Hi There! :)
An equilibrium constant is not changed by a change in pressurea. True
b. False
False! :P
        
                    
             
        
        
        
Answer:
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
Explanation:given values
Half life of lipase t_1/2 = 8 min x 60s/min = 480 s
Rate constant for first order reaction
k_d = 0.6932/480 = 1.44 x 10^-3 s-1
Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L
rate of hydrolysis Vm0 = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3
K_m = 5mmol/L
time take is given by 
![t= -\frac{1}{K_d}ln[1-\frac{K_d}{V_m_0}(k_mln\frac{s_0}{s}+(s_0-s))]](https://tex.z-dn.net/?f=t%3D%20-%5Cfrac%7B1%7D%7BK_d%7Dln%5B1-%5Cfrac%7BK_d%7D%7BV_m_0%7D%28k_mln%5Cfrac%7Bs_0%7D%7Bs%7D%2B%28s_0-s%29%29%5D)
all values are given and putting these value we get 
t=1642.83 secs
which is equal to 
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
 
        
             
        
        
        
Answer:
10.58 ft
Explanation:
Force, F = 1.4 N
Diameter of membrane = 7.4 mm 
radius of membrane, r = 7.4 / 2 = 3.7 mm = 3.7 x 10^-3 m
Area, A = 3.14 x (3.7 x 10^-3)^2 = 4.3 x 10^-5 m^2
Density, d = 1.03 x 10^3 kg/m^3
Pressure at depth, P = h x d x g 
Let h be the depth.
Pressure = force / Area 
h x 1.03 x 10^3 x 9.8 = 1.4 / (4.3 x 10^-5)
h = 3.225 m = 10.58 ft
Thus, the depth of water is 10.58 ft.
 
        
             
        
        
        
Answer:
since small stone has less mass so the gravitational pull of the earth is lesser in case of this but this is not for the bigger stone as the gravitational pull of the earth is greater...
PLEASE MARK BRAINLIEST!!!!!
 
        
             
        
        
        
Answer:
See description
Explanation:
With the given information we have:

the interval is ![[0,\pi ]](https://tex.z-dn.net/?f=%5B0%2C%5Cpi%20%5D)
now the mass  has the given expression:
 has the given expression:

we will use the formula for a line integral and let:

therefore we have:

we solve the integral:
