Answer:
The answer to your question is Final volume = 58.37 ml
Explanation:
Data
density = 8.96 g/cm³
mass = 75 g
volume of water = 50 ml
Process
1.- Calculate the volume of copper
Density = mass / volume
Solve for volume
Volume = mass / density
Substitution
Volume = 75/8.96
Simplification
Volume = 8.37cm³ or 8.37 cm³
2.- Calculate the new volume of water in the graduated cylinder
Final volume = 50 + 8.37
Final volume = 58.37 ml
Answer: My opinion is the second one because a lot of humans uses technology and technilogy replaces a lot of stuff.
Explanation:
Answer:
8.2 x 106^-11
Explanation:
To begin this problem you must remember the basic rule of scientific notation, which is, must be between 1-10. .000000000082 is much smaller than 1. However by moving the decimal 11 spots to the right, we can make it 8.2
Continue to move the decimal to the right until the value is in the 1-10 range. Make sure to count the moves to the right.
Once the decimal is in the right spot count the spots moved.
Since the number is wayyy smaller than the answer given the number will be negative 10^-11, in order to make it what is was before.
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.
