(a) Interpretation: The boiling point of water in o M (new temperature scale) needs to be calculated if the melting point of mercury is 0 o M or − 38.9 o C and boiling point of mercury is 100 o M or 356.9 o C. Concept introduction: The conversion factor required to determine the results in different or desired scale by equivalizing different units. The temperature of the sample at given scale degree M can be identified as follows: ( T calculated ( in o M ) – T lower ( in o M ) ) ( T higher ( in o M ) – T lower ( in o M ) ) = ( T calculated ( in o C ) – T lower ( in o C ) ) ( T higher ( in o C ) – T lower ( in o C ) )
(a) Interpretation: The boiling point of water in o M (new temperature scale) needs to be calculated if the melting point of mercury is 0 o M or − 38.9 o C and boiling point of mercury is 100 o M or 356.9 o C. Concept introduction: The conversion factor required to determine the results in different or desired scale by equivalizing different units. The temperature of the sample at given scale degree M can be identified as follows: ( T calculated ( in o M ) – T lower ( in o M ) ) ( T higher ( in o M ) – T lower ( in o M ) ) = ( T calculated ( in o C ) – T lower ( in o C ) ) ( T higher ( in o C ) – T lower ( in o C ) )
The boiling point of water in oM (new temperature scale) needs to be calculated if the melting point of mercury is 0 oM or − 38.9 oC and boiling point of mercury is 100 oM or 356.9 oC.
Concept introduction:
The conversion factor required to determine the results in different or desired scale by equivalizing different units.
The temperature of the sample at given scale degree M can be identified as follows:
(Tcalculated(in oM)– Tlower(in oM))(Thigher( in oM)– Tlower(in oM)) = (Tcalculated( in oC)– Tlower(in oC))(Thigher(in oC)– Tlower( in oC))
Interpretation Introduction
(b)
Interpretation:
The absolute zero temperature in o M needs to be determined if melting point of mercury is 0 o M or − 38.9 o C and boiling point of mercury is 100 o M and 356.9 o C.
Concept introduction:
The conversion factor required to determine the results in different or desired scale by equivalizing different units.
The temperature of the sample at given scale degree M can be identified as follows:
(Tcalculated(in oM)– Tlower(in oM))(Thigher( in oM)– Tlower(in oM)) = (Tcalculated( in oC)– Tlower(in oC))(Thigher(in oC)– Tlower( in oC))
Given that a theoretical yield for isolating Calcium Carbonate in this experiment would be 100%. From that information and based on the results you obtained in this experiment, describe your success in the recovery of calcium carbonate and suggest two possible sources of error that would have caused you to not obtain 100% yield.
Results are attached form experiment
5) Calculate the flux of oxygen between the ocean and the atmosphere(2 pts), given that:
(from Box 5.1, pg. 88 of your text):
Temp = 18°C
Salinity = 35 ppt
Density = 1025 kg/m3
Oxygen concentration measured in bulk water = 263.84 mmol/m3
Wind speed = 7.4 m/s
Oxygen is observed to be about 10% initially supersaturated
What is flux if the temperature is 10°C ? (2 pts) (Hint: use the same density in your calculations). Why do your calculated values make sense (or not) based on what you know about the relationship between gas solubility and temperature (1 pt)?