What exactly is the question
Answer:
V = 15.9512 dm³
Explanation:
Given data:
Pressure = P = 1.37 atm
Temperature = T= 315 K
Number of moles of nitrogen= n = 0.845 mol
Volume = V = ?
Formula:
PV = nRT
Now we will put the values in equation.
V = nRT/ P
V = ( 0.845 mol× 0.0821 dm³.atm.K⁻¹.mol⁻¹ × 315 K) / 1.37 atm
V = 21.853 dm³. atm/ 1.37 atm
V = 15.9512 dm³
Answer:
For CH3CH2OH the OH stretch between 3550-3200 cm-1
For CH3CH2NH2 a pair of bands at 3500 and 3400 cm-1 respectively corresponding to the N-H stretch
Explanation:
CH3CH2OH is ethanol. This is a primary alcohol characterized by the by the -OH (hydroxyl) group. A strong broad IR absorption in the 3550 - 3200 cm-1 corresponds to the O-H stretch and indicates that the compound could be an alcohol or it has an OH functional group.
CH3CH2NH2 is ethylamine. This is a primary amine characterized by the -NH2 (amino) group. The presence of a primary amine shows up as a pair of medium strength absorption features corresponding to the N-H stretch, one at around 3500 cm-1 and the other around 3400 cm-1.
Pretty sure it's 5.25Explanation:
