Mass = Density × Volume
= 30.0 mg / mL × 375 mL
= 11250 mg
= 11.25 g
∴ the total mass of insulin in the bottle is 11.25 g (11250 mg)
Answer:
a): not necessarily due to London Dispersion Forces and dipole-dipole interactions.
b): not necessarily due to London Dispersion Forces.
Explanation:
There are three major types of intermolecular interaction:
- Hydrogen bonding between molecules with H-O, H-N, or H-F bonds and molecules with lone pairs.
- Dipole-dipole interactions between all molecules.
- London dispersion forces between all molecules.
The melting point of a substance is a result of all three forces, combined.
Note that the more electrons in each molecule, the stronger the London Dispersion Force. Generally, that means the more atoms in each molecule, the stronger the London dispersion force. The strength of London dispersion force between large molecules can be surprisingly strong.
For example,
(water) molecules are capable of hydrogen bonding. The melting point of
at
is around
. That's considerably high when compared to other three-atom molecules.
In comparison, the higher alkane hexadecane (
, straight-chain) isn't capable of hydrogen bonding. However, under a similar pressure, hexadecane melts at around
above the melting point of water. The reason is that with such a large number of atoms (and hence electrons) per molecule, the London dispersion force between hexadecane molecules could well be stronger than that the hydrogen bonding between water molecules.
Similarly, the dipole moments in HCl (due to the highly-polar H-Cl bonds) are much stronger than those in hexadecane (due to the C-H bonds.) However, the boiling point of hexadecane under standard conditions is much higher (at around
than that of HCl.
Answer:
Explanation:
The most common example is the molar volume of a gas at STP (Standard Temperature and Pressure), which is equal to 22.4 L for 1 mole of any ideal gas at a temperature equal to 273.15 K and a pressure equal to 1.00 atm.
Answer: The atomic mass of E is 16
Explanation:Please see attachment for explanation
<span>Answer:
W must be 5-chloro-2-methylpentane. It can give only 4-methy-1-pentene (Y) upon dehydrohalogenation:
X must be 4-chloro-2-methylpentane. Dehydrohalogenation yields both Y and 4-methyl-2-pentene. (Z)</span>