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Gatorade

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violent_anger said:
.43 moles C
1.3 moles H
.21 moles O

More like C2H6O

You divide all the mole amounts by the smallest mole amount, in this case .21 moles.
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Ok, got it thank you!

Edit: So I have another question. In chemical reactions, if you have a solid as a reactant and the other is an aqueous solution, does the solid change anything? For instance, I have the problem:
ZnS (s) + HBr (aq)

What would the products be and why?

Ah shit I got another question. In a chemical reaction, can both products be aqueous solutions? Or would there not be any reaction?
 
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Fuego6

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Here is some of my homework

Adventure games have declined in popularity in the last 10 years. Why do you think this has happened? What would you change about the content or structure in adventure games to incite new interest in this genre?


Strategy games don’t always need to have a military backdrop. Discuss three settings or scenarios not related to the military that could be incorporated successfully into a strategy game.
 
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Ams Tram Gram

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It may be a while, but I'll post here whenever I take the follow-up to my Differential Calculus class in Cegep.
 
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LOL, the only thing homework-related for me is a 5 page paper on Beowulf and his choices as a hero. Doubt any of you could help me much.

And I'll try to help here, but it's been a while since I've been in high school. I'll know more about English and History than Science and Math.
 
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violent_anger said:
There is no way that's right.

1 mole of O has a mass of 16g, and 1 mole of C has a mass of 12g. so 1 mole of CO2 would have a mass of 44G. To find the mass of C in CO2, multiply it by 12/44. 19.10 g of CO2 X 12/44 would be about 5.209 g of C.

Mass of H is about 1.304, if you do the same thing.
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Totally wrong. Oxygen is O2, making the atomic mass 32 U and one mole weighs 32 grams. Second of all, they react in this way: 2 mole of C with 2 mole of H and 3 mole of O2. His first number was the closest. If he used the way i said he should use then that number is right.
 
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Gatorade

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ikkuh2 said:
Totally wrong. Oxygen is O2, making the atomic mass 32 U and one mole weighs 32 grams. Second of all, they react in this way: 2 mole of C with 2 mole of H and 3 mole of O2. His first number was the closest. If he used the way i said he should use then that number is right.
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Alright well idk I already handed it in. It's fine though.
 
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very differentiable
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Well, it's just homework, as long ass you learn from your book when you get the test, you'll be fine. And having a hard time with calculus?
 
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very differentiable
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So I have another question. In chemical reactions, if you have a solid as a reactant and the other is an aqueous solution, does the solid change anything? For instance, I have the problem:
ZnS (s) + HBr (aq)

What would the products be and why?

Ah shit I got another question. In a chemical reaction, can both products be aqueous solutions? Or would there not be any reaction?
Click to expand...

Didn't see those until now

1. HBr is a strong acid, so it will completely dissolve, yielding bit H3O+ ions and Br- ions. If ZnS would be the only material along with water it wouldn't dissolve, but you'll get Br-ions and ZnBr dissolves easily so you will get dissolved Zn(2+) ions and S(2-)ions. Two H3O+ and one S(2-) can form H2S. Because H2s is a weak acid, it would only partially dissolve if it would be dissolved. In this case a part of the H3O+ stays dissolved and a part would react and form H2S which will solidify. This would give the following products: Zn(2+) (dissolved), Br- (dissolved), H3O+ (dissolved) and H2S (solid).

The numbers between the brackets shows the charge the ions have.

2. Both can be dissolved in water but they don't have too. NaF completely dissolves in water, as does CaCl2. Ca and F however don't dissolve in water, so if you would add a NaF solution To a CaCl2 solution, the Ca and F React and Solidify. An example of a gas reacting with a solution is this: NaOH disolved reacts With HCl, which is a gas. Adding The gas with the solution you would get H2O and a NaCl solution.
 
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Gatorade

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ikkuh2 said:
Didn't see those until now

1. HBr is a strong acid, so it will completely dissolve, yielding bit H3O+ ions and Br- ions. If ZnS would be the only material along with water it wouldn't dissolve, but you'll get Br-ions and ZnBr dissolves easily so you will get dissolved Zn(2+) ions and S(2-)ions. Two H3O+ and one S(2-) can form H2S. Because H2s is a weak acid, it would only partially dissolve if it would be dissolved. In this case a part of the H3O+ stays dissolved and a part would react and form H2S which will solidify. This would give the following products: Zn(2+) (dissolved), Br- (dissolved), H3O+ (dissolved) and H2S (solid).

The numbers between the brackets shows the charge the ions have.

2. Both can be dissolved in water but they don't have too. NaF completely dissolves in water, as does CaCl2. Ca and F however don't dissolve in water, so if you would add a NaF solution To a CaCl2 solution, the Ca and F React and Solidify. An example of a gas reacting with a solution is this: NaOH disolved reacts With HCl, which is a gas. Adding The gas with the solution you would get H2O and a NaCl solution.
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Wow you really know your stuff. Thanks again.
 
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violent_anger

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Totally wrong. Oxygen is O2, making the atomic mass 32 U and one mole weighs 32 grams. Second of all, they react in this way: 2 mole of C with 2 mole of H and 3 mole of O2. His first number was the closest. If he used the way i said he should use then that number is right.
Click to expand...
...Are you assuming all the O comes from the compound, and not the air?
 
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i dont have to go back to school for like another week :D!

But this thread is great i'll sure to post here
 
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It clearly says he begins with a mixture of C, H and O2. 10 grams of the mixture reacts and in this case we assume it reacts in a vacuum, he might be off a little like a gram or so but it's pretty close. Half of the mixture being C is very unlikely, 24 U of C reacts with 92 U of O2, almost four times as much.
 
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violent_anger

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You don't assume a vacuum, it's freaking combustion. 10 grams of reactant can't make 30 grams of product, so about 20 grams comes from O2 in the air.

And it doesn't contain O2, just regular O. Whether its diatomic or not doesn't matter in a compound.

What you're suggesting for the empiracle formula would be about 90 hydrogen and 30 oxygen for each carbon.
 
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very differentiable
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Go back to school, now, you make up random gibberish and faulty arguments. He gave the beginning conditions, 10 grams of a mixture of C, H and O2. It's explicitly stated it's complete combustion, only from that 10g. He could've misread the reaction products, fact is your way gives some weird numbers.

Oxygen is always O2. Single O does not exist because only radicals form O and they react straight away into other products, that's the most famous property of radicals. And it's not 90H with 30O2. The reaction for full combustion is 2C+2H+3O2 gives 2 CO2 and 2H2O. 2 atoms of C react with three molecules Of O2, 2*12 U with 3*32 U makes 24 U with 96 U.

If you show me more of your knowledge of chemistry you have the right to speak, until then, don't deny the facts.
 
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violent_anger

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A compound contains three elements carbon, hydrogen, and oxygen. Combustion of a 10.0 g sample of that compound gives 19.10 g of CO2 and 11.74 g of H2O.
Click to expand...

As conservation of mass states, matter cannot be destroyed or created. In what universe does 10.0 g of a compound with nothing added produce 30.84 g of matter?



And my solution gave C2H6O, which is pretty common.
 
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