(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))
Draw the major product of this reaction. Ignore inorganic
byproducts.
N
O'
1. H2O, pyridine
2. neutralizing work-up
く
8
W
a
Ideal Gas Law Practice
Name
If you need a refresher on Ideal Gas Law, go back to your Ideal Gas Law Reading Assignment from last
week!
On all of the following, you'll need to make sure to convert pressures to atm and convert temperatures to
Kelvin in order to be able to use the R gas constant on your equation sheet!
Given: Ideal Gas Law =
then P= pressure
V = volume
R= ideal gas consent
PV=nRT
namount of substance
n=PV/TR
P=nRT/V
I = temperature
V=nRT/P
T=PV/nR
R=PV/nT
1. What pressure is required to contain 0.023 moles of nitrogen gas in a 4.2 L container at a
temperature of 20.°C?
2. Oxygen gas is collected at a pressure of 123 kPa in a container which has a volume of 10.0 L. What
temperature must be maintained on 0.500 moles of this gas in order to maintain this pressure? Express
the temperature in degrees Celsius.
3. How many moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a
temperature of 100. °C? After determining the number of moles,…
1. The following conversion includes chemistry we have covered very recently, some
chemistry from last term, and chemistry from the first chapter of this unit. Provide curly
arrows and an explanation for this mechanism. Use the reagents in the order given. You
do not need any other reagents.
1. NaOEt
OEt
2.