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3 years ago

7

Which image is cohesion stronger than adhesion and which image is adhesion stronger than cohesion

1 answer:

matrenka [14]3 years ago

5 0

Image C is adhesion stronger and Image D is cohesion stronger

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Which type of monomer combines and forms polypeptides?

irga5000 [103]

Option A is the correct answer

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Show that the Joule-Thompson Coefficient is zero for ideal gas.

melomori [17]

Answer:

Joule-Thomson coefficient for an ideal gas:

$\mu_{J.T} = 0$

Explanation:

Joule-Thomson coefficient can be defined as change of temperature with respect to pressure at constant enthalpy.

Thus,

$\mu_{J.T} = \left [\frac{\partial T}{\partial P} \right ]_H$

Also,

$H= H (T,P)$

$Differentiating\ it,$

$dH= \left [\frac{\partial H}{\partial T}\right ]_P dT + \left [\frac{\partial H}{\partial P}\right ]_T dT$

Also, $C_p$ is defined as:

$C_p = \left [\frac{\partial H}{\partial T}\right ]_P$

$So,$

$dH= C_p dT + \left [\frac{\partial H}{\partial P}\right ]_T dT$

Acoording to defination, the ethalpy is constant which means $dH = 0$

$So,$

$\left [\frac{\partial H}{\partial P}\right ]_T = -C_p\times \left [\frac{\partial T}{\partial P}\right ]_H$

$Also,$

$\mu_{J.T} = \left [\frac{\partial T}{\partial P}\right ]_H$

$So,$

$\left [\frac{\partial H}{\partial P}\right ]_T =-\mu_{J.T}\times C_p$

For an ideal gas,

$\left [\frac{\partial H}{\partial P}\right ]_T = 0$

So,

$0 =-\mu_{J.T}\times C_p$

Thus, $C_p$ ≠0. So,

$\mu_{J.T} = 0$

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3 years ago

Which is a form of kinetic energy?

hodyreva [135]

The answer is c) Electrical energy

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3 years ago

Calculate the density of air at 100 Deg C and 1 bar abs. Use the Ideal Gas Law for your calculation and give answer in kg/m3. Us

madam [21]

Answer:

$d=0.92\frac{kg}{m^{3}}$

Explanation:

Using the Ideal Gas Law we have $PV=nRT$ and the number of moles n could be expressed as $n=\frac{m}{M}$ , where m is the mass and M is the molar mass.

Now, replacing the number of moles in the equation for the ideal gass law:

$PV=\frac{m}{M}RT$

If we pass the V to divide:

$P=\frac{m}{V}\frac{RT}{M}$

As the density is expressed as $d=\frac{m}{V}$ , we have:

$P=d\frac{RT}{M}$

Solving for the density:

$d=\frac{PM}{RT}$

Then we need to convert the units to the S.I.:

$T=100^{o}C+273.15$

$T=373.15K$

$P=1bar*\frac{0.98atm}{1bar}$

$P=0.98atm$

$M=28.9\frac{kg}{kmol}*\frac{1kmol}{1000mol}$

$M=0.0289\frac{kg}{mol}$

Finally we replace the values:

$d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}$

$d=9.2*10^{-4}\frac{kg}{L}$

$d=9.2*10^{-4}\frac{kg}{L}*\frac{1L}{0.001m^{3}}$

$d=0.92\frac{kg}{m^{3}}$

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3 years ago

a solution must be at a higher temperature than a pure solvent to boil. what colligative property can be employed to achieve thi

Kazeer [188]

Boiling-point elavation.

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