4:16 am

Unread postby Rhiannon » Mon Jun 16, 2003 8:36 pm

GuanYu wrote:Can anybody help with my geography essay?


Depends on what kind of help you need. :wink: Just post your questions/concerns/etc, we'll help out as we can.
"For us to have self-esteem is truly an act of revolution and our revolution is long overdue."
— Margaret Cho
User avatar
Rhiannon
Joy & Oblivion
Joy & Oblivion
 
Posts: 5268
Joined: Sat Nov 23, 2002 10:10 pm

Unread postby Shi Jing Xu » Sun Jul 13, 2003 1:59 am

More help with Chem ^_^

Alrighty, here is the question (this one doesnt give an answer....sadly).


38.) An excited hydrogen atom emits light with a frequency of 1.141 X 10ˉ14 (14th power) Hz to reach the energy level for which n=4. In what principal quantum level did the electron begin?

Now, this is what I did. I figured that it was necessary to find the wavelength first, using the equation λν=c or λ= c/ν.

So my work was (c being the speed of light and ν being the frequency)
λ= 2.9979 x 10ˉ8 (8th power) meters per second/
1.141 X 10ˉ14 (14th power)sˉ
λ= 2.63 x 10ˉ22 (22nd power) meters [with all the other units canceling]

Now, using the wavelength, speed of light, and Planck's constant (which is 6.625 x 10ˉ34 (-34th power) J·s) we will find the change in energy (ΔE).

ΔE=h(c/λ)

SO,
ΔE= 6.625 x 10ˉ34 (-34th power) J·s (2.9979 x 10 (8th power)) meters per second/2.63 x 10 (22nd power) meters
ΔE= 7.55 X 10 ˉ4 J

So, I figured that I would use the equation below in order to find the change in energy of n=4 (which is the 4th principal quantum #) and subtract that from the change in energy (which was 7.55 X 10 ˉ4 J) in order to find the initial quantum number. This is the first equation I thought to use:

ΔE= -2.178 x 10ˉ18 (18th power) J (Z²/n²) [where Z= nuclear charge and n= principal quantum number]

So therefore, I had
ΔE= -2.178 x 10ˉ18 (18th power) J · (1²/4²)
ΔE= -1.36 X 10ˉ14 (14th power) J

Now, to find the initial energy of the initial quantum number, we use this equation:

ΔE= energy of final state - energy of initial state (or in this case)
ΔE - energy of final state = energy of initial state (just rearranged to find initial state)

So, I had
7.55 X 10ˉ4 (4th power) J - (1.36 X 10ˉ14 (14th power) J)= initial state energy

So, I did it, and I got 7.55 X 10ˉ4 (4th power) J
Seem odd? I had the same answer for the initial state energy as I did for the energy change between them and I can't figure out why!! NOTE: I am not using the same calculator as I usually am seeing as how mine was stolen (it was a graphing calculator TI-86), and for some reason, doing the practice problems in the book (with the answer) even gives the wrong answer. I mean, I do the same work it shows and everything and my scientific notation is off by about 30 decimal places and they are in the correct modes and everything. Any help would be appreciated because I am stuck. Even after I find the right energy for the intitial energy state I dont know how to find the initial quantum number.
"As if you could kill time without injuring eternity." ~ Henry David Thoreau
User avatar
Shi Jing Xu
Hermione G
 
Posts: 1659
Joined: Sat Jun 15, 2002 6:05 pm
Location: The Middle Way

Next

Return to Literature, Academics, and Philosophy

Who is online

Users browsing this forum: No registered users and 3 guests

Copyright © 2002–2008 Kongming’s Archives. All Rights Reserved