I would say that honestly reporting experimental findings is an example of using good science because science is definitely about honesty in accepting experimental findings and realizing that one has to face up to the consequences and develop things from there rather than wishing the outcome was different.
Answer:
1-Ethyl-3-methylidenecyclopentane
Step-by-step explanation:
Formula = C₈H₁₄. An alkane has formula C₈H₁₈. ∴ X contains 2 double bonds, 2 rings, or 1 ring and 1 double bond.
X absorbs only 1 mol of hydrogen. ∴ X contains 1 ring and 1 double bond.
Hydrogenation gives 1-ethyl-3-methylcyclopentane.
Ozonolysis gives formaldehyde, so X must contain a =CH₂ group.
Hydrogenation of X converted the =CH₂ to -CH₃.
X is 1-ethyl-3-methylidenecyclopentane.
You can see the reactions in the image below.
The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:
ΔS = nCln(T₂/T₁)
n = number of moles
C = molar heat capacity
T₂ = final temperature
T₁ = initial temperature
We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:
[(3 moles)(273.15 K) + (1 mole)(373.15 K)]/(4 moles) = 298.15 K
Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:
ΔS = (3 moles)(75.3 J/Kmol)ln(298.15/273.15)
ΔS = 19.8 J/K
ΔS = (1 mole)(75.3 J/Kmol)ln(298.15/373.15)
ΔS = -16.9 J/K
Now we combine the entropy change of each portion of water to get the total entropy change for the system:
ΔS = 19.8 J/K + (-16.9 J/K)
ΔS = 2.9 J/K
The entropy change for combining the two temperatures of water is 2.9 J/K.
The answers is C because they are pulling with different amount of forces opposite ways and there is more force towards the left.
