Response:
of 
can be found in 39.5 grams of .
Clarification:
Atomic weights: P= 31, F= 19,
The molar mass equals 1 atomic weight of P + 5 atomic weights of
F= 31+5 × 19
= 31+95
=126 g/mole
The number of moles in 39.5 gm of
equals 
= 
=0.3134 moles
1 mole of any substance encompasses
0.3131 moles comprises 0.3134
Thus, of can be found in 39.5 grams of .
Solution:
The gas's new temperature is 604K
Justification:
Assuming standard temperature and pressure, we can determine the gas's temperature using the ideal gas law;
Step 1: Formulate the general gas law equation
P1V1/T1 = P2V2/T2
Step 2: Insert the values, converting as needed to standard units.
P1 = 0.800 atm
V1 = 0.180 L
T1 = 29°C = 273 + 29 = 302K
P2 = 3.20 atm
V2 = 90 mL = 90 * 10^-3 L = 0.09 L
Step 3: Solve for T2
The new gas temperature T2 is calculated as:
T2 = P2V2T1/(P1V1)
T2 = 3.20 * 0.09 * 302 / (0.800 * 0.180)
T2 = 86.976 / 0.144
T2 = 604K
The gas's new temperature is 604K.
Response:
The conclusion to your inquiry is Pressure 1 = 1.73 atm
Clarification:
Data provided
Volume 1 = 5 l
Pressure 1 =?
Volume 2 = 12 l
Pressure 2 = 0.72 atm
Procedure
To resolve this issue, Boyle's law is applied
Pressure 1 x Volume 1 = Pressure 2 x Volume 2
-Finding Pressure 1
Pressure 1 = Pressure 2 x Volume 2 / Volume 1
-Replace values
Pressure 1 = 0.72 x 12 / 5
-Calculating
Pressure 1 = 8.64/5
-Final outcome
Pressure 1 = 1.73 atm
