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  1. V

    Empire: Total War

    Empire: Total War Exclusive Preview: The Road to Independence, Multiplayer, and Hands-on Naval Combat - PC News at GameSpot GameSpot Video: Empire: Total War Road to Independence Trailer And the demo Empire: Total War™ Demo on Steam Giggity.
  2. V

    Help/Support ► which road.....

    5 Things You Think Will Make You Happy (But Won't) | Cracked.com think about it
  3. V

    Help/Support ► ...that suicide is painless. It brings so many....

    So the thread title...It's like a song or something?
  4. V

    Help/Support ► ITT: Milia help

    wait wait, like, these are those little bumps on the eyelid, right?
  5. V

    Help/Support ► Homework Help

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

    heh, you need twice as much HCl as Mg(OH)2, so that's false.
  7. V

    Help/Support ► Homework Help

    Nope. Parenthesis, man, everything is grouped.
  8. V

    Help/Support ► Homework Help

    The X intercept means when a line hits the X axis. Confusingly enough, when the line hits the X axis is actually when Y is zero. It's total BS, I know. so put in zero for Y, and you get 0 = -4/5x - 4. add 4 to both sides, 4 = -4/5x. Now you multiply both sides by -5/4. -4/5 X -5/4 equals 1...
  9. V

    Help/Support ► Homework Help

    that last part, -1/2m - 1/2m simplifies to -2/2m, which again simplifies to -m. As for the first part, -1/2m X -1/2m. -1/2 X -1/2 = 1/4, and m X m is m^2. so, 1/4m^2 1/4m^2 - m I usually do my math by breaking equations into smaller pieces.
  10. V

    Help/Support ► Official Family Problems Thread.

    My family is just...always making rice. Help?
  11. V

    Help/Support ► Homework Help

    Yes it is a single agent. That's why it's called a compound.
  12. V

    Help/Support ► Homework Help

    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.
  13. V

    Help/Support ► Homework Help

    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...
  14. V

    Help/Support ► Homework Help

    I did read it and i'm still sure you're wrong. .2 grams carbon couldn't possibly make 19 grams of CO2.
  15. V

    Help/Support ► I have this very strange problem

    obviously, you need to do something to show you notice the fart. i suggest you start coughing like crazy whenever it happens, maybe glare at her.
  16. V

    Help/Support ► Homework Help

    ...Are you assuming all the O comes from the compound, and not the air?
  17. V

    Help/Support ► Homework Help

    .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.
  18. V

    Help/Support ► Homework Help

    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.
  19. V

    Help/Support ► Homework Help

    that shit is easy. all you need to know is that you have unlimited O2 from the air, and a small amount in the compound. from 19.10 g of CO2 and 11.74 g of H2O, you need to find out the mass of C and H respectively. add them together, then subtract that number from the 10g, and that'll give you...
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    Help/Support ► Who Am I really?

    who you are will tell you what cause to join ...what was the question again?

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