Answer:
![K=\frac{[CaO][CH_{4}][H_{2}O ]^{2} }{[CaCO_{2}][H_{2}]^4 }](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BCaO%5D%5BCH_%7B4%7D%5D%5BH_%7B2%7DO%20%5D%5E%7B2%7D%20%20%7D%7B%5BCaCO_%7B2%7D%5D%5BH_%7B2%7D%5D%5E4%20%20%7D)
Explanation:
The equilibrium expression is the K value equal to the product of the concentrations of the products over the product of the concentrations of the reactants. If there is a coefficient in front of the compound, raise the molecule to that power.
Since K is big, more product is expected. This is because of mathematic principles. A large numerator with a small denominator will produce a large number.
The percent composition of each element can be calculated as follows:
% composition = (mass of element / total mass) * 100
The total mass of the quarter is given to be 5.670 grams
Mass of Cu = 5.198 grams
Mass of Ni = 0.472 grams
Substitute in the above equation to get the mass percentage of each element as follows:
% of Cu = (5.198/5.670) * 100 = 91.675%
% of Ni = (0.472/5.670) * 100 = 8.325%
Explanation:
According to the given data, we will calculate the following.
Half life of lipase
= 8 min x 60 s/min
= 480 s
Rate constant for first order reaction is as follows.
=
Initial fat concentration
= 45
= 45 mmol/L
Rate of hydrolysis
= 0.07 mmol/L/s
Conversion X = 0.80
Final concentration (S) =
= 45 (1 - 0.80)
= 9
or, = 9 mmol/L
It is given that
= 5mmol/L
Therefore, time taken will be calculated as follows.
t = ![-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7BK_%7Bd%7D%7Dln%5B1%20-%20%5Cfrac%7BK_%7Bd%7D%7D%7BV%7D%7BK_%7BM%7D%20ln%20%28%5Cfrac%7BS_%7Bo%7D%7D%7BS%7D%29%20%2B%20%28S_%7Bo%7D%20-%20S%29%5D)
Now, putting the given values into the above formula as follows.
t =
= ![-\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s }{K_{M} ln (\frac{45 mmol/L }{9 mmol/L }) + (45 mmol/L - 9 mmol/L )]](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B1.44%20%5Ctimes%2010%5E%7B-3%7Ds%5E%7B-1%7D%7Dln%5B1%20-%20%5Cfrac%7B1.44%20%5Ctimes%2010%5E%7B-3%7Ds%5E%7B-1%7D%7D%7B0.07%20mmol%2FL%2Fs%0A%7D%7BK_%7BM%7D%20ln%20%28%5Cfrac%7B45%20mmol%2FL%0A%7D%7B9%20mmol%2FL%0A%7D%29%20%2B%20%2845%20mmol%2FL%20-%209%20mmol%2FL%0A%29%5D)
= 
= 27.38 min
Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.
1.785714286 moles
The number of moles (n) for nitrogen is: [ n=50.0÷28.0 ] = 1.785714286 moles.
Jim, because he ran a greater distance in the same time :)
By the way, this is a maths question