(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))
(a) Normally the human body can endure a temperatureof 105°F for only short periods of time without permanentdamage to the brain and other vital organs. What is thistemperature in degrees Celsius? (b) Ethylene glycol isa liquid organic compound that is used as an antifreezein car radiators. It freezes at −11.5°C. Calculate itsfreezing temperature in degrees Fahrenheit. (c) Thetemperature on the surface of the sun is about 6300°C.What is this temperature in degrees Fahrenheit?
A graduated cylinder (measuring cylinder) contains 27 mL of water at 25 °C and the
mass of the water is 26.92 g. A sphere of iron with a diameter of 18.0 mm is added to
the water in the graduated cylinder. The total mass of the water and the iron is 50.92 g.
(The volume of a sphere is given by the expression: V = (4/3)ar³).
(i) Calculate the total volume of the water and the iron in the graduated cylinder.
(ii) Determine the density of iron at 25 °C? Show your calculations.
(iii) Without doing any calculations, what do you think the density of 48.00 g of iron is
at the same temperature? Explain your answer.
A 25.5 g sample of a metal was placed into water in a gradu- ated cylinder. The metal sank to the bottom, and the water level rose from 15.7 mL to 25.3 mL. What is the identity of the metal?
(a) Tin (density = 7.31 g/cm3)(b) Lead (density = 11.34 g/cm3)(c) Silver (density = 10.49 g/cm3)(d) Aluminum (density = 2.64 g/cm3)