Answer:
d = 0.793 g/L
Explanation:
Given data:
Density of fluorine gas = ?
Pressure of gas = 0.554 atm
Temperature of gas = 50 °C (50+273.15K = 323.15 K)
Solution:
Formula:
PM = dRT
M = molar mass of gas
P = pressure
R = general gas constant
T = temperature
d = PM/RT
d = 0.554 atm × 37.99 g/mol / 0.0821 atm.L /mol.K × 323.15 K
d = 21.05 atm.g/mol/26.53 atm.L /mol
d = 0.793 g/L
The atomic mass would be 28.08535 amu. Multiply 27.9769 by .92297 = 25.803. Multiply 28.9765 by .046832 to get 1.357. Multiply 29.9738 by .03872 to get .925351136. Add 25.803 + 1.357 + .03872 to get 28.08535 amu
A. Potassium oxide
B. Calcium chloride
C. Magnesium nitride
D. Sodium hypochlorite
E. Potassium nitrate
Answer:
Explanation has been given below.
Explanation:
- Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
- Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
- First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
- Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
- Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
- Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
128 ml is the voume of the balloon if the temperature of the gas increases to 320.0k.
Explanation:
given that:
T1 (initial temperature) = 300K
V1 ( initial volume) = 120ml
T2 (final temperature) = 320 K
V2 (final volume) = ?
Pressure remained constant throughout the process.
From the equation
= 
Since pressure is constant the equation will be:
= 
V2 = 
Putting the values in the above formula:
V2 = 
= 128 ml
128 ml is the volume of the gas if temperature increases from 3OO K to 320k
