Answer:
Explanation:
The rather high specific activity of intracellular sucrase towards sucrose is known to be optimal at pH 6.0 and at a temperature of about 30°C. And thus, we can say that the optimum pH for sucrase activity is exactly at 6. Also, it's behaviour is said to be decreasing with increasing acidic and increasing alkallinic values.
Answer:
Joule-Thomson coefficient for an ideal gas:

Explanation:
Joule-Thomson coefficient can be defined as change of temperature with respect to pressure at constant enthalpy.
Thus,
![\mu_{J.T} = \left [\frac{\partial T}{\partial P} \right ]_H](https://tex.z-dn.net/?f=%5Cmu_%7BJ.T%7D%20%3D%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20T%7D%7B%5Cpartial%20P%7D%20%5Cright%20%5D_H)
Also,


![dH= \left [\frac{\partial H}{\partial T}\right ]_P dT + \left [\frac{\partial H}{\partial P}\right ]_T dT](https://tex.z-dn.net/?f=dH%3D%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20T%7D%5Cright%20%5D_P%20dT%20%2B%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_T%20dT)
Also,
is defined as:
![C_p = \left [\frac{\partial H}{\partial T}\right ]_P](https://tex.z-dn.net/?f=C_p%20%3D%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20T%7D%5Cright%20%5D_P)

![dH= C_p dT + \left [\frac{\partial H}{\partial P}\right ]_T dT](https://tex.z-dn.net/?f=dH%3D%20C_p%20dT%20%2B%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_T%20dT)
Acoording to defination, the ethalpy is constant which means 

![\left [\frac{\partial H}{\partial P}\right ]_T = -C_p\times \left [\frac{\partial T}{\partial P}\right ]_H](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_T%20%3D%20-C_p%5Ctimes%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20T%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_H)

![\mu_{J.T} = \left [\frac{\partial T}{\partial P}\right ]_H](https://tex.z-dn.net/?f=%5Cmu_%7BJ.T%7D%20%3D%20%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20T%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_H)

![\left [\frac{\partial H}{\partial P}\right ]_T =-\mu_{J.T}\times C_p](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_T%20%3D-%5Cmu_%7BJ.T%7D%5Ctimes%20C_p)
For an ideal gas,
![\left [\frac{\partial H}{\partial P}\right ]_T = 0](https://tex.z-dn.net/?f=%5Cleft%20%5B%5Cfrac%7B%5Cpartial%20H%7D%7B%5Cpartial%20P%7D%5Cright%20%5D_T%20%3D%200)
So,

Thus,
≠0. So,

Answer:
See explanation
Explanation:
Copper II oxide is a base but not an alkali. An alkali is a soluble base. Since Copper II oxide is not soluble in water then it is not an alkali.
Let us recall that the change of colour of litmus with an alkali requires the presence of water. In the absence of water, solid Copper II oxide does not turn red litmus paper blue.
The ability to turn red litmus paper blue is commonly observed with alkalis and Copper II oxide is not an alkali.
Also recall that since Copper II oxide is not soluble, hydroxide ions are absent hence Copper II oxide does not turn red litmus paper blue.
Inhalation is the process of taking air into the lungs. For this to occur, the air pressure inside the lungs must be lower than that of the external atmosphere as air flows from areas of higher pressure to lower pressure. <span>Exhalation is the process of expelling air out of the lungs. For this to occur, the air pressure inside the lungs must be higher than that of the external atmosphere as air flows from areas of higher pressure to ones of lower pressure.</span>
Answer:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Explanation:
The silver nitrate, AgNO₃, dissolves in water as follows:
AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)
The Ag⁺ reacts with Cl⁻ producing AgCl(s), a white insoluble salt. The net ionic equation that describes the formation of the precipitate is:
<h3>Ag⁺(aq) + Cl⁻(aq) → AgCl(s)</h3><h3 /><h3 />