Answer:
The compound which are boh soluble in water and hexane is
B. Ethanol and 1-propanol
Explanation:
The compounds ethanol and 1-propanol are soluble in both hexane and water.
It is soluble in water as both consists of polar end due to hydrogen bonding present in the -OH functional group.
and both are soluble in hexane as they contain a non polar end and the alliphatic hydrocarbon chain in them.
The solubility of alcohols varies in increasing order as the hydrocarbon chain increases. And becaue of this it becomes more non polar.
Non polar properties decreases for branched molecules.
so, the correct option is ethanol and 1-propanol.
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.
Answer:
1.73 atm
Explanation:
Given data:
Initial volume of helium = 5.00 L
Final volume of helium = 12.0 L
Final pressure = 0.720 atm
Initial pressure = ?
Solution:
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
P₁ × 5.00 L = 0.720 atm × 12.0 L
P₁ = 8.64 atm. L/5 L
P₁ = 1.73 atm
Answer:
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Answer:
Physical properties are used to observe and describe matter. Physical properties can be observed or measured without changing the composition of matter. These are properties such as mass, weight, volume, and density.
