You can take two liquids of different densities (how much mass is in a given volume) and pour them into a funnel. An example is oil and water. When the mixture settles, the denser liquid will be at the bottom, and drips through the funnel first. This is a separation that you can just let occur naturally.
Answer:
about 79% (79.04369332 to be exact)
Explanation:
Percent composition=(Molar mass of element x amount of it)/Molar mass of compound x 100
Br= 3 x 79.9/303.25 x100=79.04369332
The black fur absorbs all the colors where the white fun reflects all the colors because we see the green grass it absorbs all the colors except green and the yellow dandelion is reflecting yellow and orange because it’s color is like a mix of them but it still absorbs the other colors
Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after 
is 0.00345 mol/L.
Explanation:
Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t =
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after is 0.00345 mol/L.
Answer:
The answer to your question is C₂HO₃
Explanation:
Data
Hydrogen = 3.25%
Carbon = 19.36%
Oxygen = 77.39%
Process
1.- Write the percent as grams
Hydrogen = 3.25 g
Carbon = 19.36 g
Oxygen = 77.39 g
2.- Convert the grams to moles
1 g of H ----------------- 1 mol
3,25 g of H ------------- x
x = (3.25 x 1) / 1
x = 3.25 moles
12 g of C ---------------- 1 mol
19.36 g of C ---------- x
x = (19.36 x 1) / 12
x = 1.61 moles
16g of O --------------- 1 mol
77.39 g of O --------- x
x = (77.39 x 1)/16
x = 4.83
3.- Divide by the lowest number of moles
Carbon = 3.25/1.61 = 2
Hydrogen = 1.61/1.61 = 1
Oxygen = 4.83/1.61 = 3
4.- Write the empirical formula
C₂HO₃
