Compute the value of the molar heat capacity at constant volume, CVCV, for CO2CO2 on the assumption that there is no vibrational energy. (Note: CO2CO2 is linear; SO2SO2 and H2SH2S are not. Recall that a linear polyatomic molecule has two rotational degrees of freedom, and a nonlinear molecule has three.)

Explanation:

Molar heat capacity at constant volume Cv  of a gas = n x .5 R where n is degree of freedom of the gas molecules

CO₂ is a linear molecule , so number of degree of freedom =  3 + 2 = 5

3 is translational and 2 is rotational degree of freedom . There is no vibrational degree of freedom given .

So Cv = 5 / 2 R

= 2.5 R .

Answer:

Step-by-step explanation:

The input force is 50 N. But it will not create not any change. No mechanical advantage is observed.

glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

Answer:

Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

Explanation:

In this case, you know:

P= 0.884 atm

V= ?

n= 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)

R=0.082

T= 328 K

Replacing in the ideal gas law:

0.884 atm×V= 0.857 moles× 0.082 ×328 K

Solving:

V= 26.07 L

The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.