Answer:
Volume of sample after droping into the ocean=0.0234L
Explanation:
As given in the question that gas is idealso we can use ideal gas equation to solve this;
Assuming that temperature is constant;
Lets 
and 
are the initial gas parameter before dropping into the ocean
and 
and 
are the final gas parameter after dropping into the ocean
according to boyle 's law pressure is inversly proportional to the volume at constant temperature.
hence,
P1=1 atm
V1=1.87L
P2=80atm
V2=?
After putting all values we get;
V2=0.0234L
Volume of sample after droping into the ocean=0.0234L
! mole of CO2 at STP has a volume of 22.4 liters
88 grams = 2 moles
so the required volume = 2*22.4 = 44.8 liters
Answer: The answer is A, A new element or different atom formed from the original two.
Hope this helps! :D
-<em>TanqR</em>
Answer: density equals 3 g/mL
Step by step explanation:
D=m/v
D=45/15
D=3
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.
