Answer:
C₁₂H₂₂O₁₁ + H₂O → C₅H₁₂O₆ + C₆H₁₂O₆
Explanation:
Chemical equation:
C₁₂H₂₂O₁₁ + H₂O → C₅H₁₂O₆ + C₆H₁₂O₆
Source of sucrose:
Sucrose is present in roots of plants and also in fruits. It is storage form of energy. Some insects and bacteria use sucrose as main food. Best example is honeybee which collect sucrose and convert it into honey.
Monomers of sucrose and hydrolysis:
Sucrose consist of monomers glucose and fructose which are join together through glycosidic bond. Hydrolysis break the sucrose molecule into glucose and fructose. In hydrolysis glycosidic bond is break which convert the sucrose into glucose and fructose. Hydrolysis is slow process but this reaction is catalyze by enzyme. The enzyme invertase catalyze this reaction.
The given reaction also completely follow the law of conservation of mass. There are equal number of atoms of elements on both side of chemical equation thus mass remain conserved.
The answer is acid rain
for study island your welcome :)
Answer: 1 mole of H2O= about 1/3 of a cup (18 mL). It is helpful ... 6.02 x 1023 H2O molecules. = 6.02 x 1023 NaCl formula unit. 1 mole C. 1 mole H2O. 1 mole
Explanation:
Answer:
Structure in attachment.
Explanation:
The oxymercuration-demercuration of an asymmetric alkene usually produces the Markovnikov orientation of an addition. The electrophile ⁺Hg(OAc), formed by the electrophile attack of the mercury ion, remains attached to least substituted group at the end of the double bond. This electrophile has a considerable amount of positive charge on its two carbon atoms, but there is more positive charge on the more substituted carbon atom, where it is more stable. The water attack occurs on this more electrophilic carbon, and the Markovnikov orientation occurs.
In hydroboration, borane adds to the double bond in one step. Boron is added to the less hindered and less substituted carbon, and hydrogen is added to the more substituted carbon. The electrophilic boron atom adds to the less substituted end of the double bond, positioning the positive charge (and the hydrogen atom) at the more substituted end. The result is a product with the anti-Markovnikov orientation.