Answer: ![-\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BBr%5E.%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Explanation:
Rate of a reaction is defined as the rate of change of concentration per unit time.
Thus for reaction:
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
![Rate=-\frac{d[Br^.]}{2dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7Bd%5BBr%5E.%5D%7D%7B2dt%7D)
or ![Rate=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Thus ![-\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BBr%5E.%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
hi <3
i believe i explained this answer properly in my last answer but it would be 4kg and 2400m as these are the SI units for these values.
hope this helps :)
They are formed from erosion and weathering.
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
Answer:
Rate = vmax k3/k2+k3
Explanation:
The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction, is referred to as Vmax.
This is usually expressed as the Km ie. Michaelis constant of the enzyme, an inverse measure of affinity. For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax.
Please kindly check attachment for the step by step solution of the given problem.
