Answer:
The products must contain the same numbers and types of atoms
Answer:
The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.
Explanation:
Given:
The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1
The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1
The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1
Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1
Hey!
u=25m/s
v=15m/s
t=2s
a=?
v-u=at
15-25=a×2
-10=2a
-10/2=a
a=-5m/s^2
Hope it helps...!!!
Answer:
The concentration of mole evil at oxygen on that day is 0.00858 mol/L
Explanation:
Here, we want to calculate the concentration of molecular oxygen
The pressure on that day is 1.0 atm
Since oxygen is at a concentration of 21%, the pressure of oxygen will be 21/100 * 1 = 0.21 atm
Now let’s calculate the concentration;
From Ideal gas law;
PV = nRT
This can be written as;
P/RT = n/V
The term n/V refers to concentration;
Let’s make substitutions now;
P = pressure = 0.21 atm
R = molar gas constant = 0.0821 L•atm/mol•k
T = temperature = 25 = 25 + 273.15 = 298.15 K
Substituting these values, we have;
n/V = C = 0.21/(0.0821 * 298.15) = 0.00858 mol/L
Answer:
v=14.14 m/s
t=1.141 s
Explanation:
Given that
h = 10 m
Initial velocity ,u = 0 m/s
We know that acceleration due to gravity g= 10 m/s²
Lets take final velocity = v
Final velocity v is given as
v² = u²+ 2 g h
v²= 0² + 2 x 10 x 10
v²= 200
v=14.14 m/s
Time taken t is given as
v= u + a t
a=acceleration
t=time
Now by putting the values in the above equation we get
14.14= 0 + 10 x t
14.14 = 10 t
t=1.141 s