Answer:
C) Temperature and Kinetic Energy.
Explanation:
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In this case, according to the generic heating curve on the attached file, it possible to see that on the point C, whereas the line is diagonal, the temperature increases, but also the kinetic energy increases because the molecules gain energy due to the increase of the temperature. It is important to say that on flat lines, like those on B and D, the phase change takes place and just the potential energy change.
In such a way, we infer that the answer is C) Temperature and Kinetic Energy.
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<u>Answer:</u> The atomic weight of the second isotope is 64.81 amu.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of atomic masses of each isotope each multiplied by their natural fractional abundance
Formula used to calculate average atomic mass follows:
.....(1)
We are given:
Let the mass of isotope 2 be 'x'
Mass of isotope 1 = 62.9 amu
Percentage abundance of isotope 1 = 69.1 %
Fractional abundance of isotope 1 = 0.691
Mass of isotope 2 = 'x'
Percentage abundance of isotope 2 = 30.9%
Fractional abundance of isotope 2 = 0.309
Average atomic mass of copper = 63.5 amu
Putting values in equation 1, we get:
![\text{Average atomic mass of copper}=[(62.9\times 0.691)+(x\times 0.309)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20copper%7D%3D%5B%2862.9%5Ctimes%200.691%29%2B%28x%5Ctimes%200.309%29%5D)
Hence, the atomic weight of second isotope will be 64.81 amu.
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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.
