A poor ratio of Oxygen will cause incomplete combustion and the consequent deposit of carbon.
All living organisms are composed of one or more cells. The cell is the basic unit of structure and organization in organisms. Cells arise from pre-existing cells.
Answer:
(1) -12 Kcal/mol
Explanation:
Our answer options for this question are:
(1) -12 Kcal/mol
(2) -13 Kcal/mol
(3) -15 Kcal/mol
(4) -16 Kcal/mol
With this in mind, we can start with the chemical reaction (Figure 1). In this reaction, <u>two bonds are broken</u>, a C-H and a Br-Br. Additionally, a C-Br and a H-Br are <u>formed</u>.
If we want to calculate the enthalpy value, we can use the equation:
<u>ΔH=ΔHbonds broken-ΔHbonds formed</u>
If we use the energy values reported, its possible to calculate the energy for each set of bonds:
<u>ΔHbonds broken</u>
<u />
C-H = 94.5 Kcal/mol
Br-Br = 51.5 Kcal/mol
Therefore:
105 Kcal/mol + 53.5 Kcal/mol = 146 Kcal/mol
<u>ΔHbonds formed</u>
C-Br = 70.5 Kcal/mol
H-Br = 87.5 Kcal/mol
Therefore:
70.5 Kcal/mol + 87.5 Kcal/mol = 158 Kcal/mol
<u>ΔH of reaction</u>
<u />
ΔH=ΔHbonds broken-ΔHbonds formed=(146-158) Kcal/mol = -12 Kcal/mol
I hope it helps!
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Answer:
a. Concave down
Linear increasing
b. Increases the reaction rate
c. The reaction approaches the saturation point of the enzyme
Explanation:
a. For the reaction with enzyme, the shape is concave down. The action of the enzyme on the preferred substrate is initially very rapid and decreases as the enzyme becomes saturated and the ratio of products to substrate increases to approach an equilibrium rate of reaction
For the reaction without enzyme, the shape is linear and increasing. Increase in the concentration of the substrate will increase the number of effective collisions that lead into product formation leading to an increased rate of the chemical reaction
b. The enzyme increases the proportion of effective combination of substrates to form the products
c. The curve of the reaction with enzyme flattens out because as the concentration of the substrate increases while that of the enzyme remains the same, the enzyme becomes saturated and less able to increase the rate of the reaction of the excess substrate.
