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Gatorade

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Alright, I know this problem is probably painful simple but I want to make sure it's right.

2 + 3
2x-5 3x-2

The directions say simplify completely. I got 12x-19 as my answer, but that doesn't seem right.

Edit:

2/2x-5 + 3/3x-2
 
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Nostalgia

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I'm a little confused as to how the 2 + 3 relates to the rest of it.
 

Gatorade

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I'm a little confused as to how the 2 + 3 relates to the rest of it.
It's a fraction. I keep fixing it, but then it resets to being smooshed like that. Try and read it with the mindset that it's a fraction. Sorry.

Edit: In these problems where you add and subtract fractions, you basically just cross multiply right?
 

very differentiable
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Al right, to add fractions the part under the line must be set the same for both if you want to add. This means you first get 2x-5=3x-2. Using the standard rules of algebra you get x=-3. This gives a division by -11 in both fractions and then adding them you get a fraction of 5/-11.
 

Gatorade

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Al right, to add fractions the part under the line must be set the same for both if you want to add. This means you first get 2x-5=3x-2. Using the standard rules of algebra you get x=-3. This gives a division by -11 in both fractions and then adding them you get a fraction of 5/-11.
Huh? That makes no sense to me.
 

very differentiable
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When you add fractions together, you must set the numbers under the line equal, which you do by multiplying both fractions until the bottom numbers match. In this case you can't multiply the fractions because of x. To set the parts under the line equal you must solve for the equation that the parts under the line are equal. Then you get the x you need, you fill it in one of the parts under the line and you get the number under the line.
 

blinkboy211

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2/2x-5 + 3/3x-2
Well since you cannot equal the bottoms you need to multiple them to the other side so you get:
2+3=(2x-5)(3x-2)
Then root it out so you get
6x(sqaured)-19x+10=5
6x(sq)-19x+5
which I don't know if there really is a answer there and don't want to work it out any further.
That could be one way to do it or you have to do it like ikkuh said, but you cannot match the sides equally from what i can see. Your LCM is 30, but really cannot match the numbers correctly. I guess it really just matters what you need done for the problem.
 

violent_anger

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The two above me were talking gibberish. Ikkuh assumed the denominators were already equal, while, Blink assumed the whole thing was = 0.

2/2x-5 + 3/3x-2
You must get the denominators to be equal. You can multiply any number by 1, and a fraction with the same number for numerator and denominator would simplify to 1.

so 2/2x-5 * 3x-2/3x-2 would be 6x-4/(2x-5)*(3x-2)
and 3/3x-2 * 2x-5/2x-5 would be 6x-15/(2x-5)*(3x-2)

6x-4/(2x-5)*(3x-2) + 6x-15/(2x-5)*(3x-2)

Since they have the same denominator, now you add them.
12x-19/(2x-5*(3x-5)

now you simplify it.
 

blinkboy211

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so 2/2x-5 * 3x-2/3x-2 would be 6x-4/(2x-5)*(3x-2)
and 3/3x-2 * 2x-5/2x-5 would be 6x-15/(2x-5)*(3x-2)

6x-4/(2x-5)*(3x-2) + 6x-15/(2x-5)*(3x-2)
funny part is that i remembered you could do that soon after i posted, but didn't want to take time to write it all out again.
 

Gatorade

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Alright I have to do a Chemistry lab report and I am really confused on one of the questions. We dissolved solid NaOH in HCl and we recorded the initial and final temperature of the solution and calculated the change in temperature (T final- T initial) and it was 9.3 degrees Celsius. The question states:

Calculate the experimental enthalpy change of this reaction in kJ per mole of NaOH.

What do I do to find this?
 

Weeaboo

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Alright I have to do a Chemistry lab report and I am really confused on one of the questions. We dissolved solid NaOH in HCl and we recorded the initial and final temperature of the solution and calculated the change in temperature (T final- T initial) and it was 9.3 degrees Celsius. The question states:

Calculate the experimental enthalpy change of this reaction in kJ per mole of NaOH.

What do I do to find this?

Because the bond enthalpies you use i calculations are mean bond enthalpies not the actual bond enthalpies of the bond in a particular molecule.

A mean bond enthalpy is the enthalpy of a bond averaged over several different compounds. This is because a bond has a slightly different bond enthalpy depending which compound you measure it in.

To get the same values for the formation of water and the combustion of hydrogen you would need the O-H bond enthalpies for the O-H bond within a water molecule not the average bond enthalpy.

At least, that's what my sister told me o_o;
 

Gatorade

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美咲 四季;3687479 said:
Because the bond enthalpies you use i calculations are mean bond enthalpies not the actual bond enthalpies of the bond in a particular molecule.

A mean bond enthalpy is the enthalpy of a bond averaged over several different compounds. This is because a bond has a slightly different bond enthalpy depending which compound you measure it in.

To get the same values for the formation of water and the combustion of hydrogen you would need the O-H bond enthalpies for the O-H bond within a water molecule not the average bond enthalpy.

At least, that's what my sister told me o_o;
Alright that just confused me 100000x more. I do appreciate the response though, even though it wasn't what I was looking for, I was looking for a specific formula, but thanks. I eventually figured it out, after texting 5 people and messaging about 10 other people on facebook.
 

Weeaboo

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Alright that just confused me 100000x more. I do appreciate the response though, even though it wasn't what I was looking for, I was looking for a specific formula, but thanks. I eventually figured it out, after texting 5 people and messaging about 10 other people on facebook.

I know it even confused me more than it should have :confused:
 

Vayne Mechanics

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Alright I have to do a Chemistry lab report and I am really confused on one of the questions. We dissolved solid NaOH in HCl and we recorded the initial and final temperature of the solution and calculated the change in temperature (T final- T initial) and it was 9.3 degrees Celsius. The question states:

Calculate the experimental enthalpy change of this reaction in kJ per mole of NaOH.

What do I do to find this?
Change in enthalpy = (sum of bond energies of reactants) - (sum of bond energies in products)

or some stupid shit like that, I forget
 
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