Answer:
B) hyperbolic curve; saturated with substrate
Explanation:
Enzymatic kinetics studies the speed of enzyme catalyzed reactions. These studies provide direct information about the mechanism of the catalytic reaction and the specificity of the enzyme. The speed of a reaction catalyzed by an enzyme can be measured with relative ease, since in many cases it is not necessary to purify or isolate the enzyme. The measurement is always carried out under the optimal conditions of pH, temperature, presence of cofactors, etc., and saturating substrate concentrations are used. Under these conditions, the reaction rate observed is the maximum speed (Vmax). The speed can be determined either by measuring the appearance of the products or the disappearance of the reagents.
Following the rate of appearance of product (or disappearance of the substrate) as a function of time, the so-called reaction progress curve is obtained, or simply, the reaction kinetics. This curve is represented by a hyperbolic curve
Answer:
3626.76dm³
Explanation:
Given parameters:
Number of moles of Nitrogen in tank = 17moles
Temperature of the gas = 34°C
Pressure on the gas = 12000Pa
Unkown:
Volume of the tank, V =?
Converting the parameters to workable units:
We take the temperature from °C to Kelvin
K = 273 + °C = 273 + 34 = 307k
Taking the pressure in Pa to atm:
101325Pa = 1atm
12000Pa = 0.118atm
Solution:
To solve this problem, we employ the use of the ideal gas equation. The ideal gas law combines three gas laws which are the Boyle's law, Charles's law and the Avogadro's law.
It is expressed as PV = nRT
The unknown is the Volume and we make it the subject of the formula
V = 
Where R is called the gas constant and it is given as 0.082atmdm³mol⁻¹K⁻¹
Therefore V =
= 3626.76dm³
0.042 moles of Hydrogen evolved
<h3>Further explanation</h3>
Given
I = 1.5 A
t = 1.5 hr = 5400 s
Required
Number of Hydrogen evolved
Solution
Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.
Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)
Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻
Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)
So at the cathode H₂ gas is produced
Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C

Q = i.t
Q = 1.5 x 5400
Q = 8100 C
mol e⁻ = 8100 : 96500 = 0.084
From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042
A=acid
B=it say neither it say it’s alkaline
C=acid
D= it say neither it say it’s alkaline
Answer:
Data is not valid
Explanation:
When two liquids having different temperatures are mixed, regardless of the volumes, the final mix temperature will ALWAYS be between the initial temperature values.
1st Law Thermo => Law of Conservation of Energy => Energy can not be created nor destroyed, only changed in form. Mixing 22°C with 75°C will NOT result in a mix having a final temperature of 80°C.
∑ΔE = 0 => (mcΔT)₁ + (mcΔT)₂ = 0
[(20g)(1cal/g·°C)(Tₓ - 22°C)] + [(80g)(1cal/g·°C)(Tₓ - 75°C)] = 0
=> 20(Tₓ - 22) + 80(Tₓ - 75) = 0
=> 20Tₓ - 440 + 80Tₓ - 75 = 0
=> 100Tₓ = 440 + 75 = 515
=> Tₓ = (515/100)°C = 51.5°C final mix temperature