The proper name for the following alkyl side group where the main carbon chain is denoted with a squiggly line is isopropyl.
In natural chemistry, an alkyl substituent is an alkane missing one hydrogen. The term alkyl is intentionally unspecific to include many viable substitutions. An acyclic alkyl has the overall formulation of CₙH₂ₙ₊₁.
An alkyl is a purposeful institution of an organic chemical that includes only carbon and hydrogen atoms, that are organized in a chain. Examples include methyl CH3 (derived from methane) and butyl C2H5 (derived from butane). they may be now not located on their own however are discovered attached to different hydrocarbons.
what is an alkyl group? Alkyl group is shaped through putting off a hydrogen atom from the molecule of alkane. Alkanes are quite regularly represented as R-H and here R stands for alkyl group. the overall method of the alkyl organization is CₙH₂ₙ₊₁. The smallest alkyl organization is CH3 referred to as methyl.
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Mitochondria is the powerhouse of the cell, but don't let this distract you from the fact that Mr. Krabs sold SpongeBob's soul for 62 cents
Answer: El carbono, que en estado sólido, puede adoptar muchas formas alotrópicas, siendo las más comunes el diamante (red tridimensional) y el grafito (láminas), aunque también puede formar nanoestructuras en forma de balón de fútbol (fullerenos) o tubos diminutos (nanotubos de carbono), entre otras posibilidades.
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The correct would be C i think :)
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Answer: The concentration of the OH-, CB = 0.473 M.
Explanation:
The balanced equation of reaction is:
2HCl + Ca(OH)2 ===> CaCl2 + 2H2O
Using titration equation of formula
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)
NB is the number of mole of base = 1 (from the balanced equation of reaction)
CA is the concentration of acid = 1M
CB is the concentration of base = to be calculated
VA is the volume of acid = 23.65 ml
VB is the volume of base = 25mL
Substituting
1×23.65/CB×25 = 2/1
Therefore CB =1×23.65×1/25×2
CB = 0.473 M.