***BRAINLIEST ASNWERRR***How many grams are in 34.2 moles of Lithium (Li)?
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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.
Learn more about temperature here:-brainly.com/question/24746268
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- Get 3 cups of water at the exact same temperature, using the thermometer to check.
- Label the cups as ‘whole’, ‘pieces’, and ‘crushed’
- Next, get something to dissolve, in this case, polident. Take one of the polident tablets and break it into 4 pieces, and set it aside.
- Take another polident tablet and this time put it into a different cup, and crush it. Set it aside.
- Keep the last tablet whole.
- Set up your stopwatch and drop the polident tablet that is whole in the cup labeled ‘whole’, starting the stopwatch at the same time.
- Watch the cup and see when the tablet is fully dissolved, then stop the stopwatch.
- Record the time in the table.
- Repeat steps 6-8 for both the ‘pieces’ and ‘crushed’ tablets.
Hope this helps! Please let me know if you need more help, or if you think my answer is incorrect. Brainliest would be MUCH appreciated. Have a great day!
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