Answer:
V2 = 894.4mL
Explanation:
P1= 124.1, V1= 578mL, P2 = 80.2kPa, V2= ?
Applying Boyle's law
P1V1 = P2V2
Substitute and simplify
124.1*578=80.2*V2
V2= 894.4mL
The volume<span> of </span>gas<span> because of the </span>increase<span> and decrease in the speed in which the molecules bounce around. ... Boyle's Law states that if temperature stays the same, the </span>amount of<span> space a </span>gas takes up will increase<span> if the </span>pressure<span> decreases. The </span>amount of gas<span> will take up less space if the </span>pressure<span> is increased. this would be the correct answer </span>
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.
