PHA 317 Pharmaceutics I Exam 1 UPDATED ACTUAL Questions and CORRECT
Answers
7^71 x 7^5 x 7^21 7^97
(10^4)^5 10^20
Any number multiplied by zero 0
Any number multiplied by Infinity Infinity
Any number divided by infinity Undefined
In y = 2.303 log y
Convert log 234 into ln (Natural Log) log y = 1/2.303 ln y = 0.4343 ln y
Answer: 0.4343 ln 234
Specific Gravity = Density sub./ Density H2O
What is the specific gravity of a substance
1 m^3 = 1000000 cm3
of density 16 Kg/m^3? Given that the
Convert 16kg/ m^3 = 1600g/ 1000000 cm^3 =0.016
density of water is 1g/cc.
0.016/1 = 0.016 g/cc
One mole of ether was confined in a 7 liter PV= nRT
vessel at 300°k. Calculate the pressure P= (1 x 0.0821 x300)/ 7
produced using the ideal gas equation. P= 3.519 atm
What pressure is necessary to keep butane 1. Convert all temperatures to Kelvin
as a liquid in a container on a day when the 2. Plug in to Equation given
storeroom temperature is 104°F. The heat of log(VP2/1)= (5318 cal/mole x (313.15-272.65))/
vaporization butane is 5318 cal/mole and (2.303x1.9872x313.15x272.65) = VP2= anti log 0.55=
the boiling point at 1 atm is -0.50°c 3. 56 atm
300 absolute temperature is equal to how 300-273.16 = 26.84°C
much degree Celsius?
What happens to the volume of a gas when The volume would be double of the original or initial volume
it's temperature is increased 2 times at
constant pressure?
Answers
7^71 x 7^5 x 7^21 7^97
(10^4)^5 10^20
Any number multiplied by zero 0
Any number multiplied by Infinity Infinity
Any number divided by infinity Undefined
In y = 2.303 log y
Convert log 234 into ln (Natural Log) log y = 1/2.303 ln y = 0.4343 ln y
Answer: 0.4343 ln 234
Specific Gravity = Density sub./ Density H2O
What is the specific gravity of a substance
1 m^3 = 1000000 cm3
of density 16 Kg/m^3? Given that the
Convert 16kg/ m^3 = 1600g/ 1000000 cm^3 =0.016
density of water is 1g/cc.
0.016/1 = 0.016 g/cc
One mole of ether was confined in a 7 liter PV= nRT
vessel at 300°k. Calculate the pressure P= (1 x 0.0821 x300)/ 7
produced using the ideal gas equation. P= 3.519 atm
What pressure is necessary to keep butane 1. Convert all temperatures to Kelvin
as a liquid in a container on a day when the 2. Plug in to Equation given
storeroom temperature is 104°F. The heat of log(VP2/1)= (5318 cal/mole x (313.15-272.65))/
vaporization butane is 5318 cal/mole and (2.303x1.9872x313.15x272.65) = VP2= anti log 0.55=
the boiling point at 1 atm is -0.50°c 3. 56 atm
300 absolute temperature is equal to how 300-273.16 = 26.84°C
much degree Celsius?
What happens to the volume of a gas when The volume would be double of the original or initial volume
it's temperature is increased 2 times at
constant pressure?
