The answer is A: Between 50 and 5,000 amino acids
The entropy of the given reactions increases (S° > 0):
- 2C2H6(g) + 7O2(g) ----> 4CO2(g) + 6H2O(g) ---> S° > 0.
- NH4Cl(s) ----> NH3(g) + HCl(g) ---> S° > 0.
<h3>What is entropy?</h3>
Entropy measure how disordered a system is. It a measure of how dispersed or random the total energy of a system is. The symbol for entropy is S.
A system in which entropy increases is one in which S° > 0.
The entropy of a system decreases when S° < 0.
Entropy of a system increases (S° > 0) if any change results in an increase in temperature, increase in number of molecules, or an increase in volume.
Considering the given systems, the entropy changes is as follows:
- 2C2H6(g) + 7O2(g) ----> 4CO2(g) + 6H2O(g) ---> S° > 0.
- 2CO2(g) + N2(g) ---->2CO(g) + 2NO(g) --> no change
- 2N2(g) + O2(g) -----> 2N2O(g) ---> decreases
- S (s,rhombic) + 2CO(g) ----> SO2(g) + 2C (s,graphite) ---> no change
- NH4Cl(s) ----> NH3(g) + HCl(g) ---> S° > 0.
In conclusion, an increase in volume and moles of substances results in entropy increase.
Learn more about entropy at: brainly.com/question/26691503
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Answer : The correct option is (B).
Explanation :
Body-centered cubic structure (bcc) :
Crystal structure : The manner in which the atoms or ions are spatially arranged.
In bcc arrangement, the atoms are present at eight corners of the cube and at the center of the cube. And the coordination number is 8.
Each unit cell has
atoms at the corner and 1 full atom at the center of the cube.
Therefore, the each unit cell has
atoms.
Image of bcc structure is shown below.
First statement
Fifth statement
Sixth statement
Answer:
c = 0.07 j/g.k
Explanation:
Given data:
Mass of sample = 35 g
Heat absorbed = 48 j
Initial temperature = 293 K
Final temperature = 313 K
Specific heat of substance = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
ΔT = 313 k - 293 K
ΔT = 20 k
Now we will put the values in formula.
48 j = 35 g × c× 20 k
48 j = 700 g.k ×c
c = 48 j/700 g.k
c = 0.07 j/g.k