It can be done. Normally the boiling point of water is 100°C. It will boil at temperature greater than 100°C more quickly. Water can be boiled at 95°C but for that the atmospheric pressure of the water should be decreased which will decrease the boiling point of water.
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Concept :</h3>
To boil water at 95°C, decrease the atmospheric pressure.
At 105°C, the water will be boiling quickly than normal at 100°C.
Explanation:
The given data is as follows.
![V_{max} = 6.8 \times 10^{-10} \mu mol/min](https://tex.z-dn.net/?f=V_%7Bmax%7D%20%3D%206.8%20%5Ctimes%2010%5E%7B-10%7D%20%5Cmu%20mol%2Fmin)
![K_{m} = 5.2 \times 10^{-6} M](https://tex.z-dn.net/?f=K_%7Bm%7D%20%3D%205.2%20%5Ctimes%2010%5E%7B-6%7D%20M)
Now, according to Michaelis-Menten kinetics,
![V_{o} = V_{max} \times [\frac{S}{(S + Km)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%20V_%7Bmax%7D%20%5Ctimes%20%5B%5Cfrac%7BS%7D%7B%28S%20%2B%20Km%29%7D%5D)
where, S = substrate concentration =
M
Now, putting the given values into the above formula as follows.
![V_{o} = V_{max} \times [\frac{S}{(S + Km)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%20V_%7Bmax%7D%20%5Ctimes%20%5B%5Cfrac%7BS%7D%7B%28S%20%2B%20Km%29%7D%5D)
![V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%206.8%20%5Ctimes%2010%5E%7B-10%7D%20%5Cmu%20mol%2Fmin%20%5Ctimes%20%5B%5Cfrac%7B10.4%20%5Ctimes%2010%5E%7B-6%7D%20M%7D%7B%2810.4%20%5Ctimes%2010%5E%7B-6%7DM%20%2B%205.2%20%5Ctimes%2010%5E%7B-6%7D%20M%29%7D%5D)
![V_{o} = 6.8 \times 10^{-10} \mu mol/min \times 0.667](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%206.8%20%5Ctimes%2010%5E%7B-10%7D%20%5Cmu%20mol%2Fmin%20%5Ctimes%200.667)
= ![4.5 \times 10^{-10} \mu mol/min](https://tex.z-dn.net/?f=4.5%20%5Ctimes%2010%5E%7B-10%7D%20%5Cmu%20mol%2Fmin)
This means that
would approache
.
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A physicist
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