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Anit [1.1K]
3 years ago
11

An increase in temperature results in A) a decrease in the required activation energy while the reaction rate remains constant.

B) an increase in reaction rate due to a decrease in the kinetic energy of the reactants. C) an increase in the rate of reaction because reactant molecules collide with greater energy. D) an increase in both the reaction rate and activation energy due to increased kinetic energy.
Chemistry
1 answer:
N76 [4]3 years ago
7 0

Answer:

C) an increase in rate of reaction because reactant molecules collide with greater energy

Explanation:

Temperature is one of the factors that affect the rate of a reaction. The rate of a reaction increases with an increase in temperature and vice versa. When the temperature of a reaction increases, the kinetic energy of the reactant molecules increases causing them to react at a faster rate.

The reactant molecules respond to an increase in temperature by colliding at a faster rate due to an increased kinetic energy between the reactant molecules.

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Complete the equation for the conversion of sucrose into glucose <br> (1)C12H22O11 + H2O
Svetlanka [38]

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.

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8 0
3 years ago
Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −
kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH  of the respective molecules given above,

ΔH  = -1270 + (-393.5) - (-1207)

ΔH  = -456.5 KJ

8 0
3 years ago
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