Answer:
<span>Carbon readily forms covalent bonds with other carbon atoms.
Explanation:
As we know approximately more than 95 % compounds, either isolated, discovered or synthesized belongs to organic compounds containing carbon atoms.
This great diversity of organic compounds is due to following facts.
1) Catenation:
Carbon has a peculiar behavior of self linkage. This self linkage of one carbon with another is called as catenation. In this way carbon can form a long chain of carbon atom. A branching can also take place when one carbon is bonded further to three of four carbon atoms.
2) Isomerism:
Secondly the carbon containing compounds show isomerism. In which molecular formula is same but structural formula is different. For example molecular formula C</span>₅H₁₂ can make following compounds,
a) n-Pentane
b) 2-Methylbutane
c) 2,2-Dimethylpropane
3) Multiple Bonds:
Carbon can form multiple bonds i.e double bond like in alkenes and triple bonds like in alkyne.
Due to these factors carbon gets very high number of opportunities to form large number of compounds.
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Answer:
a) C6H5COOH + H2O ↔ H3O+ + C6H5COO-
b) [ H3O+ ] = 2.517 E-3 M
c) pH = 2.599
Explanation:
a) balanced equation:
C6H5COOH + H2O ↔ H3O+ + C6H5COO-
⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5
mass balance:
0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)
charge balance:
[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant
⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)
b) (2) in (1):
⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]
⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]
⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5
⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0
⇒ [ H3O+ ] = 2.517 E-3 M
c) pH = - log [ H3O+ ]
⇒ pH = - Log ( 2.517 E-3 )
⇒ pH = 2.599
Answer:
Since the noble gases are unreactive or inert, they are safe to use. Helium is used to fill balloons and airships, because it is much lighter that air and it will not catch fire. Neon is used in advertising signs. It will give red glow, but the color can be changes by mixing it with other gases.
Explanation:
To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:
45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present
Hope this answers the question. Have a nice day.
