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A little Problem...

So my Maths teacher set us homework. It's Percentage profit/loss. In class the only example she gave us was a grid. In the textbook it has equations she didn't specify how we should set it out so i'm confused T_T

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Original post by Zeetingman
So my Maths teacher set us homework. It's Percentage profit/loss. In class the only example she gave us was a grid. In the textbook it has equations she didn't specify how we should set it out so i'm confused T_T

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Can u take pic of ur text book page I wanna see the equations if u don't mind :smile:

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Are you sure that's the right thing? That doesn't look like maths :nah:
Original post by Zeetingman
Here are some photos of the work


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Cool I know these equations how do u want me to help u? do u want me to tell u how to balance equations
Do u want me to help u in maths too as u mentioned in ur previous post??

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Reply 5
Forget the Maths.
Explain simply how to work it out.
And tips on working it out!!!

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Original post by Zeetingman
Forget the Maths.
Explain simply how to work it out.
And tips on working it out!!!

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Some of these are displacement reactions. Just swap the cations in each compound.
This is just balancing equations. The number of each type of atom on the left side of the equation has to equal the number of that type of atoms on the right.
Some of the questions involve electrons, in which case you use the same rules, also with the number of electrons on the left equals the number of electrons on the right.
Original post by Zeetingman
Forget the Maths.
Explain simply how to work it out.
And tips on working it out!!!

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Have u heard that mass is neither created nor destroyed. Use this to balance equations
equation:
H2SO4+NaOH-----Na2SO4+H20
so we balance this by considering I element at a time so the first one is H see before the reaction there are 3 atoms of H but after the reaction there are only 2 which means it is not balance. We balance this by adding 2 in front of H20 so this is going to be 2H20. Then for S there is 1 atom before the reaction and 1 atom after the reaction which means it is balance. Now O as u can see there are 5 atoms of O before the reaction but after the reaction they turned to 6 because we added 2 in front of H20 which means that there are 2 atoms of O present in this compound. Now as for Na there is one atom before the reaction but after the reaction there are 2 atoms so you simply write 2 in front of NaOH making it 2NaOH.
So the equation is going to be
H2SO4+2NaOH----Na2SO4+2H2O

I used this equation from the pic u send me of ur textbook. Anything else? :smile:
(edited 9 years ago)
Original post by Fallen99
Have u heard that mass is neither created nor destroyed.

According to relativity, mass is created and destroyed all the time.
In this case, you would use the principle that an atom can't change elements within a purely chemical reaction.

The rest of it is correct.
Original post by morgan8002
According to relativity, mass is created and destroyed all the time.
In this case, you would use the principle that an atom can't change elements within a purely chemical reaction.

The rest of it is correct.

Isn't it the conversation of mass where it say mass can neither be created nor destroyed. I might be wrong but can u explain how plz?
Original post by Fallen99
Isn't it the conversation of mass where it say mass can neither be created nor destroyed. I might be wrong but can u explain how plz?

Sorry I meant the mass associated with matter could change, in the context of chemical reactions. Mass itself can't be destroyed. An atom/molecule with greater bonds but the same particles will have less mass.
Why do you add two to the H?
When you can add one??
Original post by Zeetingman
Why do you add two to the H?
When you can add one??

Because there are more atoms of H on the left than on the right. You add the two in an attempt to balance it. The job isn't complete at that stage though, because there are now more H on the right than the left.
Original post by Zeetingman
Why do you add two to the H?
When you can add one??

No u cant add 1 u knw if i put 2 infront of H20 to make it 2H2O this means that the big 2 is multiplying by the small 2 it means H have 4 atoms as 2*2 is 4. U dont put 1 cuz multiplying by 1 won't work :smile:
Original post by morgan8002
Sorry I meant the mass associated with matter could change, in the context of chemical reactions. Mass itself can't be destroyed. An atom/molecule with greater bonds but the same particles will have less mass.



An atom/molecule with greater bonds but the same particles will have less mass.
I don't get this sorry :frown:
Thank you so much!!
The textbook doesn't expain stuff that good. I like things simplified. I think I should try some myself. There are more stuff but I don't wanna waste your time again thanks for the help
Original post by Fallen99

An atom/molecule with greater bonds but the same particles will have less mass.
I don't get this sorry :frown:

When the particles are arranged in a way where they have bonds, the molecule will have a smaller potential energy than when the particles are separate. Energy is given off when bonds are formed. As E=mc2 E = mc^{2}, the particles will also have a smaller mass.

You don't need to know this.
Original post by Zeetingman
Thank you so much!!
The textbook doesn't expain stuff that good. I like things simplified. I think I should try some myself. There are more stuff but I don't wanna waste your time again thanks for the help

I have time u can ask me more glad I could help u nad make sense to u :smile:
Original post by morgan8002
When the particles are arranged in a way where they have bonds, the molecule will have a smaller potential energy than when the particles are separate. Energy is given off when bonds are formed. As E=mc2 E = mc^{2}, the particles will also have a smaller mass.

You don't need to know this.

Yeah I haven't learned this yet may be next year :smile: thanks for explaining though :smile:
Original post by Fallen99
Yeah I haven't learned this yet may be next year :smile: thanks for explaining though :smile:

It's not GCSE. I think it is touched on at the end of A2 physics, but I'm not sure.

It is interesting though.

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