What is the correct IUPAC name for S2F8?
1 answer:
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A curve of temperature vs. time for the entire heating process.
The sample is heated up to 100.°C, therefore, the heat and time required to heat the sample to its boiling point, the heat and time required to boil the sample, and the heat and time required to heat the sample from its boiling point to 100.°C are needs to be calculated.
i ) Calculating the heat and time required to heat the sample to its boiling point:
Boiling point = 85°C
C(liquid) = 2.5 J/g °C
The heat required up to melting the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from -20°C to 85°C can be calculated as,
Therefore, T f = 85°C and T i = - 20°C
Plug in the values in the specific heat formula to calculate the heat energy required to heat the sample to its melting point,
q3 = 25 g × 2.5 J/g °C × [85 - (-20)]°C
= 25 J/°C ×[85+20]°C
= 6562.5 J
The total heat energy required for heating the sample from initial temperature to boiling point is:-
q1 + q2 + q3 = 500 J + 4500 J + 6562.5 J
= 11562.5 J
The Rate of heating = 450 J/min
450. J = 1 min
11562.5 J = ? min
11562.5 J × 1min/450 J = 25.69 min
ii) Calculating the heat and time required to boil the sample:
∆H Vap = 500 J/g
The boiling is the phase change from liquid to gas at 85°C, therefore, the heat required to boil the sample can be determined
q4= m × ∆Hvap
= 25 g × 500 J/g
= 12500 J
Thus, total heat required to this phase change is q1 + q2 + q3 + q4 = 500 J + 4500 J +6562.5 J + 12500 J = 24062.5 J
The Rate of heating = 450 J / min
450 J = 1 min
24062.5 J = ? min
24062.5J × 1min / 450 J = 53.47 min
iii) Calculating the heat and time required to heat the sample from its boiling point to 100°C
C gas = 0.5 J / g °C
The heat required to boil the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from 85°C to 100°C can be calculated as,
Therefore, T f = 100.°C and T i = 85°C
q5 = 25 g × 0.5 J / g °C × [100 - 85] °C
= 25 J / °C ×15 °C
= 187.5 J
The total heat energy required for heating the sample from initial temperature to 100°C is
q1 + q2 + q3 + q4 + q5 = 500 J + 4500 J + 2625J + 12500 J + 187.5 J
=24250 J
The Rate of heating = 450 J / min
450. J = 1 min
24250 J=? min
Thus, heating the sample to 100.°C takes a total of 53.89 min.
iv) Draw a curve of temperature vs. time for the entire heating process:-
Temperature °C Temperature K Heat energy (J) Time (min)
-40 °C 233 0 0
-20 °C 253 500 1.11
Melting -20 °C 253 5000 11.11
85 °C 358 11562.5 25.69
Boiling 85 °C 358 24062.5 53.475
100 °C 373 24250 53.89
Hence, the graph for the result is in the image.
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Answer:
1. pH = 1.23.
2.
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
Whereas the pKa is:
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:
Which is also shown in net ionic notation.
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