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Giggity.
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...
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...
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.
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.
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...
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.
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...