(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))
"Water gas" is an industrial fuel composed of a mixture of carbon monoxide and hydrogen gases. When this
fuel is burned, carbon dioxide and water result. From the information given below, write a balanced equation
and determine the enthalpy of this reaction:
CO(g) + O2(g) → CO₂(g) + 282.8 kJ
H2(g) + O2(g) → H₂O(g) + 241.8 kJ
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4. Calculate AG for the following reaction at 25°C. Will the reaction occur (be spontaneous)? How do you
know?
NH3(g) + HCl(g) → NH4Cl(s)
AH=-176.0 kJ
AS-284.8 J-K-1
true or false
The equilibrium constant for this reaction is 0.20.
N2O4(g) ⇔ 2NO2(g)
Based on the above, the equilibrium constant for the following reaction is 5.
4NO2(g) ⇔ 2N2O4(g)
Chapter 1 Solutions
General Chemistry: Principles and Modern Applications (11th Edition)