Which of the following types of energy does an object store in the bonds of atoms and molecules? (5 points)

What part of the microscope is the letter k?

snow_tiger [21]

Answer:

if you are asking k then the round one is condenser

if not then its a stage clip

6 0

3 years ago

The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal

spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0

4 years ago

The kingdom, protista, includes

lyudmila [28]

<span>The kingdom, protista’s characteristics are that the organism (not a plant, animal or fungus) are: unicellular however some are multicellular like algae, are heterotrophic or autotrophic, others lives in water while some live in moist areas or human body, have a nucleus, cellular respiration is primarily aerobic, some are pathogenic (e.g. causing Malaria) and reproduction is mitosis or meiosis. This kingdom includes: Sacordinians – pseudopods (e.g. Amoeba, Foraminiferans<span>.)</span>, Zooflagellates – flagellates (e.g. Trypanosoma gambiense), Ciliaphorans – ciliates (e.g. paramecium) and Sporozoans (e.g. Plasmodium).</span>

3 0

3 years ago

Read 2 more answers

What is the practical use of scientific knowledge to solve everyday problems and improve lives?

Aleks04 [339]

A. Research is the answer

5 0

4 years ago

Instructions:

never [62]

Answer:

See explanation

Explanation:

A. Constitutional or structural isomers have the same molecular formula but different structural formulas.

B. Conformational isomers are compounds having the same atom to atom connectivity but differ by rotation about one or more single bonds.

C. Stereo isomers are compounds having the same molecular mass and atom to atom connectivity but different arrangement of atoms and groups in space.

I. Enantiomers are stereo isomers (optical isomers particularly) that are non-superimposable mirror images of each other.

II. Diasteromers are optical isomers that are not mirror images of each other.

Both diasteromers and enantiomers are types of optical isomers which in turn is one of the types of stereo isomers.

Stereo isomers differ from conformational isomers in that the arrangement of atoms in stereo isomers is permanent while conformational isomers results from free rotations in molecules about single bonds.

7 0

3 years ago