How to produce sugar from sugarcane

2 answers:

4 0

<h3>Answer:</h3><h3>ok</h3><h3 /><h3 /><h3>Step By Step Explanation:</h3><h3>Cane sugar processing consists of the following steps: sugar cane is crushed, the juice is heated and filtered, then sent to a series of crystallization steps to create crystals of raw sugar, followed by centrifugation to remove any remaining juice or syrup.</h3>

Schach [20]2 years ago

3 0

Answer:

Cane sugar processing consists of the following steps: sugar cane is crushed, the juice is heated and filtered, then sent to a series of crystallisation steps to create crystals of raw sugar, followed by centrifugation to remove any remaining juice or syrup

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Kemmi Major does some experimental work on the combustion of sucrose: C12H22O11(s) 12 O2(g) → 12 CO2(g) 11 H2O(g) She burns a 0.

Pavlova-9 [17]

Answer: $5.81\times 10^6J/mol$

Explanation:

Heat of combustion is the amount of heat released when 1 mole of the compound is completely burnt in the presence of oxygen.

$C_{12}H_{22}O_{11}(s)+12O_2\rightarrow 12CO_2(g)+11H_2O(g)$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.05392g}{342g/mol}=1.577\times 10^{-4}moles$

Thus $1.577\times 10^{-4}moles$ of sucrose releases = 916.6 J of heat

1 mole of sucrose releases = $\frac{916.6}{1.577\times 10^{-4}}\times 1=5.81\times 10^6J$ of heat

Thus ∆H value for the combustion reaction is $5.81\times 10^6J/mol$

6 0

3 years ago

A catalyst lowers the activation energy for both the forward and the reverse reactions in an equilibrium system, so it has no ef

rosijanka [135]

Answer:

True.

Explanation:

Catalyst increases the rate of the reaction without affecting the equilibrium position.

Catalyst increases the rate via lowering the activation energy of the reaction.

This can occur via passing the reaction in alternative pathway (changing the mechanism).

The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).

in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.