The oxidation state of the compound Mn (ClO4)3 is to be determined in this problem. For oxygen, the charge is 2-, the total considering its number of atoms is -24. Mn has a charge of +3. TO compute for Mn, we must achieve zero charge overall hence 3+3x-24=0 where x is the Cl charge. Cl charge, x is +7.
Oxygen and Hydrogen would most likely form a covalent bond that is polar, or a polar covalent bond. Due to the electronegativity difference between the 2 elements, unequal sharing of the valence electrons will occur, electrons being in closer proximity to Oxygen and farther away from Hydrogen. Resulting in the characteristic partial positive and negative charges to appear for the respective elements.
Ethane is an alkane. Methane is also an alkane and is considered to be the simplest alkane. The difference is ethane has only 2 carbon. That carbon has 6 hydrogen attached to it. So what we do is we multiply the moles of ethane by the number of hydrogen (by dimension analysis) resulting to 82.68 moles H.
Answer:
The answer to your question is: 17.26% of carbon
Explanation:
Data
CxHy = 0.2121 g
BaCO₃ = 0.6006 g
Molecular mass BaCO₃ = 137 + 12 + 48 = 197 g
Reaction
CO₂ + Ba(OH)₂ ⇒ BaCO₃ + H₂O
Process
1.- Find the amount of carbon in BaCO₃
197 g of BaCO₃ --------------- 12 g of Carbon
0.6006 g ---------------- x
x = (0.6006 x 12) / 197
x = 0.0366 g of carbon
2.- Calculate the percentage of carbon in the organic compound
0.2121 g of organic compound --------------- 100%
0.0366g -------------- x
x = (0.0366 x 100) / 0.2121
x = 17.26%
